Heat capacity and modes of heat transfer: Physics unit 10 notes (FBISE class 10)

10.1Specific heat capacityOften examined

Why does water take time to boil, and why does a metal spoon feel colder than a wooden one at the same temperature? Questions like these lead to specific heat capacity. When the same heat is given to equal masses of different materials, their temperatures do not rise by the same amount. Specific heat capacity measures this difference between materials.

Specific heat capacity (specific heat): The amount of heat absorbed or lost to change the temperature of $1 kg$ of a substance by $1$ kelvin or $1$ degree Celsius.

c = \frac{Q}{m Δ T}

10.1

Here $c$ is the specific heat of the material, $Q$ is the amount of heat absorbed or lost, $m$ is the mass of the substance and $Δ T = T_{final} - T_{initial}$ is the change in its temperature. Making $Q$ the subject gives the form used to find the heat needed for a given temperature change:

Q = m c Δ T

Units

The SI unit of specific heat is joule per kilogram per kelvin ( $J k g^{- 1} K^{- 1}$ ). Other units in use are $J k g^{- 1}^{\circ} C^{- 1}$ and $J g^{- 1}^{\circ} C^{- 1}$ . For example, the specific heat of aluminium is $900 J k g^{- 1} K^{- 1}$ . In the other units it is written $900 J k g^{- 1}^{\circ} C^{- 1}$ and $0.9 J g^{- 1}^{\circ} C^{- 1}$ .

Heat capacity

The summary of the unit also defines the heat capacity of a body. It is the heat needed to raise the temperature of the whole body, whatever its mass, by $1^{\circ} C$ or $1 K$ . Putting this beside Eq. 10.1 shows how the two ideas are related: heat capacity belongs to a particular body, specific heat to a material.

C = \frac{Q}{Δ T} = m c

unit: J per degree Celsius, written $J^{\circ} C^{- 1}$ or $J K^{- 1}$

Materials with low and high specific heat

Some materials need only a small amount of heat to change their temperature, such as metals (for example aluminium, copper, iron and diamond). These have a low specific heat. Such materials are good conductors of heat. They cannot hold heat for a long time, so they heat up and cool down quickly. For this reason they are used in heat sinks.

Other substances, like water, need a large amount of heat to raise their temperature by one kelvin. They have a large specific heat: they absorb or lose a large amount of heat for a small change in temperature. They cannot be easily heated up or cooled down, which means they can retain heat. Such materials are used for controlling temperature in different devices. For example, water has a large specific heat, so it is used in cars as engine coolant.

Table 10.1: Specific heats of selected substances.
Material	Specific heat ( $J k g^{- 1} K^{- 1}$ )
Aluminium	$910$
Copper	$387$
Glass	$840$
Gold	$129$
Iron / steel	$452$
Lead	$128$
Silver	$235$
Ethanol	$2450$
Glycerin	$2410$
Mercury	$139$
Water	$4190$
Air	$721$
Carbon dioxide	$638$
Oxygen	$651$
Steam ( $10 0^{\circ} C$ )	$1520$

Example 10.1: specific heat of a metal

A

200 g

sample of a solid metal is heated from

2 5^{\circ} C

to

7 5^{\circ} C

. If

3500 J

of heat energy is required to raise the temperature of the metal, what is the specific heat of the metal?

Given

$m = 200 g = 0.2 kg$
$Δ T = 7 5^{\circ} C - 2 5^{\circ} C = 5 0^{\circ} C = 50 K$
$Q = 3500 J$

To find

specific heat $c$

Change in temperature in kelvin and in degrees Celsius is the same number, so $Δ T = 50 K$ . The mass must be in kilograms for the SI unit.
Write the formula for specific heat.
$c = \frac{Q}{m Δ T}$
Substitute the values.
$c = \frac{3500 J}{( 0.2 kg ) ( 50 K )} = \frac{3500}{10} J k g^{- 1} K^{- 1}$

c = 350 J k g^{- 1} K^{- 1}

Example 10.2: heat absorbed by aluminium

A half kilogram sample of aluminium (specific heat capacity

c = 900 J k g^{- 1} K^{- 1}

) is heated from room temperature of

2 5^{\circ} C

to

10 0^{\circ} C

. How much heat is absorbed by the aluminium to raise its temperature?

Given

$m = 0.5 kg$
$c = 900 J k g^{- 1} K^{- 1}$
$T_{initial} = 2 5^{\circ} C$
$T_{final} = 10 0^{\circ} C$

To find

heat absorbed $Q$

Find the change in temperature.
$Δ T = 10 0^{\circ} C - 2 5^{\circ} C = 7 5^{\circ} C = 75 K$
Specific heat is $c = \frac{Q}{m Δ T}$ , so the heat absorbed is $Q = m c Δ T$ .
$Q = m c Δ T$
Substitute the values.
$Q = (0.5 kg) \times (900 J k g^{- 1} K^{- 1}) \times (75 K) = 33750 J$

Q = 33750 J = 33.75 kJ

Worked example: the same heat in water and in aluminium

A

500 g

block of aluminium and

500 g

of water are both at

2 0^{\circ} C

. Each receives

21 kJ

of heat. Take

c = 900 J k g^{- 1} K^{- 1}

for aluminium and

4200 J k g^{- 1} K^{- 1}

for water. Find the final temperature of each and explain why a graph of temperature against heat supplied has a steeper line for aluminium.

Given

$m = 500 g = 0.5 kg$
$Q = 21 kJ = 21000 J$
$c_{Al} = 900 J k g^{- 1} K^{- 1}$
$c_{w} = 4200 J k g^{- 1} K^{- 1}$
$T_{i} = 2 0^{\circ} C$

To find

final temperature of each

Change the units first: mass to kilograms and heat to joules.
Use $Q = m c Δ T$ and make $Δ T$ the subject.
$Δ T = \frac{Q}{m c}$
Water.
$Δ T_{w} = \frac{21 000}{0.5 \times 4200} = \frac{21 000}{2100} = 10 K \Rightarrow T = 20 + 10 = 3 0^{\circ} C$
Aluminium.
$Δ T_{Al} = \frac{21 000}{0.5 \times 900} = \frac{21 000}{450} = 46.7 K \Rightarrow T = 20 + 46.7 = 66. 7^{\circ} C$
Check: $m c Δ T$ for aluminium is $0.5 \times 900 \times 46.7 = 21000 J$ , which equals the heat supplied.
The temperature rise per joule is $Δ T / Q = 1/ (m c)$ . This is the gradient of the graph. Aluminium has the smaller $m c$ ( $450$ against $2100 J K^{- 1}$ ), so its line is more than four times steeper.

Temperature against heat supplied

0

2

4

6

8

10

12

14

16

18

20

22

10

20

30

40

50

60

70

Q (kJ)

T (^{\circ} C)

66.7

30

aluminium

water

Water and aluminium,

0.5 kg

each, both starting at

2 0^{\circ} C

. The gradient of each line is

1/ (m c)

: the line for aluminium is steeper because its

c

is smaller.

T_{w} = 3 0^{\circ} C, T_{Al} = 66. 7^{\circ} C

10.1.1Uses of large specific heat of waterOften examined

Water has a large specific heat ( $4200 J k g^{- 1} K^{- 1}$ ), and this has important uses in daily life.

A. Temperature variation in land and coastal areas

The specific heat of dry soil ( $800 J k g^{- 1} K^{- 1}$ ) is smaller than that of water. It is about five times smaller. This is why land heats up quickly in summer, with temperatures rising from about $4 5^{\circ} C$ to $5 0^{\circ} C$ , and also cools down quickly in winter, falling from $5^{\circ} C$ to $0^{\circ} C$ . In coastal areas the temperature variation is small: in Karachi the average temperature varies only from $1 9^{\circ} C$ to $3 2^{\circ} C$ .

Oceans and large lakes absorb a lot of heat from the Sun during the summer and keep it stored in them. In winter they release heat into the atmosphere. This keeps the temperature moderate between summer and winter in coastal areas.

B. Maintaining stability of ocean and lake temperatures

Lakes and oceans heat up and cool down slowly because of the high specific heat of water. This prevents sudden changes in the temperature of the lake or ocean and so provides a suitable environment for aquatic life.

C. Human body temperature regulation

The human body contains approximately $60%$ water, and a healthy body keeps its temperature at about $3 7^{\circ} C$ . Because the water in the body has a high specific heat, the body absorbs or gives out a lot of heat with only a small change in its temperature. This helps the body to hold its temperature steady.

D. Cooling systems and heat exchangers

Water is used as a coolant in power plants, in different industrial processes and in the cooling systems of automobiles (radiators), as shown in Figure 10.1. Water takes a large amount of excess heat from the machines and effectively transfers it away using a heat exchanger. This prevents overheating of the machinery and maintains its high efficiency.

Figure 10.1: Cooling system of a vehicle. The water pump pushes coolant (coloured) through the engine, where it takes up heat. The hot coolant passes the thermostat and the upper radiator hose to the radiator core, where the fan and the air flow carry the heat away. The cooled coolant returns through the lower radiator hose to the pump.

E. Cooking

Water is good for cooking because of its high specific heat. It heats up slowly and distributes heat evenly, so food cooks without burning. Hot drinks like tea and coffee stay hot for a long time.

10.1.2 ASpecific heat by the method of mixturesOften examined

There are different experiments for finding the specific heat of solids and liquids. The most common methods are the method of mixtures and the electrical heating method. The first of these is described here and the second in the next section.

A. Method of mixtures

These experiments use a calorimeter. When a hot sample (for example a solid) is added to the water in the calorimeter, heat is exchanged between the sample, the water and the calorimeter. By the law of conservation of energy, the heat lost by the hot bodies is equal to the heat gained by the cold bodies. After measuring the change in temperature of the water in the calorimeter and the masses of the substances, an equation is formed and, after careful analysis, the specific heat of the sample is calculated.

i. Measurement of the specific heat of a solid

Apparatus: a Bunsen burner or hot plate, a copper calorimeter with lid and stirrer and an insulating cover, a metallic solid cylinder (for example brass or copper), a weighing machine, water, a beaker, a laboratory stand and two thermometers.

Key variables: the independent variable is the temperature of the metal block, and the dependent variable is the energy supplied to the metal block by heating.

Figure 10.2: Measuring the specific heat of a solid by the method of mixtures. (a) The temperature

T_{1}

of the water in the calorimeter. (b) The temperature

T_{2}

of boiling water, which is also the temperature of the solid cylinder suspended in it. (c) The final temperature

T_{3}

of the mixture after the hot solid is placed in the calorimeter.

Procedure:

Take an empty calorimeter of known specific heat $c_{c}$ (its heat capacity is $m_{c} c_{c}$ ). Measure the mass $m_{c}$ of the calorimeter with the stirrer and lid.
Fill the calorimeter with enough water to completely submerge the solid later. Weigh ( $m$ ) the calorimeter with water. The difference of masses, $m - m_{c}$ , is the mass of water $m_{w}$ in the calorimeter.
Place the calorimeter in its insulating cover and insert a thermometer in it. Measure the temperature $T_{1}$ of the water (equal to the temperature of the surroundings), as in Fig. 10.2 (a).
Measure the mass of the given solid with a weighing machine and note it down as $m_{s}$ .
Boil water in a beaker with the solid submerged. Measure the boiling-water temperature $T_{2}$ . In this step make sure that the solid reaches the same temperature as the boiling water, as in Fig. 10.2 (b).
Remove the solid from the boiling water, shake off the excess water and quickly place it in the calorimeter. Cover the calorimeter immediately.
Stir the water continuously with the stirrer so that heat (and temperature) is distributed uniformly.
Measure the final temperature of the mixture once it becomes constant on reaching equilibrium. Record it as $T_{3}$ , as in Fig. 10.2 (c).
The rise in temperature of the water and calorimeter is $T_{3} - T_{1}$ . The fall in temperature of the solid is $T_{2} - T_{3}$ .
By the principle of calorimetry (the law of conservation of energy), the heat given by the solid in cooling from $T_{2}$ to $T_{3}$ is equal to the heat gained by the water and calorimeter in rising from $T_{1}$ to $T_{3}$ .

Derivation Specific heat of a solid by the method of mixtures

Aim. Find an expression for the specific heat $c_{s}$ of the solid from the measured masses and temperatures.

Assumptions.

No heat is lost to the surroundings (the calorimeter has an insulating cover and a lid).
The solid, the water and the calorimeter all reach the same final temperature $T_{3}$ .
The solid has reached the temperature $T_{2}$ of the boiling water before it is moved.

The solid cools from $T_{2}$ to $T_{3}$ , so it loses heat.
$Q_{s} = m_{s} c_{s} (T_{2} - T_{3})$
The water warms from $T_{1}$ to $T_{3}$ , so it gains heat.
$Q_{w} = m_{w} c_{w} (T_{3} - T_{1})$
The calorimeter also warms from $T_{1}$ to $T_{3}$ and gains heat.
$Q_{c} = m_{c} c_{c} (T_{3} - T_{1})$
Heat lost by the solid equals heat gained by the water and the calorimeter.
$Q_{s} = Q_{w} + Q_{c} or m_{s} c_{s} (T_{2} - T_{3}) = m_{w} c_{w} (T_{3} - T_{1}) + m_{c} c_{c} (T_{3} - T_{1})$
Both gains have the same temperature change, so take $(T_{3} - T_{1})$ as a common factor.
$m_{s} c_{s} (T_{2} - T_{3}) = (m_{w} c_{w} + m_{c} c_{c}) (T_{3} - T_{1})$
Divide both sides by $m_{s} (T_{2} - T_{3})$ to make $c_{s}$ the subject.
$c_{s} = \frac{( m _{w} c _{w} + m _{c} c _{c} ) ( T _{3} - T _{1} )}{m _{s} ( T _{2} - T _{3} )}$

Result

c_{s} = \frac{( m _{w} c _{w} + m _{c} c _{c} ) ( T _{3} - T _{1} )}{m _{s} ( T _{2} - T _{3} )} (10.2)

Putting the measured masses of the solid, water and calorimeter, the specific heats of water and calorimeter and the temperatures $T_{1}$ , $T_{2}$ and $T_{3}$ into Eq. 10.2 gives the specific heat of the solid.

ii. Measurement of the specific heat of a liquid

The same procedure is followed, but now a solid of known specific heat $c_{s}$ is used and the liquid has the unknown specific heat $c_{l}$ . The law of conservation of energy now reads

m_{s} c_{s} (T_{2} - T_{3}) = m_{l} c_{l} (T_{3} - T_{1}) + m_{c} c_{c} (T_{3} - T_{1})

Rearranging to get the term with $c_{l}$ alone,

m_{l} c_{l} (T_{3} - T_{1}) = m_{s} c_{s} (T_{2} - T_{3}) - m_{c} c_{c} (T_{3} - T_{1})

and dividing by $m_{l} (T_{3} - T_{1})$ gives

c_{l} = \frac{m _{s} c _{s} ( T _{2} - T _{3} ) - m _{c} c _{c} ( T _{3} - T _{1} )}{m _{l} ( T _{3} - T _{1} )}

10.3

Example 10.3: specific heat of an unknown solid

A

150 g

piece of an unknown solid cylinder is heated to

10 0^{\circ} C

and then placed into a calorimeter containing

200 g

of water at

2 5^{\circ} C

. The calorimeter has a mass of

50 g

and a specific heat capacity of

0.9 J g^{- 1}^{\circ} C^{- 1}

. The final temperature of the system (solid, water and calorimeter) is

3 0^{\circ} C

. Assume no heat is lost to the surroundings. Calculate the specific heat capacity of the solid.

Given

$m_{s} = 150 g$
$m_{w} = 200 g$
$m_{c} = 50 g$
$T_{1} = 2 5^{\circ} C$
$T_{2} = 10 0^{\circ} C$
$T_{3} = 3 0^{\circ} C$
$c_{w} = 4.18 J g^{- 1}^{\circ} C^{- 1}$
$c_{c} = 0.9 J g^{- 1}^{\circ} C^{- 1}$

To find

specific heat $c_{s}$ of the solid (in $J g^{- 1}^{\circ} C^{- 1}$ )

The solid is hot and the water and calorimeter are cold, so use the principle of calorimetry, Eq. 10.2.
$c_{s} = \frac{( m _{w} c _{w} + m _{c} c _{c} ) ( T _{3} - T _{1} )}{m _{s} ( T _{2} - T _{3} )}$
Work out the temperature changes. The water and calorimeter rise from $2 5^{\circ} C$ to $3 0^{\circ} C$ . The solid falls from $10 0^{\circ} C$ to $3 0^{\circ} C$ .
$T_{3} - T_{1} = 30 - 25 = 5^{\circ} C, T_{2} - T_{3} = 100 - 30 = 7 0^{\circ} C$
Substitute the values. Grams are used throughout, so no conversion is needed.
$c_{s} = \frac{{( 200 ) ( 4.18 ) + ( 50 ) ( 0.9 )} ( 5 )}{( 150 ) ( 70 )}$
Work out the top line and the bottom line.
$c_{s} = \frac{( 836 + 45 ) \times 5}{10 500} = \frac{4405}{10 500}$

c_{s} = 0.42 J g^{- 1}^{\circ} C^{- 1} (= 420 J k g^{- 1} K^{- 1})

Worked example: final temperature when hot iron meets cold water

A

250 g

piece of iron at

10 0^{\circ} C

is dropped into

500 g

of water at

2 0^{\circ} C

in an insulated container. Neglect the container. Take

c = 452 J k g^{- 1} K^{- 1}

for iron and

4200 J k g^{- 1} K^{- 1}

for water (Table 10.1 and Section 10.1.1). Find the final temperature of both. Why is it so much closer to

2 0^{\circ} C

than to

10 0^{\circ} C

?

Given

$m_{Fe} = 0.25 kg$
$c_{Fe} = 452 J k g^{- 1} K^{- 1}$
$T_{Fe} = 10 0^{\circ} C$
$m_{w} = 0.5 kg$
$c_{w} = 4200 J k g^{- 1} K^{- 1}$
$T_{w} = 2 0^{\circ} C$

To find

final temperature $T$

Let the common final temperature be $T$ . The iron falls from $10 0^{\circ} C$ to $T$ and the water rises from $2 0^{\circ} C$ to $T$ . Heat lost by the iron equals heat gained by the water.
$m_{Fe} c_{Fe} (100 - T) = m_{w} c_{w} (T - 20)$
Work out the two products $m c$ (the heat capacities of the bodies).
$113 (100 - T) since 0.25 \times 452 0.5 \times 4200 = 2100 (T - 20) = 113 = 2100$
Expand both sides.
$11300 - 113 T = 2100 T - 42000$
Collect the terms in $T$ on one side and the numbers on the other.
$53300 = 2213 T$
Solve for $T$ .
$T = \frac{53 300}{2213} = 24. 1^{\circ} C$
Check with heat: iron loses $113 \times (100 - 24.08) = 8580 J$ and water gains $2100 \times (24.08 - 20) = 8570 J$ . The two agree (the small difference is rounding).
The water's heat capacity ( $2100 J K^{- 1}$ ) is about $19$ times that of the iron ( $113 J K^{- 1}$ ). The iron has to lose many joules to change the water only a little, so the final temperature stays close to the water's starting temperature.

Hot iron in cold water

0

1

2

3

4

5

6

7

8

10

20

30

40

50

60

70

80

90

100

time (schematic)

T (^{\circ} C)

T_{3} = 24. 1^{\circ} C

iron

water

The two temperatures move towards the same final value

T_{3}

. The curves are schematic (the time scale is not given): the iron cools by about

76 K

while the water warms by only about

4 K

.

T = 24. 1^{\circ} C

10.1.2 BSpecific heat by the electrical heating method

B. Electrical heating method

In this method the amount of heat transferred to a material and the change in its temperature are measured. The heat comes from an electrical heater, so it can be found from the electrical energy supplied.

i. Measurement of the specific heat of a solid

Apparatus: a solid block or cylinder, for example aluminium, with a hole in it for the heater and a smaller hole for a thermometer, an immersion heater, a voltmeter, an ammeter and a thermometer.

Figure 10.3: Measurement of the specific heat of a solid by the electrical method. (a) The immersion heater (coloured) and the thermometer in their holes in the block, wrapped in cotton wool. (b) The heater is joined to the supply through an ammeter (in series) and a voltmeter (across the heater).

Procedure:

Measure the mass of the solid block.
Place an electrical immersion heater into the central hole at the top of the weighed metal block. Connect the heater with an ammeter to measure the current and with a voltmeter to measure the voltage (potential difference) applied on the heater, as shown in Fig. 10.3.
Place the thermometer into the smaller hole. Add a few drops of oil into the hole to make sure that the thermometer is surrounded by hot material.
Insulate the block by wrapping it with cotton wool.
Record the initial temperature of the block.
Connect the heater to the power supply and turn it on for ten minutes. Record the current and the potential difference.
After ten minutes, turn the heater off. Record the highest temperature that the block reaches (the temperature reading will rise even after the heater is turned off).
Calculate the heat energy generated by the heater.

Result: calculate the specific heat of the solid block using the formula

c=\dfrac{Q}{m\,\Delta T}

10.4

Q=V\,I\,t

V in volt, I in ampere, t in second, Q in joule

ii. Measurement of the specific heat of a liquid

Apparatus: an immersion heater, a voltmeter, an ammeter, a thermometer, water (or another liquid) and a calorimeter.

Figure 10.4: Measurement of the specific heat of a liquid by the electrical method. The heating resistor (coloured) is immersed in the liquid in an insulated container. The ammeter reads the current through it and the voltmeter the voltage across it, from the 12\,\mathrm{V} DC supply.

Procedure:

Measure the mass of the empty calorimeter (m_c).
Add some liquid, for example water, in the calorimeter and measure the mass of the calorimeter plus liquid (m). The difference of the masses, m-m_c, gives the mass of the liquid m_l.
Measure the initial temperature T_i of the calorimeter plus liquid using a thermometer.
Place the electrical heater inside the liquid and connect the ammeter and voltmeter with it. Switch it on. Measure the voltage and the current passing through the heater, as shown in Fig. 10.4.
The temperature starts to rise. When it has risen by 10^\circ\mathrm{C}, switch off the heater, stir the liquid well and measure the highest temperature T_f of the liquid. Then find the temperature difference between the highest and the initial temperature.
Apply the law of conservation of energy: the energy supplied is equal to the heat gained by the liquid plus the heat gained by the calorimeter.

Q_{\text{heater}}=Q_l+Q_c=m_lc_l(T_f-T_i)+m_cc_c(T_f-T_i)

Rearranging this equation to make c_l the subject gives the result:

c_l=\dfrac{Q_{\text{heater}}-m_cc_c(T_f-T_i)}{m_l(T_f-T_i)}

10.5

By putting in the measured values of the heat energy produced by the heater, the masses of the liquid and the calorimeter and the initial and final temperatures, the specific heat of the liquid can be found.

Worked example: electrical heating of a metal block

A 0.50\,\mathrm{kg} metal block is heated by an immersion heater connected to a 12\,\mathrm{V} supply. The ammeter reads 5.0\,\mathrm{A} and the heater is on for 5.0 minutes. The temperature of the block rises from 20^\circ\mathrm{C} to 60^\circ\mathrm{C}. Find the specific heat of the metal and say which metal of Table 10.1 or Section 10.1 it could be. Would the true value be larger or smaller than your answer if some heat escaped from the block?

Given

V = 12\,\mathrm{V}
I = 5.0\,\mathrm{A}
t = 5.0\ \text{min}=300\,\mathrm{s}
m = 0.50\,\mathrm{kg}
\Delta T = 60-20=40\,\mathrm{K}

To find

specific heat c

The heat given to the block equals the electrical energy supplied. The time must be in seconds.
Q=VIt=12\times5.0\times300=18\,000\,\mathrm{J}
Use c=\dfrac{Q}{m\,\Delta T}.
c=\dfrac{18\,000}{0.50\times40}=\dfrac{18\,000}{20}=900\,\mathrm{J\,kg^{-1}\,K^{-1}}
This is the value the book gives for aluminium (900\,\mathrm{J\,kg^{-1}\,K^{-1}}), so the block is probably aluminium.
If some heat escaped, the block would gain less than 18\,000\,\mathrm{J}, so the real heat that raised its temperature is smaller than the Q used. The true specific heat is therefore smaller than 900. This is why the block is wrapped in cotton wool.

c=900\,\mathrm{J\,kg^{-1}\,K^{-1}}\ (\text{aluminium});\ \text{heat loss makes the measured value too large}

10.2Transfer of heat

Heat transfer occurs when thermal energy is transferred from one object to another because of a difference in temperature. The transfer continues until both objects reach the same temperature, which is known as thermal equilibrium. There are three methods of heat transfer: conduction, convection and radiation.

The three modes of heat transfer (from the unit summary and Sections 10.2.1 to 10.2.3).
Mode	How the heat travels	Where it occurs	Medium needed?
Conduction	By direct contact of the particles of the medium, which do not change their positions	In solids, because of a difference in temperature	Yes
Convection	By the motion of the particles themselves	In liquids and gases, because of the density difference between different regions	Yes (a fluid)
Radiation	By electromagnetic waves (infrared radiation)	Anywhere, including a vacuum	No

10.2.1Thermal conductionOften examined

The top edges of a cooking pot get extremely hot when they are touched after it has been on the stove for a while, even though they are not getting heat directly. The heat is transferred to these top edges by conduction. In the same way, when vegetables are fried in a pan (Fig. 10.5), heat is transferred from the flame to the pan and then to the vegetables by conduction.

This method of heat transfer takes place in solids. In conduction, heat is transferred from atom to atom. Metals are good conductors of heat because they have a large number of free electrons that travel faster across the metal when it is heated. That is why heat energy is transferred quickly in metals.

Thermal conduction: The method of transfer of heat due to collisions of atoms or molecules and motion of free electrons in solids, from the hot part to the cold part.

Figure 10.6: Thermal conductors. Materials are placed from better conductors (left) to better insulators (right).

How conduction takes place

When one end of a rod is heated, the atoms at this end start vibrating with a larger amplitude. These atoms hit their neighbouring atoms and pass heat energy to them (Fig. 10.7). Now these atoms vibrate with a larger amplitude and hit their nearby atoms, and the process continues. In this manner heat energy is transferred between the two ends.

When a copper rod and a wooden stick are heated simultaneously, the other end of the copper rod heats up more quickly than the other end of the wooden stick. This is because metallic rods such as copper have a lot of free electrons. In comparison with vibrating atoms, these free electrons get energy from the hot end and carry it quickly from one end of the rod to the other. So metals transfer heat energy more quickly than other solids such as wood and glass.

Figure 10.7: Mechanism of conduction in a metal rod. At the hot end the atoms vibrate with a larger amplitude (marks at their sides) and collide with slower neighbours. The free electrons (coloured) pick up energy at the hot end and carry it quickly through the metal to the cold end.

Experiment: good and bad conductors

Apparatus: three long thin strips of different metals (for example copper, iron and aluminium rods), Vaseline or wax, drawing pins and a Bunsen burner.

Control variables: length of the metal rods, diameter of the rods and position in the Bunsen flame.

Figure 10.8: Good and bad conductors. The rods are heated together at the common end. The heat flows along each rod and melts the wax, so the drawing pin drops. The shorter the time for a pin to drop, the better the conductor.

Method:

Apply small dots of wax at the ends of the metal rods and fix the drawing pins in them.
Position the other end of the metal rods into a Bunsen flame. Make sure that heat can reach every metal rod in the same amount.
As the heat flows through the metal rods, the wax will melt and the pin will drop. Record the time taken for the wax to melt and the drawing pin to drop off. The fastest time shows the best conductor of heat, because good conductors allow heat to flow fast through them and cause the wax to melt soon.

Result: the drawing pin falls off the copper rod first, followed by the aluminium.

Conclusion: copper conducts better than aluminium, while aluminium conducts better than iron and brass.

10.2.2ConvectionOften examined

When water is heated on a stove, the water can be seen moving. The water molecules at the bottom receive heat first. According to the kinetic molecular theory, the intermolecular forces between the water molecules decrease as the temperature rises, and the distance between the molecules increases. This makes the water at the bottom less dense. This less dense water moves from the bottom to the top after taking heat energy. The dense cold water at the top sinks to fill the gap and gains heat energy in its turn. In this way the molecules of the water move themselves to transfer heat energy, because of the differences in density between two parts of the fluid (Fig. 10.9).

Convection: The method of transfer of heat due to bulk movement of molecules in fluids. It can also be defined as the transfer of heat energy due to the transfer of the fluid itself.

Convection takes place only in liquids and gases. It cannot take place in solids.

Experiments on convection

Experiment 1 (Fig. 10.10). Put some potassium permanganate crystals at the bottom of a beaker filled with water. When the beaker is heated, a stream of potassium permanganate travels upward and shows a convection current. The water at the bottom of the beaker heats up, expands and becomes less dense. This less dense water rises and carries heat energy from the bottom to the top of the container. Then cold (denser) water flows downward from the sides and replaces the hot water. This process continues, and heat is transferred continuously because of the continuous flow of the water.

Figure 10.10: Formation of a convection current. Water heated at the bottom rises through the middle of the beaker carrying the dye, spreads out at the top and sinks down the sides.

Experiment 2 at home (Fig. 10.11). Use blue and red food colours. Use blue food colour to make blue ice cubes. Take water in a transparent bowl or box and put the blue ice cubes on one of its sides. Add warm water on the other side. Place an ink pot filled with red food colour on the warm-water side at the bottom of the container. The red colour on the warmer side rises first and then moves towards the cooler blue ice side. The cold water, being denser, sinks and its place is filled by the red-coloured water. This process continues and shows the transfer of heat due to the motion of molecules.

Worked example: drawing and explaining a convection current

A beaker of water has a crystal of potassium permanganate at the bottom. (a) Draw the path of the dye when the beaker is heated under the crystal, and explain in terms of density why the water circulates. (b) What happens if the same beaker is heated only near the top of the water instead? Why?

(a) The flame heats the water at the bottom first. This water expands, so the same mass takes more volume and its density falls.
The less dense hot water is pushed upward by the denser cold water around it. It rises through the middle of the beaker and carries the dye with it.
At the top the water spreads out and loses heat. It is now cooler and denser, so it sinks down the sides of the beaker and returns along the bottom to be heated again. This continuous circulation is a convection current, drawn in part (a) of the figure.
(b) If the heat is supplied near the top, the hot water is already at the top and is less dense than the water below it. There is nothing to make it sink and nothing to make the cold water rise, so no convection current forms.
The cold water at the bottom is then warmed only very slowly by conduction. Water is a poor conductor (Fig. 10.6), so the lower part stays cold for a long time.

(a) Heated at the bottom, the water circulates and the whole beaker warms. (b) Heated near the top, the hot water stays on top, the cold water stays below and there is no circulation.

\text{Heated from below: hot water rises and cold water sinks, so a current forms. Heated from the top: no current.}

10.2.2 AConvection and marine life

Convection is a crucial process that helps to support marine life in the oceans. It involves the movement of seawater driven by differences in temperature and salinity (the saltiness of the water) across the globe (Fig. 10.12). Salty water is denser than fresh water. In some areas seawater becomes saltier and denser, especially where a lot of water evaporates because of hot weather.

Warm water is less dense than cold water. When the Sun heats the surface of the ocean, the water at the surface becomes warmer and less dense. This warm water rises and spreads out across the surface. Cold water is denser than warm water, so in polar regions the seawater cools down, becomes denser and sinks to the ocean depths. These movements of water cause convectional currents. Figure 10.12 of the book shows such currents across the whole globe, with heat released to the atmosphere where the loops come to the surface.

Upwelling: The upward movement of sea water.
Downwelling: The downward movement of sea water.

Figure 10.13: Upwelling and downwelling. (a) Where winds blow along a coast, surface water is pushed away and colder, nutrient-rich water rises from below to replace it. (b) In polar waters the surface water cools, becomes denser and sinks.

Distribution of food

Upwelling typically occurs along coastlines where winds blow parallel to the shore. As surface water is pushed away from the coast, it is replaced by water rising from below. This deeper water is usually colder and richer in nutrients (containing nitrogen, phosphate and organic materials) than the surface water. Phytoplankton (a tiny plant-like organism) uses these nutrients and sunlight to grow and multiply (Fig. 10.14). Phytoplankton is the food source for many marine creatures, from small ones (zooplankton) to large ones (whales).

Upwelling feeds the ocean food chain

Winds blowing parallel to the shore push the surface water away from the coast.
Colder water from below rises to replace it (upwelling). It is rich in nutrients.
Phytoplankton uses the nutrients and sunlight to grow and multiply.
Zooplankton, fish and other creatures feed on the phytoplankton, and larger animals such as whales feed on them.

Distribution of oxygen

Marine animals, like fish, need oxygen to live. Most oxygen is dissolved in the top water, which is exposed to the air. Convection causes this water to go deep into the ocean. Without convection, deep waters would lack oxygen, making it difficult for marine life to survive.

Distribution of heat

Because of upwelling, warm water rises and takes heat to the top. Convection distributes heat throughout the ocean, which helps to stabilise the temperature. The movement of warm and cold water around the globe also affects climate patterns, which can affect the habitats of marine life.

10.2.2 BLand breeze and sea breezeOften examined

A breeze blows continuously on the coastal areas because of convection. A breeze during the day is called a sea breeze and a breeze during the night is called a land breeze.

Sea breeze

During the day, land heats up more quickly than the sea because of its low specific heat. Therefore the air above the land becomes hot, expands and becomes less dense. This hot air above the land rises, and cool air above the sea moves towards the land to fill the gap.

Land breeze

At night, the temperature of the land falls more quickly than that of the water (water can maintain its temperature for a long time because of its high specific heat capacity). Because of this, the air above the sea is hotter than the air above the land. Therefore the hot air above the water rises and cool air from the land moves towards the sea to fill the gap.

Figure 10.15: Sea breeze and land breeze. (a) By day the land is hotter than the sea, so cool air blows from the sea to the land at the surface and warm air returns above. (b) At night the sea is hotter than the land, so the flow reverses.

Sea breeze and land breeze compared.
	Sea breeze	Land breeze
When	During the day	During the night
Which surface is hotter	Land (low specific heat, heats quickly)	Sea (high specific heat, cools slowly)
Air that rises	Hot air above the land	Hot air above the sea
Air that fills the gap	Cool air from the sea moves to the land	Cool air from the land moves to the sea

These breezes blow continuously across the coastal areas and help to maintain a moderate climate there (Fig. 10.15). These natural wind patterns help to cool coastal areas during the day and moderate temperatures at night, resulting in a more temperate and stable climate. The sea breeze carries moisture from the ocean, influencing coastal humidity and precipitation patterns. The continuous blowing of breezes also exchanges gases from land to sea and from sea to land, which prevents stagnation of gases in the region.

10.2.2 CThermals and the flight of birds

During the day the Sun heats the land unevenly. Dark surfaces and concrete structures absorb more heat than other areas such as grass or water. The air above these hot areas becomes less dense and rises from the ground. This rising hot air, surrounded by the cooler air around it, forms an air column called a thermal.

Thermal: A column of rising hot air, surrounded by cooler air, formed over ground that the Sun has heated more than its surroundings.

Birds like eagles, hawks and vultures are expert thermal riders. These birds have large and wide wings. By spreading their wings and tilting them slightly upwards, they ride on this rising air (Fig. 10.16). To stay within the thermal in the rising air, they fly in circles. They glide between thermals and maintain their high altitude without flapping their wings for a long period of time.

Figure 10.16: Birds on thermals. Warm air rises in a column over hot ground. A bird circles inside the column to stay in the rising air and gains height, then glides to the next thermal. Colder air sinks between the columns.

10.2.2 DHurricanes and cyclones

Hurricanes form over warm ocean waters. The warm water causes evaporation, and this moist air rises because of convection. As it rises, it cools and forms clouds by condensation. This condensation releases latent heat, which warms the surrounding atmosphere and makes the air there lighter. As a result the air continues to rise, lowering the atmospheric pressure at the surface. This decrease in pressure increases the rate of evaporation and causes more warm, moist air from the ocean to move up. The Earth’s rotation causes this rising air to spin, forming a cyclonic system. As the system gathers strength, it can develop into a hurricane with high wind speeds.

Formation of a hurricane

Warm ocean water evaporates, and the moist air rises by convection.
The rising air cools and forms clouds by condensation.
The condensation releases latent heat, which warms the surrounding air and makes it lighter, so it keeps rising.
The surface pressure falls, so more evaporation takes place and more warm, moist air from the ocean moves up.
The Earth’s rotation makes the rising air spin and a cyclonic system forms.
As the system gathers strength it can become a hurricane with high wind speeds.

How a cyclone forms. Warm ocean water evaporates, the moist air rises by convection and condenses into clouds, which lowers the pressure at the surface. More warm moist air flows in and rises, and the Earth’s rotation makes the whole system spin.

Cyclone: A large revolving tropical storm caused by winds blowing around a central area of low atmospheric pressure.
Hurricane: A tropical system with winds that have reached a constant speed of 74 miles per hour or more.

10.2.3Thermal radiationOften examined

Heat transfer through electromagnetic waves does not require any physical contact. For example, the Sun heats the Earth through radiation, and you feel warmth from a fireplace even at a distance. Figure 10.17 of the book shows three sources of thermal radiation: the Sun over dry, cracked land, burning wood and hot iron.

Radiation: The process of transfer of heat in the form of electromagnetic radiation (infrared radiation).

Heat transfer through conduction and convection requires a medium. Heat transfer through radiation does not require a material medium. There is a vacuum between the Earth and the Sun, but heat (infrared radiation, also known as IR radiation) and light still reach the Earth.

Light is the fastest entity in nature. All forms of electromagnetic radiation (Fig. 10.18), including heat, move at the speed of light. Therefore radiation is the fastest method for transferring heat. Every object emits infrared radiation, but hotter objects emit a greater amount of it. For example, the Sun emits a large amount of radiation, whereas less hot objects like the human body emit a small amount of radiation.

Figure 10.18: Solar spectrum (sketch). The Sun’s radiation spreads over the ultraviolet, visible and infrared regions. The book marks 47\,\% of it as visible light and 2\,\% as ultraviolet.

The three modes compared.
	Conduction	Convection	Radiation
Medium needed	Yes (a solid)	Yes (a liquid or gas)	No, it can cross a vacuum
How the heat moves	Collisions of atoms and free electrons	Bulk movement of the fluid	Electromagnetic waves (infrared)
Example in this unit	Metal rod heated at one end	Water heated from below	Sun heating the Earth, fireplace

10.2.3 AEffect of different surfaces on radiationOften examined

All objects absorb as well as radiate heat energy in the form of infrared radiation. The amount of thermal radiation emitted by an object depends on different factors: the colour of its surface, the texture of its surface and its surface area.

Mini lab: which insulating material is best?

Materials: three identical hot water bottles, three different insulating materials (for example bubble wrap, foam sheet and newspaper), thermometers and hot water.

Fill the three hot water bottles with hot water (around 80^\circ\mathrm{C} to 90^\circ\mathrm{C}).
Wrap each bottle with a different insulating material: bubble wrap (conduction insulation), foam sheet (convection insulation) and newspaper (radiation insulation).
Measure and record the initial temperature of the water in each bottle using a thermometer.
Place the bottles in a room with a consistent temperature (around 20^\circ\mathrm{C} to 25^\circ\mathrm{C}).
Measure and record the temperature of the water in each bottle every 15 minutes for an hour.
Compare the temperature changes in each bottle to see which insulating material performed best.

Surface colour of the object

Different coloured surfaces have different absorption and emission abilities, as shown in Table 10.2. A good absorber is a good emitter, but it is a bad reflector. For example, wearing black clothes on a sunny day absorbs more heat and warms us. Black-coloured surfaces are good emitters, which is why laptop chargers are made of black-coloured surfaces: it helps them to cool quickly. Similarly, the worst absorber is also the worst emitter, but it is a good reflector. Shiny silver surfaces are good reflectors, which is why car windscreen shades are made from silver-coloured surfaces.

Table 10.2: Effects of radiation on different surface colours.
Colour	Absorbing	Emitting	Reflecting
Black	Best	Best	Worst
Dull / dark	Good	Good	Bad
White	Bad	Bad	Good
Shiny silver	Worst	Worst	Best

Texture of the surface

Texture is the quality of something that can be known by touch, or the degree to which something is rough or smooth, soft or hard. Rough, dull and dark surfaces are good absorbers and emitters of radiation, but they are the worst reflectors. Smooth and shiny surfaces are bad absorbers and emitters of radiation, but they are good reflectors. Before painting the walls of houses, they are made smooth and usually painted white, so that they become good reflectors and poor absorbers. This helps to control the temperature of the houses.

Surface area of the object

At the same temperature, a hot body with a big surface area emits more radiation than a body with a small surface area. Therefore heat radiators installed in houses, car radiators and similar devices have a large surface area, which helps them to radiate quickly. For example, an infrared heater (Fig. 10.19), which primarily emits infrared waves, has a large plate area so that it can radiate more heat.

Mini lab: which surface absorbs better?

Materials: two similar cans (one with a shiny silver surface and the other painted dull black) with lids, thermometers and a radiant heater. Pour an equal amount of water into the two cans. Place both cans at the same distance from an electric heat source and insert thermometers into the cans as shown in the figure. Two experiments can be conducted with this set-up:

Turn on the heater. After ten minutes the water in the can with the dull black surface will be significantly hotter than the water in the other can.
Keep the heater off and pour boiled water into the cans. After ten minutes the water in the can with the dull black surface will be much cooler than the water in the other can.

Mini lab set-up: two similar cans of water, one with a shiny silver surface and one painted dull black, stand at the same distance from a radiant heater. A thermometer in each can shows its water temperature.

Conclusion: the first experiment shows that the dull black surface is the good absorber and the shiny silver surface is the good reflector of heat. The second shows that the dull black surface is also the better emitter.

Worked example: heating a black can and a shiny can

In the two-can experiment each can holds 200\,\mathrm{g} of water at 20^\circ\mathrm{C} and both cans are the same distance from the radiant heater. After ten minutes the dull black can reads 34^\circ\mathrm{C} and the shiny silver can reads 26^\circ\mathrm{C}. (a) Find the heat gained by the water in each can. (b) How many times more radiation has the black can absorbed? (c) Sketch the temperature against time for each can and say what the result shows. (d) In the second experiment boiled water is poured into the two cans. Which water cools faster and why? Take c=4200\,\mathrm{J\,kg^{-1}\,K^{-1}} for water and neglect the mass of the cans and any heat lost to the surroundings.

Given

m = 200\,\mathrm{g}=0.2\,\mathrm{kg}
c = 4200\,\mathrm{J\,kg^{-1}\,K^{-1}}
T_i = 20^\circ\mathrm{C}
T_{\text{black}} = 34^\circ\mathrm{C}
T_{\text{silver}} = 26^\circ\mathrm{C}

To find

heat gained Q by each can
ratio of the heat absorbed
the two temperature-time graphs

(a) Find each rise in temperature.
\Delta T_{\text{black}}=34-20=14\,\mathrm{K},\qquad \Delta T_{\text{silver}}=26-20=6\,\mathrm{K}
Use Q=mc\,\Delta T for each can. First mc=0.2\times4200=840\,\mathrm{J\,K^{-1}}.
Q_{\text{black}}=840\times14=11\,760\,\mathrm{J}
Silver can.
Q_{\text{silver}}=840\times6=5040\,\mathrm{J}
(b) Divide the two. The mass and the specific heat are the same, so the ratio is just the ratio of the temperature rises.
\dfrac{Q_{\text{black}}}{Q_{\text{silver}}}=\dfrac{11\,760}{5040}=\dfrac{14}{6}\approx2.3
The black can has absorbed about 2.3 times as much radiation as the shiny silver can. The dull black surface is a good absorber. The silver surface reflects most of the radiation that falls on it, so it absorbs less (it is a good reflector).
(c) With steady heating each temperature rises at a constant rate, so each graph is a straight line from 20^\circ\mathrm{C}. The black line is steeper: 14/10=1.4\,^\circ\mathrm{C} per minute against 6/10=0.6\,^\circ\mathrm{C} per minute.
(d) Now the cans are filled with boiled water and the heater is off. The dull black can is a good emitter as well as a good absorber, so its water loses heat by radiation faster and cools more quickly. The shiny silver can is a bad emitter, so its water stays hot for longer.

Black can and silver can near a heater

The black can warms more than twice as fast. A good absorber is also a good emitter and a bad reflector.

Q_{\text{black}}=11\,760\,\mathrm{J},\ Q_{\text{silver}}=5040\,\mathrm{J},\ \text{ratio}\approx2.3

Summary of the factors that affect radiation.
Surface feature	Better absorber and emitter	Better reflector
Colour	Black or dark	White or shiny silver
Texture	Rough and dull	Smooth and shiny
Surface area (same temperature)	Large area emits more	Small area emits less

10.2.3 BEffect of temperature on the rate of radiation

All objects absorb as well as radiate heat. The rate of radiation emitted by an object is directly proportional to the temperature of the object. High-temperature objects radiate more than low-temperature bodies. The Sun has a high temperature and a large surface area, so it emits a large amount of radiation, while a cooking stove has a low temperature and a smaller surface area than the Sun, so it emits less radiation.

When an object is absorbing more heat than radiating, its temperature will rise. When the object is emitting more radiation than absorbing, its temperature will drop, that is, it will cool down. An object will remain at the same temperature if it is absorbing and radiating heat at an equal rate. This state is called thermal equilibrium.

Absorption, emission and temperature.
Absorbing and emitting	What happens to the temperature
Absorbs more than it emits	Rises
Emits more than it absorbs	Falls (the object cools down)
Absorbs and emits at the same rate	Stays the same (thermal equilibrium)

A hot cup of tea is placed on a table. Initially its temperature is higher than that of its surroundings, so it emits radiation at a faster rate. Its temperature will decrease. After a while its temperature will reach the temperature of the surroundings, that is, it will reach thermal equilibrium.

A hot cup of tea reaches thermal equilibrium

The hot tea emits more than it absorbs, so it cools quickly at first and then more slowly. A cold drink absorbs more than it emits and warms up. Both end at the temperature of the surroundings (thermal equilibrium). The curves are schematic.

10.2.3 CLeslie cube

The Leslie cube is used to demonstrate how different surface textures affect the emission of thermal radiation. It has four sides with different surfaces, usually black and white painted surfaces, a matt black or dark-coloured surface and a shiny metal surface. It is a hollow cube with a lid at its top. Hot water is usually used to heat the interior of the cube. An infrared (IR) thermometer is used as the detector to measure the amount of IR radiation emitted from each surface of the cube.

Figure 10.20: Leslie cube seen from above. The hollow cube is filled with hot water and its four sides have different surfaces. An infrared detector placed at the same distance from each side in turn measures the radiation that side emits.

The experiment on the Leslie cube follows this procedure.

Boiling water is added to the Leslie cube for heating it.
Wait until the cube reaches the desired temperature.
Measure the quantity of radiation emitted from each surface by placing the IR thermometer at the same distance from each one in turn. (The infrared thermometer detects the infrared radiation emitted by each surface. A higher radiation translates to a higher temperature reading.)

The surfaces that are better at emitting radiation show a higher temperature reading on the infrared thermometer, while the surfaces that are bad at emitting radiation have a lower reading. After performing the experiment, these conclusions follow:

The black surface is the best absorber and emitter but the worst reflector.
A dark-coloured surface is a good absorber and emitter but a bad reflector.
A white surface is a good reflector but a bad absorber and emitter.
A shiny metal (shiny silver) surface is the best reflector but the worst absorber and emitter.

10.3 ACooking in a kitchen pan

When you place a kitchen pan on a stove burner, heat from the burner conducts through the base of the pan (Fig. 10.21). The metal of the pan absorbs the heat energy because of direct contact with the hot surface. The hot metal surface transfers the heat energy to the food in contact with it by conduction. This heat also travels to the whole surface of the pot and the handle. This explains why the handles are made of materials that are poor heat conductors, such as plastic or wood.

When heating water in a pan or pot on the stove, convection currents occur. As the water near the bottom of the pan heats up, it becomes less dense and rises, while cooler water from the top sinks. This circulation helps to distribute the heat evenly and speeds up the boiling process.

Heat from the flame also reaches the pot through the radiation process. In outdoor grilling or broiling in an oven, heat radiates from the hot charcoal or heating element directly onto the surface of the metal pan. The pan absorbs the radiant heat, which is then transferred to the food inside the pan. So kitchen pans can use all the heat transfer methods for cooking (Fig. 10.21).

Figure 10.21: Heat transfer in a kitchen pan. Heat reaches the food by conduction through the metal base, spreads by convection currents in the water, and also travels from the flame to the pan by radiation. It conducts along the handle too.

The three modes in cooking.
Mode	Where it happens in the pan
Conduction	From the burner through the metal base to the food, and along the handle
Convection	In the water or other liquid: hot, less dense liquid rises and cooler liquid sinks
Radiation	From the flame, charcoal or oven heating element to the surface of the pan

10.3 BHeating a room by convection

In winter, gas heaters are commonly used in Pakistan to keep rooms warm. When a gas heater is ignited, it heats the air surrounding it. The heated air becomes less dense and rises, generating convection currents. As the hot air rises, cooler air moves in to take its place near the heater. This cooler air is then heated by the heater, and rises. The air that has risen eventually cools down, becomes denser and sinks back towards the heater. This continuous cycle of hot air rising and cool air sinking creates a convection current that helps to distribute heat throughout the room, gradually raising the overall temperature.

Figure 10.22: Heating a room by convection. Air heated by the gas heater rises, spreads along the ceiling, cools and sinks on the far side, and the cooler air returns along the floor to the heater.

Point to ponder: why is the freezer at the top of a refrigerator?

The freezer is located at the top of a refrigerator to take advantage of convection. The cold air from the freezer travels down and cools the bottom of the refrigerator. The warm air from the bottom rises and gets cooled again in the freezer. Also, the freezer is placed above the hot compressor, which is usually located near the bottom of the refrigerator.

Why the freezer is at the top: the cold air from the freezer sinks and cools the compartment below, while the warm air from the bottom rises to the freezer and is cooled again.

10.3 CMechanism of a household hot-water system

The most common hot water system is the storage tank water heater (Fig. 10.23). Cold water enters your house and splits into hot and cold lines. The storage tank holds this cold water and heats it with an electric element or a gas burner controlled by a thermostat. Hot water naturally rises to the top of the tank. When you turn on a hot water tap, the hot water from the top flows out. As hot water is used, cooler water from the bottom refills the tank and is reheated. This cycle keeps a steady supply of hot water available.

Figure 10.23: Household hot-water system (storage tank type). Cold water enters at the bottom and is heated by the element or burner. The hot water rises to the top, from where it flows out to the tap.

10.3 DMaintaining liquid temperature: the thermos flaskOften examined

Wrapping an insulator around a hot object, or around a container holding a hot or cold liquid, can reduce the flow of heat. This property of insulators helps to maintain the temperature of liquids. It is used in thermos flasks (also called vacuum flasks), as shown in Fig. 10.24.

A thermos flask has a double-walled construction with a vacuum between its inner and outer walls. With a vacuum in the double walls, conduction of heat from the liquid through the outer wall is not possible. Transfer of heat is also not possible through the vacuum. Convection requires the movement of air for heat flow, so convection currents cannot be produced. Some heat is lost due to radiation. To minimise heat loss due to radiation, the inner walls of the flask are silvered. This reflects a significant portion of the radiant heat back towards the liquid inside, minimising heat loss.

Some of the heat can still flow out by conduction through the material of the flask and from the lid. Therefore the interior of thermos flasks is now made of stainless steel, which has a low conductivity, and the lids are made of plastic, which usually traps air for better insulation.

Figure 10.24: Thermos (vacuum) flask. The silvered inner wall is coloured. The vacuum between the double walls stops conduction and convection, the silvering reflects radiation, and the plastic stopper and the support limit the rest.

How each part of a thermos flask helps.
Feature of the flask	Which transfer it reduces
Double walls with a vacuum between them	Conduction and convection (there is no matter in the gap to conduct heat or to form convection currents)
Silvered inner walls	Radiation (the shiny coating reflects the radiant heat back towards the liquid)
Stainless steel interior (low conductivity)	Conduction through the material of the flask
Plastic lid that traps air	Conduction and convection through the lid

Worked example: why a thermos keeps tea hot

A thermos flask has double walls with a vacuum between them, silvered inner walls, a plastic lid and a stainless steel interior. (a) Name the mode of heat transfer that each feature reduces and say how. (b) A student says that the vacuum stops all heat loss from the tea. Is this correct? (c) Air leaks slowly into the gap between the walls. Which modes of transfer become possible again?

(a) Match each feature to the mode that it blocks. Conduction and convection both need matter, and a vacuum has none. Radiation can cross a vacuum, so it needs a different cure.
The vacuum between the double walls stops conduction and convection through the walls. The silvered inner walls reflect the radiant heat back into the liquid, which reduces loss by radiation.
The stainless steel interior has a low conductivity, and the plastic lid traps air, so both reduce the conduction that remains through the material of the flask and the lid.
(b) No. The vacuum stops conduction and convection across the gap, but radiation does cross a vacuum, and some heat also leaks by conduction through the material of the flask and the lid. This is why the walls are silvered and the interior and the lid are made of poor conductors. A thermos slows heat loss, it does not stop it.
(c) The air in the gap can now carry heat by conduction (air is a poor conductor, but not a zero conductor) and by convection currents between the two walls. The silvering still reduces radiation. The tea therefore cools faster than before.

Answer to part (a).
Feature	Mode reduced
Vacuum between the walls	Conduction and convection
Silvered inner walls	Radiation
Stainless steel interior and plastic lid	Conduction through the flask and the lid

\text{vacuum: conduction and convection;\ silvering: radiation;\ steel and plastic lid: leftover conduction}

10.3 EReducing thermal energy transfer in buildings

Proper thermal insulation in homes and other buildings maintains comfort, improves energy efficiency and reduces heating and cooling costs. Thermal insulation in winter slows down the heat loss from the warm interior to the colder outside environment, and it reduces the need for heating systems. In summer, it slows down the heat gain from the hot outside environment into the cooler interior, keeping the building cooler and reducing the need for air conditioning.

Thermal insulation can be achieved in many ways:

To reduce conduction and convection, use double wall glazing windows with a vacuum or air between the walls (Fig. 10.25 (a)).
Use foam and similar materials in the gaps in walls or doors.
A false ceiling traps air, which acts as insulation between the roof and the ground (Fig. 10.25 (b)).
Use thick curtains.
Proper orientation of the home and shading during the summer reduce the heat entering the home.
Proper ventilation also helps to reduce the temperature inside the home during summer.

Figure 10.25: Reducing thermal energy transfer in buildings. (a) A double glazing window has an air space (or vacuum) between two panes of glass, held by a spacer that contains desiccant. (b) A false ceiling traps a layer of air between the roof and the room.

10.3 FMeasuring temperature with infrared thermometers

All objects emit heat radiation in the form of infrared radiation. It is not visible, but it can be sensed because it carries heat. Heat coming from the Sun, heaters or hot plates has infrared radiation in large amount. It is this radiation that infrared thermometers detect to measure the temperature of objects.

When it is unsafe or impractical to make direct physical contact, infrared thermometers are used. For example, during the Corona pandemic, infrared thermometers were routinely used (Fig. 10.26). They are also used to determine the temperatures of very hot surfaces, such as those found in furnaces and industries, in food storage facilities and in hazardous places such as those containing radioactive materials.

Using an infrared thermometer

Carefully read the instructions in the device manual.
Power on the infrared thermometer and adjust its scale (degree Celsius or degree Fahrenheit).
Point the thermometer sensor to the target surface. Hold the thermometer steadily at a suitable distance mentioned in the manual (Fig. 10.26).
Press and hold the trigger (a button) until the reading on the display screen stabilises.
Once you hear a beep or see a stable reading on the display screen, release the trigger and note down the temperature.

Precautions

Each infrared thermometer has a distance-to-spot (D:S) ratio. This ratio tells you the diameter of the target area (whose temperature is to be measured) relative to the distance from the target. For example, a D:S ratio of 12:1 means that for every 12 units of distance from the object, the thermometer will measure a spot with a diameter of 1 unit. So, if you hold the thermometer 12 inches away from the surface, it will measure the temperature of a circle roughly 1 inch in diameter.
Avoid taking measurements of small objects, moving objects, wet objects, reflective surfaces and rough surfaces.
Keep comparing its accuracy, off and on, with a mercury thermometer.

Distance-to-spot ratio 12:1 (not to scale). At 12 units from the surface the thermometer measures the temperature of a circle of 1 unit in diameter.

10.4Greenhouse effectOften examined

Just like all objects, the Earth also absorbs and emits radiation. The Earth absorbs the high-energy radiation from the Sun and emits low-energy radiation back into the atmosphere. But water vapour, carbon dioxide, methane and the other greenhouse gases that form the atmosphere absorb these heat radiations emitted by the Earth and re-emit them back to the Earth. Because of this the Earth warms, which is necessary for life to sustain on it. This is called the greenhouse effect.

Greenhouse effect: The process through which heat is trapped near the Earth’s surface by substances known as greenhouse gases.

If the Earth had no atmosphere, it would have emitted all the absorbed radiations and its temperature would have dropped to -180^\circ\mathrm{C} at night (equal to the temperature of the Moon at night). At such a low temperature, life would not have existed on the Earth.

Figure 10.27 (left): Trapping heat by greenhouse gases. Sunlight reaches the Earth and some is reflected into space. The Earth emits infrared radiation, which the greenhouse gases absorb and re-emit back towards the Earth, so the heat is trapped near the surface.

Figure 10.27 (right): What happens to the sunlight that reaches the Earth. Outlined bars are returned to space (6+20+5=31\,\%). Shaded bars are absorbed (20+49=69\,\%).

10.4Global warmingOften examined

Human activities have increased the concentration of greenhouse gases in the atmosphere. These activities are:

the burning of fossil fuels (coal, wood, oil and natural gas);
a large number of vehicles on the roads, generating smoke;
many industrial and agricultural activities, which produce methane and nitrogen oxide;
deforestation, which reduces the number of trees that can absorb carbon dioxide.

Because of this, the amount of thermal radiation trapped by the greenhouse gases has increased. This has increased the average temperature of the Earth. This gradual increase in the temperature of the Earth’s atmosphere and oceans is called global warming.

Global warming: The long-term heating of the Earth’s surface observed since the pre-industrial period (between 1850 and 1900) due to human activities, primarily fossil fuel burning.

From human activity to global warming

Burning fossil fuels, vehicles, industry and agriculture add greenhouse gases to the air, and deforestation removes the trees that absorb carbon dioxide.
The concentration of greenhouse gases in the atmosphere increases.
More thermal radiation from the Earth is trapped, so the greenhouse effect becomes stronger.
The average temperature of the atmosphere and oceans rises: global warming.

This higher temperature is affecting the ecosystem, weather patterns (for example, the duration of the seasons is changing) and human health. Global warming brings effects like flooding, hurricanes, extreme rainfall and droughts. The next section discusses their causes.

10.4 A to FEffects of global warming on extreme weatherOften examined

A. Flooding

The increased temperature of the Earth is causing fast melting of glaciers, which is increasing the sea levels. Rising sea levels are a threat to coastal area life as they increase flooding and erosion of land.

Point to ponder: why are global fires increasing?

Climate models predict that by 2050 global fires could increase by 19\,\% compared to 2015. There are three main reasons for the increase in global fires:

Hotter, drier conditions. A key factor is rising temperatures. A warmer climate means more evaporation, leading to drier soil and vegetation. This creates more flammable fuel for fires to spread quickly.
Longer fire seasons. As temperatures rise, the window for hot and dry weather (prime fire season) gets longer. This extends the period when wildfires are most likely to ignite and spread.
Changes in precipitation patterns. Rainfall patterns are also predicted to shift because of climate change. Some areas may experience less rain, further worsening drought conditions and increasing fire risk.

B. Hurricanes

Global warming heats up the oceans, giving hurricanes more energy. This makes them stronger, with faster winds and heavier rain. Warm air can hold more moisture, so hurricanes can drop even more rain.

C. Rainfall

Rising temperature due to global warming has increased evaporation, causing greater moisture in the atmosphere and resulting in more intense rainfall in areas that already experience significant rainfall.

D. Droughts

Because of the greater evaporation rate, the water in soil and plants is decreasing. This is causing the land to get drier and making it harder for plants to acquire enough water to grow healthily.

Figure 10.28: Droughts and extreme rainfall. Global warming increases evaporation. A dry area becomes drier, and a wet area, with more moisture in the air, gets more intense rain.

E. Wildfires

Increased temperatures due to global warming are drying out more trees, vegetation and bushes, increasing the amount of fuel available for fires. If a fire starts due to lightning or human activity, it can quickly spread over a large area and become uncontrollable. This results in huge amounts of smoke, significantly decreasing air quality.

Mini lab: carbon footprint of daily transportation

A carbon footprint measures the greenhouse gas emissions from activities like fossil fuel burning, deforestation and industrial processes. It can be reduced by unplugging devices, using energy-efficient technologies and adopting renewable energy sources. The ecological footprint measures the resources needed to support a lifestyle, including land, water and waste absorption. A larger footprint means a higher environmental impact, so reducing it helps to protect the planet. Carbon footprints can be estimated using simple calculations.

Mini lab activity: carbon footprint of the journey to school

A car emits approximately 0.2\,\mathrm{kg} of \mathrm{CO_2} per kilometre travelled. A student travels 5\,\mathrm{km} to school every day by car. Calculate the daily and the weekly carbon dioxide emission from this transportation (assume 5 school days per week).

Given

\text{emission} = 0.2\,\mathrm{kg\,CO_2\,km^{-1}}
\text{distance} = 5\,\mathrm{km\ per\ day}
\text{days} = 5\ \text{per week}

To find

daily \mathrm{CO_2} emission
weekly \mathrm{CO_2} emission

Daily emission is the emission per kilometre multiplied by the distance.
\text{daily}=5\,\mathrm{km}\times0.2\,\mathrm{kg\,CO_2/km}=1\,\mathrm{kg\,CO_2/day}
Weekly emission is the daily emission multiplied by the number of school days.
\text{weekly}=1\,\mathrm{kg/day}\times5\,\mathrm{days}=5\,\mathrm{kg\,CO_2/week}

\text{daily}=1\,\mathrm{kg\,CO_2},\ \text{weekly}=5\,\mathrm{kg\,CO_2}

Class activity: every student calculates the carbon footprint of his or her own daily transportation and compares the result with classmates. Discuss actions to lower your carbon footprint.

F. Winter storms

Winter storms are events like snow, sleet and freezing rain with strong winds. Because of the increased moisture in the atmosphere, there is heavier snowfall in winter.

Effects of global warming (Section 10.4 A to F).
Effect	Cause linked to global warming
Flooding	Fast melting of glaciers raises sea levels
Hurricanes	Warmer oceans give them more energy; warm air holds more moisture
Rainfall	More evaporation puts more moisture in the air
Droughts	More evaporation dries the soil and plants
Wildfires	Drier trees and vegetation give more fuel; longer fire seasons
Winter storms	More moisture in the atmosphere gives heavier snowfall

Hurricane: A tropical system with winds that have reached a constant speed of 74 miles per hour or more.
Heat wave: A period of abnormally hot weather generally lasting more than two days, which can occur with or without high humidity.
Flooding: An overflow of water onto normally dry land.
Drought: A prolonged dry period (deficiency of precipitation) in the natural climate cycle that can occur anywhere in the world.

10.4Fighting global warming and recent events in Pakistan

Steps must be taken to fight the effects of global warming. Deforestation must stop and forestation must increase on the globe, because forests are a natural source that absorbs carbon dioxide. This can reduce the amount of carbon dioxide in the atmosphere. Energy-efficient sources that release no or a minimum of greenhouse gases must be found. Besides this, it is necessary to adapt to the conditions of global warming as well, by making resilient infrastructure that can sustain extreme weather events.

Stop deforestation and increase forestation.
Use energy-efficient sources that release no or minimum greenhouse gases.
Adapt to global warming: build resilient infrastructure that can sustain extreme weather events.

Information: recent heat waves and floods in Pakistan

Heat waves, May to June 2022. A devastating heat wave gripped Pakistan and India with temperatures exceeding 51^\circ\mathrm{C} (123.8^\circ\mathrm{F}). Scientists linked this event to global warming.

Devastating floods, 2022. Severe monsoon rains and rapid glacial melt, both influenced by climate change, caused disastrous floods in the whole country, especially in the provinces of Sindh, major areas of Khyber Pakhtunkhwa and Balochistan, and some areas of Punjab.

Information: communities at higher risk

The book also shows a picture of communities at higher risk of health issues related to climate change, with possible remedies. Dark text in the picture gives steps that can be taken to reduce the dangers those communities face, while white text highlights the risks.

10.5Structure of the Earth

The structure of the Earth is divided into three parts: core, mantle and crust (Fig. 10.29).

The core is the central part of the Earth. It is the hottest part. Its temperature is about 6000^\circ\mathrm{C}. It is mainly composed of iron and nickel. The core is so hot because of different factors: left-over heat from the formation of the Earth, radioactive decay of elements like uranium and thorium inside the core, and the extra pressure applied by the Earth’s gravity. The inner core is the portion of the core that is solid. The core’s solid form is maintained by the extreme pressure. The molten portion of the core is called the outer core. It is molten because the pressure there has dropped. The temperature of the outer core is about 4000^\circ\mathrm{C}.

The mantle lies between the core and the crust. It is the thickest part. It floats on the outer core. It is composed of silicate minerals mainly containing iron and magnesium. It is solid but shows plasticity (permanent change in shape of a solid material) under the high pressure due to gravity.

The crust is the outer part of the Earth’s structure. It is composed of silicate materials. It is divided into continental crust and oceanic crust.

Figure 10.29: Structure of the Earth. From outside to inside: crust, mantle, outer core and inner core. The thin crust and the uppermost mantle together form the lithosphere.

The layers of the Earth, from the book’s description.
Layer	What it is	Notes
Crust	Outer part, made of silicate materials; continental crust and oceanic crust	Thin layer. With the upper-most mantle it forms the lithosphere
Mantle	Between core and crust; silicate minerals mainly containing iron and magnesium	The thickest part; solid but shows plasticity; floats on the outer core
Outer core	Molten part of the core, because the pressure has dropped	About 4000^\circ\mathrm{C}
Inner core	Solid part of the core, kept solid by the extreme pressure	Core: mainly iron and nickel, about 6000^\circ\mathrm{C}

Lithosphere: The layer between the crust and the upper mantle (the crust and the uppermost mantle, as labelled in Fig. 10.29).
Asthenosphere: The layer that lies just beneath the lithosphere and plays a critical role in plate tectonics.
Magma: Molten or partially molten rock located beneath the Earth’s crust. It forms from the partial melting of the mantle and crust due to various geological processes.
Geothermal activity: A group of natural heat transfer processes, caused by the presence of excess heat in the subsurface of the affected area, occurring on the Earth’s surface.

10.5.1Flow of magma beneath the Earth

There is an extremely high temperature, around 2000^\circ\mathrm{C} to 6000^\circ\mathrm{C}, between the core and the crust. This intense heat within the mantle can cause some solid rock to partially melt, forming magma. This molten material, magma, is less dense than the surrounding solid rock.

Heat is transferred from the core to the mantle by conduction, because there is a temperature difference between the core and the mantle. In the mantle, magma gets heat, becomes less dense and rises up, that is, it transfers heat by convection. Hot magma rises up and cold magma sinks. This process develops a convectional current (Fig. 10.30). In this way magma flows from the bottom of the mantle towards the crust.

Figure 10.30: Flow of magma beneath the Earth. Hot magma rises under the oceanic ridge and moves sideways, cools and sinks. These convection currents drag the plates, giving ridge push at the ridge and slab pull at the ocean trench.

Hotter magma flows more easily than colder magma. When magma flows, it melts the rocks on the way, which may change its composition. When magma cools down, it solidifies and forms rocks deep underground. If magma finds a path to the surface through vents or fissures, it erupts as lava, shaping volcanic landscapes.

Heat flow beneath the Earth.
Where	Mode of heat transfer	Why
Core to mantle	Conduction	There is a temperature difference between the core and the mantle
Inside the mantle	Convection	Hot magma is less dense and rises, cold magma sinks

10.5.2Volcanic eruptions

Magma rises from the mantle because of convectional currents and the pressure difference in the mantle and crust. It is stored in the crust at a place called the magma chamber.

As magma rises, its pressure decreases. This allows the dissolved gases such as water vapour, carbon dioxide and other volatiles to be released, and it increases the internal pressure of the magma chamber. When this pressure exceeds the strength of the covering rock, the magma fractures it and creates pathways (main and secondary vents) for itself to reach the crust. This leads to a forceful volcanic eruption through the crater (main opening), driven by high-pressure gases and magma (Fig. 10.31). The hot magma that flows out during an eruption is called lava. A volcanic eruption depends upon factors like magma composition, temperature, amount of gas and interaction with water.

Figure 10.31: Volcanic eruption. Magma rises from the magma chamber through the conduit and the vents. Gases, ash and lava leave through the crater, and the hot magma that flows out is lava (coloured).

How a volcanic eruption happens

Magma rises from the mantle by convection currents and pressure differences, and is stored in the magma chamber in the crust.
As it rises, its pressure decreases and dissolved gases (water vapour, carbon dioxide and others) are released, which increases the pressure in the chamber.
When this pressure is greater than the strength of the rock above, the magma fractures the rock and makes main and secondary vents.
A forceful eruption occurs through the crater, driven by gases and magma. The magma that flows out is lava.

10.5.3Movement of tectonic plates

Tectonic plates are very large, rigid parts of the Earth’s crust and uppermost mantle. These plates float on the molten mantle (the semi-fluid asthenosphere). They interact with other tectonic plates and generate geothermal activities like earthquakes, volcanic eruptions and so on.

The movement of a tectonic plate is basically due to the process of magma convection currents in the mantle. Convectional currents of magma drag the plates along with them and cause horizontal movement in the tectonic plates. Similarly, the sinking of dense and cold magma due to convection applies a pull on the tectonic plates and helps them to move (Fig. 10.32).

Figure 10.32: Movement of tectonic plates caused by the motion of magma. Convection currents in the asthenosphere drag the plates (frictional drag) and the sinking cold slab at the trench pulls the plate along.

The movement of tectonic plates is very slow, a few centimetres per year. Because of their movement, tectonic plates can move apart (diverge or spread), come together (converge), or slide past each other (transform faulting).

The three kinds of plate movement, seen from above: plates move apart (diverge or spread), come together (converge) or slide past each other (transform faulting).

Information: the major tectonic plates

There are seven major tectonic plates:

Pacific Plate
North American Plate
South American Plate
Eurasian Plate
African Plate
Indo-Australian Plate (it is considered as two separate tectonic plates, that is, the Indian plate and the Australian plate)
Antarctic Plate

Apart from these there are eight minor plates (the Arabian plate, the Caribbean plate, the Indian plate and others) and many micro plates (the Rivera plate, the Somali plate and others) as well.

10.5.4Earth’s core maintaining its temperature

Scientists believe that the Earth’s formation involved accretion, where gravity pulls matter together to create celestial bodies. After the Sun formed, gravity combined dust, rocks, ice and gases to form massive objects like the Earth. According to the Giant Impact Hypothesis, a planet named Theia collided with the Earth about 4.4 billion years ago, resulting in the destruction of Theia, the partial melting of the Earth and the scattering of debris into space. The early atmosphere was mainly hydrogen and helium. As the crust solidified, volcanic eruptions and asteroid impacts introduced water, water vapour, ammonia and carbon dioxide. Over time the Earth evolved to support life. Today the Earth’s core remains extremely hot, around 6000^\circ\mathrm{C}. The following factors keep the Earth’s core at this temperature.

Accretion: A process in which particles of matter, such as dust and gas, gradually come together due to gravitational attraction to form large objects, like planets, moons and so on.
Debris: The fragments of rock, dust and other material that were scattered into space as a result of the collision between Theia and the Earth.

A. Residual heat from formation

During the formation of the Earth (the accretion process), a large number of massive objects, for example planetesimal objects and asteroids, collided with it. The kinetic energy from these collisions was converted into heat, and a large amount of this heat energy has been retained in the Earth’s core.

B. Radioactive decay

There are certain radioactive elements present in the Earth’s core, like uranium-238, thorium-232 and potassium-40. During the decay of these radioactive elements, a large amount of energy is released in the form of heat. This process is continuously contributing to the temperature of the core. Because of the decay process, the amount of these radioactive elements is also decreasing, and that is why the rate of heat generation from radioactivity has decreased slightly.

C. Gravitational compression

During the formation of the Earth, its materials were compressed under the force of gravity. This increased the pressure and hence the temperature in the core. In this way gravitational compression becomes a constant source of heat to the core.

D. Crust as an insulator

The Earth’s outer layers act as a thick layer of insulation. This slows down the rate at which heat escapes from the core into space. So the crust helps to maintain the high temperature of the core for a longer period.

Why the Earth’s core stays hot.
Factor	How it keeps the core hot
Residual heat from formation	Kinetic energy of collisions during accretion was changed to heat and is still retained
Radioactive decay	Decay of uranium-238, thorium-232 and potassium-40 releases heat continuously (the rate has decreased slightly)
Gravitational compression	Compression by gravity raises the pressure and temperature and is a constant source of heat
Crust as an insulator	The thick outer layers slow the escape of heat from the core to space

The mechanisms that generate heat are gradually slowing down, allowing heat to escape from the Earth. However, this process is so slow that the core has managed to maintain its temperature for the past 4 billion years.

Formula sheet

Specific heat capacity

c=\dfrac{Q}{m\,\Delta T}

c: specific heat of the material; Q: heat absorbed or lost; m: mass; \Delta T=T_{\text{final}}-T_{\text{initial}}: change in temperature (Eq. 10.1)

Units: \mathrm{J\,kg^{-1}\,K^{-1}} (also \mathrm{J\,kg^{-1}\,{}^\circ C^{-1}} and \mathrm{J\,g^{-1}\,{}^\circ C^{-1}})

Heat absorbed or lost

Q=mc\,\Delta T

Q: heat; m: mass; c: specific heat; \Delta T: change in temperature

Unit: \mathrm{J}

Heat capacity of a body

C=\dfrac{Q}{\Delta T}=mc

C: heat needed to raise the temperature of the whole body by 1\,\mathrm{K}

Units: \mathrm{J\,K^{-1}} (or \mathrm{J\,{}^\circ C^{-1}})

Specific heat of a solid (method of mixtures)

c_s=\dfrac{(m_wc_w+m_cc_c)(T_3-T_1)}{m_s\,(T_2-T_3)}

m_s,m_w,m_c: masses of solid, water and calorimeter; c_w,c_c: specific heats of water and calorimeter; T_1: initial temperature of water and calorimeter; T_2: initial temperature of the solid; T_3: final temperature of the mixture (Eq. 10.2)

same unit as c_w and c_c

Specific heat of a liquid (method of mixtures)

c_l=\dfrac{m_sc_s(T_2-T_3)-m_cc_c(T_3-T_1)}{m_l\,(T_3-T_1)}

m_l: mass of the liquid; c_s: known specific heat of the solid; other symbols as in Eq. 10.2 (Eq. 10.3)

same unit as c_s and c_c

Heat from the electric heater

Q=V\,I\,t

V: voltage across the heater; I: current; t: time for which it is switched on

Units: \mathrm{J} (volt, ampere, second)

Specific heat of a liquid (electrical method)

c_l=\dfrac{Q_{\text{heater}}-m_cc_c(T_f-T_i)}{m_l\,(T_f-T_i)}

Q_{\text{heater}}: heat supplied by the heater; m_l,m_c: masses of liquid and calorimeter; c_c: specific heat of the calorimeter; T_i,T_f: initial and final temperatures (Eq. 10.5)

Unit: \mathrm{J\,kg^{-1}\,K^{-1}}

Key terms

Heat capacity: The amount of heat required to raise the temperature of any amount of a substance by 1^\circ\mathrm{C} or 1\,\mathrm{K}.
Specific heat capacity: The amount of heat absorbed or lost to change the temperature of 1\,\mathrm{kg} of a substance by 1\,\mathrm{K} or 1^\circ\mathrm{C} (the unit summary of the book says one gram, with \mathrm{J\,g^{-1}\,K^{-1}}).
Conduction: The mode of transfer of heat due to direct contact of medium particles, which do not change their positions. It occurs in solids because of a difference in temperature.
Convection: The mode of transfer of heat due to the motion of particles. It occurs in liquids and gases because of the density difference between different regions.
Radiation: The mode of transfer of heat due to electromagnetic waves. It does not require a material medium for its motion.
Thermal equilibrium: The state in which two objects (or an object and its surroundings) are at the same temperature, or an object absorbs and radiates heat at equal rates, so that its temperature stays the same.
Land breeze: The breeze that blows from the land. It normally occurs during the night and early morning.
Sea breeze: The breeze that blows from the massive water bodies. It normally occurs during the summer and spring season (by day).
Thermal: A column of rising hot air, surrounded by cooler air, that birds and gliders ride to gain height.
Greenhouse effect: The process through which heat is trapped near the Earth’s surface by substances known as greenhouse gases.
Global warming: The long-term heating of the Earth’s surface observed since the pre-industrial period (between 1850 and 1900) due to human activities, primarily fossil fuel burning.
Cyclones: Large revolving tropical storms caused by winds blowing around a central area of low atmospheric pressure.
Hurricane: A tropical system with winds that have reached a constant speed of 74 miles per hour or more.
Heat wave: A period of abnormally hot weather generally lasting more than two days, which can occur with or without high humidity.
Flooding: An overflow of water onto normally dry land.
Drought: A prolonged dry period (deficiency of precipitation) in the natural climate cycle that can occur anywhere in the world.
Accretion: A process in which particles of matter, such as dust and gas, gradually come together due to gravitational attraction to form large objects, like planets, moons and so on.
Debris: The fragments of rock, dust and other material that were scattered into space as a result of the collision between Theia and the Earth.
Geothermal activity: A group of natural heat transfer processes, caused by the presence of excess heat in the subsurface of the affected area, occurring on the Earth’s surface.

Multiple choice questionsp. 40–41

1

Why is water used in radiators of automobile as coolant?

AIt is easily available
BIt is low cost or free
CIt has large specific heatAnswer
DIt has oxygen

Why

Water has a large specific heat, 4200\,\mathrm{J\,kg^{-1}\,K^{-1}}. A large specific heat means water absorbs a large amount of heat from the engine for only a small rise in its temperature, so it carries away much heat before it boils.

Availability (A) and low cost (B) are practical points, not the physical reason. Oxygen (D) does not help cooling, and water is not chosen for it.

2

Which of the following situations is the best example of conduction?

AA metal spoon becomes hot when placed in boiling waterAnswer
BWarm air rising near a heater
CSunlight warming the surface of the Earth
DA microwave oven heating food

Why

Conduction is the transfer of heat through direct contact between particles of a solid, from the hot part to the cold part. Heat from the boiling water passes along the metal spoon to the part that is not in the water. This is conduction.

Warm air rising near a heater (B) is convection, because the hot fluid itself moves. Sunlight (C) reaches the Earth by radiation through the vacuum of space, and the microwave oven (D) heats food by electromagnetic waves, so neither needs contact.

3

Which combination of heat transfer methods would be dominant when you place your hand near, but not touching, a fire?

AConduction and radiation
BConvection and conduction
CRadiation and convectionAnswer
DConduction and insulation

Why

Your hand is not touching the fire, so conduction is not possible. The infrared radiation from the fire reaches the hand directly through the air, which is radiation.

The hot air above the fire rises and carries warmth to the hand, which is convection. Conduction (A, B) needs contact, and insulation (D) is not a method of heat transfer.

4

What is symbol and what is unit for the heat capacity of an object?

A\mathrm{J\,{}^\circ C^{-1}}Answer
B\mathrm{J\,kg^{-1}\,K^{-1}}
C\mathrm{J\,kg\,K}
D\mathrm{J\,kg^{-1}\,K}

Why

The heat capacity of an object is C=\dfrac{Q}{\Delta T}, so its unit is joule per kelvin, \mathrm{J\,K^{-1}}. A change of 1 K equals a change of 1 ^\circC, so the same unit is written \mathrm{J\,{}^\circ C^{-1}}. Option A is correct.

Option B, \mathrm{J\,kg^{-1}\,K^{-1}}, is the unit of specific heat capacity, which is per kilogram. Options C and D contain kilogram in the wrong place and are not units of heat capacity.

5

If the same amount of heat energy is supplied to equal masses of water and copper, why does the temperature of copper increase faster?

ACopper has a lower specific heat capacityAnswer
BWater is a poor conductor of heat
CConvection in water dissipates heat energy quickly
DRadiation from water is stronger

Why

For the same heat Q and mass m, the rise in temperature is \Delta T=\dfrac{Q}{mc}. Copper has the smaller specific heat (387\,\mathrm{J\,kg^{-1}\,K^{-1}} against 4190\,\mathrm{J\,kg^{-1}\,K^{-1}} for water), so the same heat gives it a larger rise in temperature. Option A is the reason.

Options B, C and D describe water losing heat, which would slow the rise in water's temperature. They do not explain why copper, which has no such losses in this comparison, heats faster.

6

The transfer of heat that takes place because of density difference in fluids is

AConduction
BRadiation
CConvectionAnswer
DInsulation

Why

When a fluid is heated, the part near the heat source becomes less dense and rises, while cooler, denser fluid sinks. This bulk movement of the fluid transfers heat, and it is convection. It happens only in liquids and gases.

Conduction (A) happens in solids through particle contact, radiation (B) needs no medium, and insulation (D) is a way of reducing heat transfer, not a mode of it.

7

Which of the following statements best explains why the Earth experiences more heat from the Sun than the Moon, despite being almost the same distance away?

AEarth is better conductor than the moon
BThe Earth has greenhouse gasesAnswer
CThe Moon reflects most of the Sun's radiation
DMoon traps heat effectively

Why

The Earth has an atmosphere containing greenhouse gases such as water vapour and carbon dioxide. These gases absorb the infrared radiation emitted by the Earth and send it back, so the Earth keeps more heat. The Moon has no atmosphere, so it loses its heat away into space, which is option B.

Conduction (A) does not carry heat from the Sun across space, and the Moon does not trap heat (D). Option C is wrong because the Moon's reflection of sunlight would reduce, not increase, the heat it receives; the question asks why the Earth gets more heat.

8

Dull black colour on a surface is the best absorber of radiation, which of the followings is the best radiator?

ADull black surfaceAnswer
BShining silver surface
CRed coloured surface
DWhite surface

Why

A good absorber is also a good emitter. Table 10.2 shows that black surfaces are the best absorbers and the best emitters, so a dull black surface is the best radiator of the options. This is option A.

Shining silver (B) is the worst absorber and emitter, and white (D) is a bad emitter because it is a good reflector. Red (C) is not treated in the book as a best emitter, so it is not the best choice.

9

How does the enhanced greenhouse effect contribute to global warming?

AIt increases the Earth's ability to reflect solar radiation
BIt traps more heat in the Earth's atmosphere, raising global temperaturesAnswer
CIt blocks ultraviolet rays from entering the atmosphere
DIt increases the Earth's rotation speed, causing heat buildup

Why

The greenhouse effect traps heat radiated by the Earth in the atmosphere. When human activities add more greenhouse gases, more heat is trapped, the average temperature of the Earth rises, and this is global warming (Section 10.4).

Option A is the opposite of what happens: reflection would cool the Earth. Options C and D are not part of the greenhouse effect described in the book.

10

What is the primary driving force behind the movement of tectonic plates?

AGravitational pull of the Moon
BSolar radiation
CMantle convection currentsAnswer
DMagnetic field of the Earth

Why

The plates are moved mainly by convection currents of magma in the mantle. Hot magma rises, cool magma sinks, and the moving magma drags the plates along, as shown in Figure 10.32.

The Moon's pull (A) produces tides, solar radiation (B) heats the surface rather than moving the plates, and the magnetic field (D) is not the driving force described in the book.

11

Which layer of the Earth is composed of partially molten rock that can flow slowly?

ALithosphere
BAsthenosphereAnswer
CMesosphere
DOuter core

Why

The asthenosphere lies just beneath the lithosphere and is partly molten rock. It is soft and ductile, so it can flow slowly, which allows the plates above it to move.

The lithosphere (A) is the rigid upper layer, the outer core (D) is molten metal (iron and nickel) rather than rock, and the mesosphere (C) is not one of the layers named in this unit.

12

Which of the following extreme weather events is most directly associated with rising sea levels?

ATornadoes
BWildfires
CHurricanesAnswer
DEarthquakes

Why

Hurricanes push a large mass of sea water ahead of them, called a storm surge. When the sea level is higher, this surge floods further inland. Among the events listed, the hurricane is the one that is most directly linked to the sea water rising onto the land, which is why it is option C.

Tornadoes (A) and wildfires (B) are not caused by sea water, and earthquakes (D) come from the movement of tectonic plates. Note: the book does not state this link in words; the answer follows from its description of flooding and erosion along coasts (Section 10.4, part A).

Constructed response questionsp. 41–42

1

This AI-generated image illustrates global warming caused by the greenhouse effect and its impact on climate change. It shows how rising temperatures due to the greenhouse effect led to more frequent and intense heatwaves and hurricanes. Answer the following questions, considering the concepts mentioned above.

Solution

(a) How do increased temperatures lead to more frequent heatwaves in cities?

A heat wave is a period of abnormally hot weather lasting more than two days (book, unit 10 opener). When the average temperature of the Earth rises because of the greenhouse effect, the normal range of temperatures moves upwards, so the abnormally hot days come more often and last longer.

Cities add to this. Roads, concrete and dark buildings are dry and dull surfaces, which absorb more radiation, and dry land has a small specific heat (800\,\mathrm{J\,kg^{-1}\,K^{-1}} for dry soil against 4190\,\mathrm{J\,kg^{-1}\,K^{-1}} for water). So city surfaces heat up quickly during the day, and the air above them stays hot. This is why cities reach 45 to 50\,^\circ\mathrm{C} in summer more easily than coastal areas (Section 10.1.1, part A).

(b) How might the intensity of future hurricanes change if global temperatures continue to rise?

Hurricanes form over warm ocean water. Warmer oceans cause more evaporation, so more warm, moist air rises by convection, and more latent heat is released when it condenses. This adds energy to the storm.

If global temperatures keep rising, hurricanes are expected to become stronger, with faster winds and heavier rain, because warmer oceans give them more energy (Section 10.4, part B). The book describes this as a stronger system rather than a change in how often hurricanes form.

(c) Explain what the glass roof of a greenhouse and carbon dioxide in Earth's atmosphere have in common in terms of heat and temperature.

Both allow the sunlight to pass through them and reach the ground, and then both make it difficult for the heat (infrared radiation) given out by the ground to escape. The heat is therefore kept inside, and the temperature rises.

In the greenhouse the glass traps the warm air inside, and in the atmosphere carbon dioxide absorbs the infrared radiation from the Earth and sends it back to the surface. Both are therefore examples of trapping heat, which is the greenhouse effect (Section 10.4).

2

This table shows the temperature of 10 g of water and 10 g of iron after being in direct sunlight for up to 60 minutes. Time (min): 0, 15, 30, 45, 60. Temperature of water (°C): 25, 26.2, 27.5, 28.8, 30. Temperature of iron (°C): 25, 35, 45, 55, 65.

Temperature of 10 g of water and 10 g of iron in direct sunlight
Time (minutes)	0	15	30	45	60
Temperature of water (^\circC)	25	26.2	27.5	28.8	30
Temperature of iron (^\circC)	25	35	45	55	65

Temperature of 10 g of water and 10 g of iron in direct sunlight
Time (minutes)	Temperature of water (^\circC)	Temperature of iron (^\circC)
0	25	25
15	26.2	35
30	27.5	45
45	28.8	55
60	30	65

Solution

(a) Create line graphs of temperature (on the y-axis) and time (on the x-axis) for both water and iron on a graph sheet. Be sure to label the substances.

The two graphs are plotted below from the table. Each measured point is marked and joined to the next, so the rise of iron is much steeper than that of water. Both substances are labelled on their lines.

Temperature of water and iron in sunlight

Temperature against time for 10 g of iron and 10 g of water. Iron rises by 40\,^\circ\mathrm{C} in an hour, water by only 5\,^\circ\mathrm{C}.

(b) Based on the data in the table, which heats up quickly: water or metal?

Iron (the metal) heats up quickly. In 60 minutes its temperature rises from 25\,^\circ\mathrm{C} to 65\,^\circ\mathrm{C}, a rise of 40\,^\circ\mathrm{C}, while the water rises only from 25\,^\circ\mathrm{C} to 30\,^\circ\mathrm{C}, a rise of 5\,^\circ\mathrm{C}.

Both samples have the same mass and receive the same sunlight, so the heat received is about the same. The large difference in temperature rise shows that iron has a much smaller specific heat, 452\,\mathrm{J\,kg^{-1}\,K^{-1}} against 4190\,\mathrm{J\,kg^{-1}\,K^{-1}} for water (Table 10.1), which is why the metal heats up quickly.

(c) Which do you think will cool more slowly: water or iron? Elaborate your answer.

Water cools more slowly. Cooling follows the same law as heating: the temperature drop is \Delta T=\dfrac{Q}{mc}, so for the same heat lost and the same mass, a substance with a large specific heat changes its temperature by a small amount.

Water has a large specific heat, so it can lose a lot of heat for only a small drop in temperature. Iron has a small specific heat, so its temperature falls quickly when it loses the same heat. This is also why water keeps its warmth for longer, as the book states in Section 10.1.

(d) When you boil water in an iron pot on the stove, which heats up faster: the iron pot or the water? Provide a reason for your response.

The iron pot heats up faster. It is in direct contact with the flame or the hot plate, and heat passes into the iron by conduction. Iron has a small specific heat, so each unit of heat it receives gives it a large rise in temperature.

The water in the pot has a large specific heat, so it takes much more heat to raise its temperature. The pot heats first and then passes heat to the water, which is why the water takes longer to reach 100\,^\circ\mathrm{C}.

Short response questionsp. 42–43

1

Why should we wear dark-coloured clothes in winter and white-coloured clothes in summer?

Solution

Winter: dark colours are good absorbers of radiation. Black is the best absorber (Table 10.2), so dark clothes take in more of the radiant heat falling on them and warm the body more.

Summer: white is a good reflector and a bad absorber. White clothes reflect most of the sunlight, so the body absorbs less heat and stays cooler.

2

In a house, geysers or water boilers are fitted on the ground floor, and still, we get warm water on the top floor without using a pump. How is it possible?

Solution

This is done by convection. The water in the tank is heated at the bottom by the burner or electric element. Heated water expands, becomes less dense and rises to the top of the tank.

Cooler, denser water from the top of the house's supply flows down into the bottom of the heater to take its place and is heated in turn. The hot water at the top of the tank then flows out through the taps on the upper floor, as in the household hot-water system of Figure 10.23. No pump is needed because the density difference drives the flow.

3

Where will you get more heat from the wood fire, 1 meter above the woods or 1 meter from the front of the woods?

Solution

You will get more heat 1 meter above the woods. The hot gases and air from the fire rise by convection, because hot air is less dense. The air directly above the fire is therefore the hottest air around the fire.

Radiation from the flames reaches you on both sides, but the warm rising air carries a large part of the heat straight upward. Note: the book does not print an answer to this question; the answer is based on the convection described in Section 10.2.2.

4

Why do crowded city areas feel hotter compared to the outskirts on a hot summer day? State the reasons for this difference.

Solution

Crowded city areas have many dark and dull surfaces, such as roads, roofs and walls, which are good absorbers of radiation (Table 10.2). They take in more of the sun's heat than the grass, soil and trees of the outskirts.

The surfaces are also dry and have a small specific heat. Dry soil has 800\,\mathrm{J\,kg^{-1}\,K^{-1}} against 4190\,\mathrm{J\,kg^{-1}\,K^{-1}} for water, so the city surfaces heat up quickly and reach high temperatures, 45 to 50\,^\circ\mathrm{C} in summer (Section 10.1.1, part A).

Cities also have fewer trees and water bodies. Water keeps the temperature moderate because of its large specific heat, so the outskirts with lakes and open land stay cooler. Vehicles and industry add greenhouse gases, which trap more heat in the city air.

5

Why is the metallic handle of a door colder than the wood of the same door when touched?

Solution

Both the handle and the wood are at the same room temperature. The metal is a good conductor of heat (Figure 10.6), so when you touch it, heat moves quickly from your hand into the metal and away along it. Your hand loses heat quickly, so the metal feels cold.

Wood is a poor conductor (a better insulator), so heat from your hand does not flow into it quickly. The wood feels warmer because your hand keeps its heat near the surface.

6

How do trees help reduce the effects of climate change, and what could happen if forests are depleted?

Solution

Trees absorb carbon dioxide, which is one of the greenhouse gases. Forests are a natural carbon dioxide absorbing source, so they reduce the amount of carbon dioxide in the atmosphere, which in turn reduces the trapping of heat and the effects of global warming (Section 10.4, part F).

If forests are cut down (deforestation), fewer trees can absorb carbon dioxide, so the concentration of greenhouse gases rises. More heat is trapped, global warming increases, and the land becomes drier, which leads to droughts and changes in the weather.

7

How does gravity contribute to the Earth's core temperature?

Solution

During the formation of the Earth, large numbers of massive objects collided with it and gave their kinetic energy to heat. Gravity pulled the materials together, and this gravitational compression increased the pressure and the temperature in the core.

The compression is a continuing source of heat to the core, since gravity still acts on the materials of the Earth (Section 10.5.4, parts A and C).

8

Why do certain gases in the atmosphere trap more heat than others?

Solution

Greenhouse gases such as water vapour, carbon dioxide and methane absorb the infrared radiation emitted by the Earth and re-emit it back towards the Earth, so the heat is trapped near the surface (Section 10.4).

The book does not compare the gases quantitatively, so the answer is limited to this statement: a gas traps more heat when it absorbs more of the infrared radiation that the Earth gives out, and re-emits it back to the surface.

9

What are the potential environmental impacts of extraction of geothermal energy?

Solution

The book does not discuss the extraction of geothermal energy, so this answer is limited to the book's description of geothermal activity (Section 10.5).

Geothermal activity releases heat from the Earth's interior. Volcanic eruptions throw out ash, lava and gases such as carbon dioxide, sulphur dioxide and water vapour (Figure 10.31), and the movement of tectonic plates causes earthquakes, which are the environmental effects the book describes.

Note. The question asks about geothermal energy extraction, which is not in this unit of the book. The answer above uses only the book's section on geothermal activity.

10

How does the specific heat capacity of different materials affect their use in cookware?

Solution

Metals used for cookware, such as aluminium, copper and iron, have a small specific heat (Table 10.1: 910, 387 and 452\,\mathrm{J\,kg^{-1}\,K^{-1}}). They need little heat to raise their temperature, so the pan heats up quickly and spreads the heat through its base and sides.

The handles are made of wood or plastic, which are poor conductors, so the heat does not reach the hand. Water has a large specific heat, so it is used where a steady temperature is needed rather than a quick rise, as in the cooking of food in boiling water (Section 10.1.1, part E).

11

Water at 20\,^\circ\mathrm{C} is sent deep underground into heated layers, where it turns into a mix of steam and hot water at 100\,^\circ\mathrm{C}. As the steam and hot water cool back down to 20\,^\circ\mathrm{C}, why does 1\,\mathrm{kg} of steam release more energy than 1\,\mathrm{kg} of hot water?

Solution

Cooling hot water from 100\,^\circ\mathrm{C} to 20\,^\circ\mathrm{C} releases Q=mc\Delta T = 1\times4190\times80 \approx 3.4\times10^{5}\,\mathrm{J}. The steam has a smaller specific heat, 1520\,\mathrm{J\,kg^{-1}\,K^{-1}} (Table 10.1), so cooling the steam by the same 80\,^\circ\mathrm{C} on its own releases much less, only about 1.2\times10^{5}\,\mathrm{J}.

The steam releases much more energy in total because it must first condense from steam to water at 100\,^\circ\mathrm{C}. Condensation releases a large amount of heat without any change of temperature. Only after this does the water cool from 100\,^\circ\mathrm{C} to 20\,^\circ\mathrm{C}. The book treats this energy as latent heat in the change of state (Unit 11), so this unit gives only the principle.

Note. Table 10.1 gives the specific heat of steam at 100 ^\circC as 1520 J kg^{-1} K^{-1}, which is the value used here. Standard tables give a higher value for steam (roughly 2000 J kg^{-1} K^{-1}), so the energy released by cooling the steam alone is underestimated; the conclusion that condensation adds most of the energy is unchanged.

Long response questionsp. 43

1

Explain the concept of specific heat capacity in matter. Discuss various applications of water based on its high specific heat capacity.

Solution

Specific heat capacity

The specific heat capacity c of a substance is the amount of heat absorbed or lost by 1 kg of the substance to change its temperature by 1 K or 1 ^\circC. It is found from the heat Q gained or lost by a mass m and the change in temperature \Delta T:

c=\dfrac{Q}{m\,\Delta T}

10.1

Its SI unit is \mathrm{J\,kg^{-1}\,K^{-1}}. Rearranged, the heat needed is Q=mc\,\Delta T. A substance with a large c needs a large amount of heat for each degree of temperature rise.

The graph shows this for 1 kg of water and 1 kg of iron. For the same heat, the iron rises in temperature about nine times as much as the water, because its specific heat is about nine times smaller.

Same heat, different temperature rise

Temperature rise of 1 kg of water (c=4200\,\mathrm{J\,kg^{-1}\,K^{-1}}) and 1 kg of iron (c=452\,\mathrm{J\,kg^{-1}\,K^{-1}}) for the same heat.

Applications of the high specific heat of water

Car engine coolant. Water absorbs a large amount of heat from the engine with only a small rise in its temperature, so it carries heat away through the radiator (Figure 10.1).
Cooling in power plants and industry. Water takes large amounts of excess heat from machines through a heat exchanger, so the machinery does not overheat.
Coastal climate. Oceans and large lakes absorb a lot of heat in summer and release it in winter, so the coastal temperature varies only between about 19 and 32 ^\circC in Karachi, while inland areas vary much more.
Aquatic life. The temperature of lakes and oceans changes slowly, which gives a stable environment for fish and other water life.
Human body. The body is about 60 % water, which keeps its temperature near 37 ^\circC.
Cooking. Water heats up slowly and distributes heat evenly, so food cooks without burning and hot drinks stay hot for a long time.

2FBISE SSC-II 2026 Annual I, Set B, Q.3

What is conduction, and how is it explained by the kinetic theory of solids? What makes metals better conductors than other solid substances.

Solution

Conduction

Conduction is the transfer of heat through a solid by collisions of atoms or molecules and by the motion of free electrons, from the hot part to the cold part. The particles do not move from place to place: each one passes its energy to its neighbour.

Explanation from the kinetic theory of solids

In a solid the atoms are held in fixed positions and vibrate about them. The kinetic theory says that a higher temperature means a larger average vibration. When one end is heated, the atoms there vibrate with a larger amplitude, collide with their neighbours and transfer part of their energy to them. The neighbours then vibrate more strongly and pass the energy on, so heat moves along the rod.

Energy passes along a metal rod by atomic collisions and by free electrons moving from the hot end to the cold end.

Why metals are better conductors

Metals have a large number of free electrons that are not bound to any one atom. These electrons move quickly through the metal. Near the hot end they gain energy from the vibrating atoms, travel fast across the metal and give the energy up to atoms at the cold end. This transfer is much faster than the collisions of atoms alone.

Other solids such as wood, glass and rubber have few free electrons, so heat moves through them only by slow atomic collisions. They are therefore poor conductors, or good insulators (Figure 10.6).

The experiment with waxed drawing pins on metal rods (Figure 10.8) shows the same result: the pin falls first from the copper rod, then from the aluminium, followed by iron and brass. So copper conducts better than aluminium, and aluminium better than iron and brass.

3

Analyze how heat is transferred through convection. Provide at least three practical examples of how convection is used in daily life.

Solution

Convection

Convection is the transfer of heat by the bulk movement of the molecules of a fluid. When a liquid or gas is heated, the fluid near the source becomes less dense, rises, and is replaced by cooler, denser fluid, which is then heated in turn. This forms a convection current that carries heat from the bottom to the top. Convection takes place only in liquids and gases, not in solids.

A fluid heated from below: the warm fluid rises in the middle, cools at the top and sinks at the sides, forming a convection current.

Practical examples

Heating a room. A gas heater heats the air next to it. The hot air becomes less dense and rises, cool air moves in to take its place and is heated, and a convection current spreads the warmth through the room (Figure 10.22).
Household hot-water system. Water heated in the storage tank becomes less dense and rises to the top, so hot water is available at the taps. Cooler water from the bottom refills the tank and is reheated (Figure 10.23).
Sea breezes and land breezes. During the day the land heats up quickly, the air above it rises and cool sea air moves in to fill the gap. At night the sea stays warmer, its air rises and cool land air moves out to the sea (Figure 10.15).

Other examples in the book are the freezer compartment placed at the top of a refrigerator, and the rising warm air that birds use to glide on thermals (Figure 10.16).

4

Analyze the process of heat transfer through radiation, explaining why it is the fastest method of heat transfer. What factors influence the rate of heat transfer by radiation?

Solution

Radiation

Radiation is the transfer of heat by electromagnetic waves, mainly infrared. It needs no material medium: heat from the Sun reaches the Earth across the vacuum of space. Every object emits infrared radiation, and hotter objects emit more of it.

Why radiation is the fastest method

Heat by radiation travels as electromagnetic waves, and these move at the speed of light. Conduction and convection depend on particles colliding or on fluid moving, which is much slower. This is why the heat of a fire reaches us almost at once, and why radiation is the fastest method of heat transfer.

Factors that affect the rate of radiation

Temperature. The rate of radiation emitted by an object is directly proportional to its temperature. The Sun, at a high temperature, emits much more radiation than the human body.
Surface colour and texture. Black and dull surfaces are the best absorbers and emitters, while shiny silver and white surfaces are the best reflectors (Table 10.2).
Surface area. At the same temperature, a body with a large surface area emits more radiation. This is why radiators and car radiators have large surfaces, and why an infrared heater has a large plate (Figure 10.19).

Effects of radiation on different surface colours (Table 10.2)
Colour	Absorbing	Emitting	Reflecting
Black	Best	Best	Worst
Dull or dark	Good	Good	Bad
White	Bad	Bad	Good
Shiny silver	Worst	Worst	Best

Note. Misprint or simplification in the book: the proportionality to temperature in Section 10.4, part B, is a simplified statement.

5

Analyze the role of greenhouse gases in global warming. How does global warming contribute to the increased severity of hurricanes, heat waves, flooding, and droughts?

Solution

Greenhouse gases and the greenhouse effect

The Earth absorbs high-energy radiation from the Sun and emits lower-energy infrared radiation back into the atmosphere. Water vapour, carbon dioxide, methane and other greenhouse gases absorb this infrared radiation and re-emit it back to the Earth. The Earth is therefore warmer than it would be without the atmosphere; without it, the night temperature would fall to about -180\,^\circ\mathrm{C}. This is the greenhouse effect (Figure 10.27).

The greenhouse gases in the atmosphere absorb infrared radiation from the Earth and send part of it back, so the heat is trapped near the surface.

Human activities increase greenhouse gases

Burning fossil fuels, smoke from vehicles, industrial and agricultural activities (which produce methane and nitrogen oxides) and deforestation have raised the amount of greenhouse gases in the atmosphere. More heat is trapped, and the average temperature of the Earth rises. This gradual rise is global warming.

How global warming increases extreme events

Heat waves. A warmer atmosphere makes abnormally hot periods more frequent and longer.
Hurricanes. Global warming heats the oceans, giving hurricanes more energy. They become stronger, with faster winds and heavier rain, because warm air holds more moisture.
Flooding. Melting glaciers raise the sea level, which increases flooding and erosion of coastal land.
Droughts. Greater evaporation dries the soil and plants, so the land becomes drier and harder for plants to grow. Changed rainfall patterns can add to this.
Extreme rainfall. More evaporation puts more moisture into the air, so areas that already have heavy rain can receive more intense rainfall.

The book also links global warming to wildfires, which spread more easily in hot, dry conditions. Reducing carbon emissions and stopping deforestation are the steps the book recommends to fight global warming.

6

Why Earth's core has extreme high temperature even after 4 billion years? Explain in detail.

Solution

Structure of the Earth

The Earth is made of the crust, the mantle and the core. The core is the hottest part, at about 6000\,^\circ\mathrm{C}. The inner core is solid because of the extreme pressure, while the outer core is molten at about 4000\,^\circ\mathrm{C} (Figure 10.29).

Factors that keep the core hot

Factor	How it keeps the core hot
Residual heat from formation	During accretion, many massive objects collided with the young Earth. Their kinetic energy turned into heat, and much of this heat has been retained in the core.
Radioactive decay	Uranium-238, thorium-232 and potassium-40 in the core release energy as heat when they decay. This heat is produced continuously, though its rate falls slowly as the elements decay.
Gravitational compression	During the formation of the Earth, its materials were pressed together by gravity. This increased the pressure and the temperature in the core, and the compression continues to give heat.
Crust as insulator	The outer layers of the Earth act as a thick insulating blanket. They slow the escape of heat from the core into space, so the core stays hot for a long time.

Each process is slow, and the heat escapes from the Earth very slowly, so the core has kept a very high temperature for the past 4 billion years.

7

How are tectonic plates formed, and what factors cause them to move? Explain these causes in detail and discuss the effects of plate movements.

Solution

Tectonic plates

Tectonic plates are very large, rigid pieces of the Earth's crust and the upper mantle, which together form the lithosphere. The plates float on the asthenosphere, a layer of soft, partly molten rock. The book names seven major plates, such as the Pacific, African and Eurasian plates. The book does not describe in detail how the plates first formed, so this answer begins from their present structure.

Causes of movement

Convection currents of magma. Heat from the core passes into the mantle. The magma there is heated, becomes less dense, and rises; cold magma sinks. These convection currents drag the plates along, causing horizontal movement.
Slab pull. Dense, cold magma that sinks in a convection current pulls the plate down with it, and this helps the plate to move.
Ridge push. At the spreading ocean ridge, new hot magma rises and pushes the plates apart (Figure 10.30).

Plates move apart at the ridge, and convection currents in the asthenosphere drag them along (Figures 10.30 and 10.32).

Effects of plate movements

The plates move only a few centimetres each year, but they have large effects. Plates can move apart (diverge or spread), come together (converge), or slide past each other (transform faulting).

Earthquakes occur where the plates interact and the rock suddenly slips.
Volcanic eruptions occur where magma reaches the surface through vents and fissures, which happens mostly near plate boundaries.
Changes to the landscape follow from the slow movement of the plates over millions of years.

Numerical response questionsp. 43–44

1

Convert specific heat of water 4180\,\mathrm{J\,kg^{-1}\,K^{-1}} into units of \mathrm{J\,g^{-1}\,{}^\circ C^{-1}}.

Solution

Given

c = 4180\,\mathrm{J\,kg^{-1}\,K^{-1}} specific heat of water

To find

c in \mathrm{J\,g^{-1}\,{}^\circ C^{-1}}

There are 1000 g in 1 kg, so divide the joules and the kilograms by 1000 to change the mass unit.
c=\dfrac{4180}{1000}\,\mathrm{J\,g^{-1}\,K^{-1}}
A change of 1 K is the same as a change of 1 ^\circC, so the temperature unit can be written in either form.
Evaluate.
c=4.18\,\mathrm{J\,g^{-1}\,{}^\circ C^{-1}}

c=4.18\,\mathrm{J\,g^{-1}\,{}^\circ C^{-1}}

Book answer. 4.18 \mathrm{J\,g^{-1}\,{}^\circ C^{-1}}

2FBISE SSC-II 2026 Annual I, Set M, Q.2 (i) OR

Calculate the amount of heat given to 25\,\mathrm{kg} of water to increase its temperature by 50\,^\circ\mathrm{C}.

Solution

Given

m = 25\,\mathrm{kg} mass of water
\Delta T = 50\,\mathrm{K} change in temperature (same in kelvin and degrees Celsius)
c = 4200\,\mathrm{J\,kg^{-1}\,K^{-1}} specific heat of water, as used in Section 10.1.1

To find

Q

The heat needed to change the temperature of a mass is Q=mc\,\Delta T.
Q=mc\,\Delta T
Substitute the values.
Q=25\times4200\times50\,\mathrm{J}
Evaluate and write the result in megajoules.
Q=5.25\times10^{6}\,\mathrm{J}=5.25\,\mathrm{MJ}

Q=5.25\,\mathrm{MJ}

Book answer. 5.25 MJ

Note. The value c=4200\,\mathrm{J\,kg^{-1}\,K^{-1}} is the one the book uses in Section 10.1.1 and in questions 4 and 5. Using 4180\,\mathrm{J\,kg^{-1}\,K^{-1}} from question 1 gives 5.23\,\mathrm{MJ}.

3

A half kg block of an unknown metal is heated from 30\,^\circ\mathrm{C} to 80\,^\circ\mathrm{C}, requiring 19\,\mathrm{kJ} energy. (a) Calculate the specific heat capacity of the metal. (b) If the same metal block is heated from 80\,^\circ\mathrm{C} to 125\,^\circ\mathrm{C}, how much additional heat energy would be needed?

Solution

Given

m = 0.5\,\mathrm{kg} half kg block of metal
\theta_1 = 30\,^\circ\mathrm{C} initial temperature
\theta_2 = 80\,^\circ\mathrm{C} final temperature for part (a)
Q = 19\,\mathrm{kJ}=1.9\times10^{4}\,\mathrm{J} heat energy for part (a)

To find

(a) c
(b) Q from 80\,^\circ\mathrm{C} to 125\,^\circ\mathrm{C}

(a) Calculate the specific heat capacity of the metal.

The temperature change is \Delta T=80-30=50\,\mathrm{K}. Rearrange c=\dfrac{Q}{m\,\Delta T}.
c=\dfrac{Q}{m\,\Delta T}
Substitute the values.
c=\dfrac{19\,000}{0.5\times50}
Evaluate.
c=760\,\mathrm{J\,kg^{-1}\,K^{-1}}

c=760\,\mathrm{J\,kg^{-1}\,K^{-1}}

(b) Additional heat energy needed to heat the same block from 80 ^\circC to 125 ^\circC.

The change in temperature is \Delta T=125-80=45\,\mathrm{K}, and the specific heat is the value found in part (a).
\Delta T=45\,\mathrm{K}
Use Q=mc\,\Delta T.
Q=0.5\times760\times45\,\mathrm{J}
Evaluate and convert to kilojoules.
Q=17\,100\,\mathrm{J}=17.1\,\mathrm{kJ}

Q=17.1\,\mathrm{kJ}

Book answer. (a) 760\,\mathrm{J\,kg^{-1}\,K^{-1}}, (b) 17.1\,\mathrm{kJ}

Note. The book prints the number as '3..' (double full stop); it is question 3.

4

Calculate the change in temperature of 5 litre of water if it absorbs 8.4\,\mathrm{MJ} of heat energy. Use specific heat of water =4200\,\mathrm{J\,kg^{-1}\,K^{-1}}.

Solution

Given

V = 5\,\mathrm{L} volume of water
\rho = 1000\,\mathrm{kg\,m^{-3}} density of water, so 1\,\mathrm{L} has a mass of 1\,\mathrm{kg}
Q = 8.4\,\mathrm{MJ}=8.4\times10^{6}\,\mathrm{J} heat absorbed
c = 4200\,\mathrm{J\,kg^{-1}\,K^{-1}} specific heat of water

To find

\Delta T

The mass of 5 litres of water is m=5\,\mathrm{kg}, since 1 litre of water has a mass of 1 kg.
m=5\,\mathrm{kg}
From Q=mc\,\Delta T, make the change in temperature the subject.
\Delta T=\dfrac{Q}{mc}
Substitute the values.
\Delta T=\dfrac{8.4\times10^{6}}{5\times4200}
Evaluate.
\Delta T=400\,\mathrm{K}=400\,^\circ\mathrm{C}

\Delta T=400\,^\circ\mathrm{C}

Book answer. 400 ^\circC

Note. The arithmetic follows the book's method and gives its answer. The result is physically unrealistic: liquid water at normal pressure boils at 100 ^\circC, so 5 litres of water cannot actually rise by 400 ^\circC without changing state. The question is best read as a practice calculation.

5

In a container, 100\,\mathrm{g} of water at 80\,^\circ\mathrm{C} is mixed with 200\,\mathrm{g} of water at 20\,^\circ\mathrm{C}. What will be the final temperature of the mixture, assuming no heat loss to the surroundings (and container)? (Use specific heat capacity of water is 4200\,\mathrm{J\,kg^{-1}\,K^{-1}} or 4.2\,\mathrm{J\,g^{-1}\,K^{-1}}.)

Solution

Given

m_1 = 100\,\mathrm{g} water at 80\,^\circ\mathrm{C}
m_2 = 200\,\mathrm{g} water at 20\,^\circ\mathrm{C}
c = 4.2\,\mathrm{J\,g^{-1}\,K^{-1}} specific heat of water

To find

final temperature \theta

Heat lost by the hot water equals heat gained by the cold water. Both masses use the same c, so c cancels.
m_1(80-\theta)=m_2(\theta-20)
Substitute the masses.
100(80-\theta)=200(\theta-20)
Expand both sides.
8000-100\theta=200\theta-4000
Collect the terms in \theta.
300\theta=12\,000
Solve.
\theta=40\,^\circ\mathrm{C}

\theta=40\,^\circ\mathrm{C}

Book answer. 40 ^\circC

6

A 200\,\mathrm{g} piece of an unknown solid cylinder is heated to 110\,^\circ\mathrm{C} and then placed into a calorimeter containing 200\,\mathrm{g} of water at 20\,^\circ\mathrm{C}. The calorimeter has a mass of 60\,\mathrm{g} and specific heat capacity of 0.9\,\mathrm{J\,g^{-1}\,{}^\circ C^{-1}}. The final temperature of the system (solid, water, and calorimeter) is 35\,^\circ\mathrm{C}. Assume no heat is lost to the surroundings. Calculate the specific heat capacity of the solid.

Solution

Given

m_s = 200\,\mathrm{g} mass of solid
\theta_s = 110\,^\circ\mathrm{C} initial temperature of solid
m_w = 200\,\mathrm{g} mass of water
\theta_1 = 20\,^\circ\mathrm{C} initial temperature of water and calorimeter
m_c = 60\,\mathrm{g} mass of calorimeter
c_c = 0.9\,\mathrm{J\,g^{-1}\,{}^\circ C^{-1}} specific heat of calorimeter
c_w = 4.18\,\mathrm{J\,g^{-1}\,{}^\circ C^{-1}} specific heat of water, as in Example 10.3
\theta_3 = 35\,^\circ\mathrm{C} final temperature

To find

c_s

By the principle of calorimetry, the heat lost by the solid equals the heat gained by the water and the calorimeter.
m_s c_s(110-35)=m_w c_w(35-20)+m_c c_c(35-20)
Heat gained by the water.
Q_w=200\times4.18\times15=12\,540\,\mathrm{J}
Heat gained by the calorimeter.
Q_c=60\times0.9\times15=810\,\mathrm{J}
Total heat gained by the water and the calorimeter.
Q_w+Q_c=13\,350\,\mathrm{J}
Heat lost by the solid, with m_s\,\Delta T=200\times75.
Q_s=200\times c_s\times75=15\,000\,c_s\,\mathrm{J}
Equate the two and solve for c_s.
c_s=\dfrac{13\,350}{15\,000}=0.89\,\mathrm{J\,g^{-1}\,{}^\circ C^{-1}}
Convert to SI units.
c_s=890\,\mathrm{J\,kg^{-1}\,K^{-1}}

c_s\approx0.89\,\mathrm{J\,g^{-1}\,{}^\circ C^{-1}}=890\,\mathrm{J\,kg^{-1}\,K^{-1}}

Book answer. 890 \mathrm{J\,kg^{-1}\,K^{-1}} or 0.89 \mathrm{J\,g^{-1}\,K^{-1}}

Note. The book does not give the water value in this question; 4.18\,\mathrm{J\,g^{-1}\,{}^\circ C^{-1}} is used from Example 10.3. Using 4.2 instead gives 0.894\,\mathrm{J\,g^{-1}\,{}^\circ C^{-1}}, which still rounds to 0.89.

6

A hot steel rod is cooled by plunging it into cold water, as shown in the book's figure (a person places a steel rod into a container of water). The steel rod has a mass of 3\,\mathrm{kg} and is initially at a temperature of 450\,^\circ\mathrm{C}. It cools to 50\,^\circ\mathrm{C} when placed in the water. The specific heat capacity of steel is 460\,\mathrm{J\,kg^{-1}\,{}^\circ C^{-1}}. Calculate the thermal energy (heat) lost by the steel rod as it cools to 50\,^\circ\mathrm{C}.

Solution

Given

m = 3\,\mathrm{kg} mass of steel rod
\theta_1 = 450\,^\circ\mathrm{C} initial temperature
\theta_2 = 50\,^\circ\mathrm{C} final temperature
c = 460\,\mathrm{J\,kg^{-1}\,{}^\circ C^{-1}} specific heat of steel

To find

Q lost by the rod

The change in temperature is \Delta T=450-50=400\,\mathrm{K}.
\Delta T=400\,\mathrm{K}
Use Q=mc\,\Delta T.
Q=3\times460\times400\,\mathrm{J}
Evaluate and convert to kilojoules.
Q=552\,000\,\mathrm{J}=552\,\mathrm{kJ}

Q=552\,\mathrm{kJ}

Book answer. 552 kJ

Note. The book numbers two questions '6'. This is the second one, so it is question 7 in order; it is given the id numerical-q7 and printed number 6, as in the book.

FBISE 2026 questions

Questions from the FBISE SSC-II 2026 (1st Annual) papers that belong to this unit, answered the way an examiner expects.

FBISE SSC-II 2026 Annual I, Set B, MCQ 51 mark

Warm air rises and cool air sinks in a heated room because of:

AReflection
BRadiation
CConvectionAnswer
DConduction

Why

Warm air is less dense than cool air, so it rises, and the denser cool air sinks to take its place. This circulation of the moving fluid itself carries heat around the room. It is convection.

Conduction (D) needs direct contact between particles of a solid, and the air particles do not pass heat along a fixed chain like that. Radiation (B) is heat carried by electromagnetic waves and does not need a moving fluid. Reflection (A) is not a method of heat transfer.

FBISE SSC-II 2026 Annual I, Set B, Q.2 (i) OR3 marks

Calculate the heat required to raise the temperature of a 2 kg substance from 40^{\circ}C to 45^{\circ}C given specific heat = 420\,\mathrm{J\,kg^{-1}K^{-1}}.

Solution

Given

m = 2\,\mathrm{kg} mass of the substance
T_i = 40^\circ\mathrm{C} initial temperature
T_f = 45^\circ\mathrm{C} final temperature
c = 420\,\mathrm{J\,kg^{-1}\,K^{-1}} specific heat

To find

heat required Q

The change in temperature is the same number in degrees Celsius and in kelvin.
\Delta T=45^\circ\mathrm{C}-40^\circ\mathrm{C}=5^\circ\mathrm{C}=5\,\mathrm{K}
Use the formula from the list printed on the paper.
Q=mc\,\Delta T
Substitute the values.
Q=2\times420\times5\,\mathrm{J}
Evaluate.
Q=4200\,\mathrm{J}=4.2\,\mathrm{kJ}

Q=4200\,\mathrm{J}=4.2\,\mathrm{kJ}

FBISE SSC-II 2026 Annual I, Set B, Q.2 (vi) OR3 marks

Why do coastal areas have moderate temperatures than inland areas? Explain briefly.

Solution

Water has a large specific heat (4200\,\mathrm{J\,kg^{-1}\,K^{-1}}), while dry land has a much smaller one (about 800\,\mathrm{J\,kg^{-1}\,K^{-1}}, five times smaller). Land therefore heats up and cools down quickly, but the sea warms and cools slowly.

The sea near a coast absorbs and stores a lot of heat from the Sun in summer and releases it in winter, so the temperature changes little between day and night and between summer and winter. In Karachi the average temperature varies only from 19^\circ\mathrm{C} to 32^\circ\mathrm{C}. Inland, the land has no such store of heat, so the temperature varies much more, up to about 45^\circ\mathrm{C} to 50^\circ\mathrm{C} in summer and down to about 0^\circ\mathrm{C} in winter. Sea and land breezes help the coast further.

FBISE SSC-II 2026 Annual I, Set B, Q.32+2+1 = 5 marks

a. Explain the mechanism of heat conduction in solids with reference to lattice vibrations. b. How heat is conducted in metallic solids through free electrons? Explain with diagram.

Solution

a. Explain the mechanism of heat conduction in solids with reference to lattice vibrations.

Conduction is the transfer of heat through a solid from the hot part to the cold part by collisions of atoms or molecules and by the motion of free electrons. The particles themselves stay in their places.

In a solid the atoms are held at fixed positions in the lattice and vibrate about them. A higher temperature means a larger average vibration. When one end of a rod is heated, the atoms at that end vibrate with a larger amplitude. They hit their neighbouring atoms and pass part of their energy to them. The neighbours now vibrate more strongly and hit the atoms next to them, and the process continues along the rod. In this way heat energy is passed from atom to atom between the two ends, without the atoms leaving their positions.

b. How heat is conducted in metallic solids through free electrons? Explain with diagram.

A metal contains a large number of free electrons that are not bound to any one atom and can move through the whole metal. At the hot end these electrons gain energy from the vibrating atoms and move quickly to the cold end. There they hit atoms and give up their energy. This is much faster than passing energy by atomic collisions alone, so metals are good conductors of heat.

Heat travels along a metal rod by vibrating atoms and by free electrons moving from the hot end to the cold end.

Wood and glass have very few free electrons, so heat moves through them only by slow atomic vibrations. They are poor conductors.

FBISE SSC-II 2026 Annual I, Set D, MCQ 51 mark

A metal cooking pan heats up quickly on a stove mainly due to:

AEvaporation
BRadiation
CConvection
DConductionAnswer

Why

The metal of the pan has many free electrons, so heat from the flame is conducted quickly through the metal base to the food. The pan heats up quickly mainly by conduction.

Convection (C) occurs in fluids and is not how heat moves through the solid metal. Radiation (B) from the flame does reach the pan, but it is a smaller part of the heating than the conduction through the metal. Evaporation (A) is not a method of heat transfer into the pan.

FBISE SSC-II 2026 Annual I, Set D, Q.2 (i) OR2+1 = 3 marks

What is meant by specific heat capacity? Give its mathematical expression.

Solution

Specific heat capacity is the amount of heat absorbed or lost to change the temperature of 1\,\mathrm{kg} of a substance by 1\,\mathrm{K} (or 1^\circ\mathrm{C}).

Mathematical expression:

c=\dfrac{Q}{m\,\Delta T}

Here Q is the heat absorbed or lost, m is the mass and \Delta T is the change in temperature. Its SI unit is \mathrm{J\,kg^{-1}\,K^{-1}}. Equivalently Q=mc\,\Delta T, the form given in the formula list of the paper.

c=\dfrac{Q}{m\,\Delta T}

FBISE SSC-II 2026 Annual I, Set D, Q.2 (vi) OR1.5+1.5 = 3 marks

How do sea and land breezes help in moderating coastal climate?

Solution

Land has a small specific heat and the sea has a large one, so by day the land becomes hotter than the sea and at night the sea stays hotter than the land.

Sea breeze (day). The air above the hot land is heated, expands and rises. Cool air from the sea moves in over the land to fill the gap. This cools the coast during the day.

Land breeze (night). The land cools faster than the water, so the air above the sea is warmer and rises. Cool air from the land moves out over the sea. This keeps the night temperature of the coast from falling too low.

These two convection currents blow continuously, cool the coast by day and moderate it at night, so the coastal climate stays moderate.

FBISE SSC-II 2026 Annual I, Set D, Q.32+3 = 5 marks

a. How is heat conducted in metallic solids through free electrons? b. Compare heat transfer in metals and non-metals.

Solution

a. How is heat conducted in metallic solids through free electrons?

A metal has a large number of free electrons that are not bound to any one atom. When one end is heated, the atoms there vibrate with a larger amplitude and the free electrons near them gain energy. These fast electrons move through the metal to the cold end and give their energy to atoms there by collisions. This carries heat across the metal much more quickly than the vibration of atoms alone.

Free electrons carry energy quickly from the hot end of a metal rod to the cold end.

b. Compare heat transfer in metals and non-metals.

Heat transfer in metals and non-metals.
	Metals	Non-metals (wood, glass, rubber)
Free electrons	A large number	Very few or none
How heat is carried	Free electrons carry energy quickly, in addition to atomic vibrations	Only by slow collisions of vibrating atoms
Rate of heat transfer	Fast	Slow
Kind of material	Good conductors of heat	Poor conductors, that is good insulators
Example	Copper rod: the far end heats up quickly	Wooden stick: the far end heats up slowly

In the experiment with waxed drawing pins on rods heated at one end, the pin falls first from copper, then aluminium, then iron and brass. Metals also feel colder to the touch than wood at the same temperature, because they carry heat away from the hand faster.

FBISE SSC-II 2026 Annual I, Set M, MCQ 51 mark

Which feature helps reduce heat loss in buildings?

AOpen ventilation
BLarge windows
CThermal insulationAnswer
DThin walls

Why

Thermal insulation (double glazing, foam in gaps, false ceilings, thick curtains) slows the flow of heat through the walls, windows and roof, so less heat is lost from a building in winter.

Open ventilation (A), large windows (B) and thin walls (D) all let heat pass more easily, so they increase heat loss.

FBISE SSC-II 2026 Annual I, Set M, Q.2 (i) OR1+2 = 3 marks

What is meant by specific heat energy capacity of a substance? Calculate amount of heat required to 25kg of water to increase its temperature by 50^{\circ}C.

Solution

(1) What is meant by specific heat energy capacity of a substance?

It is the amount of heat absorbed or lost to change the temperature of 1\,\mathrm{kg} of the substance by 1\,\mathrm{K} (or 1^\circ\mathrm{C}). It is given by c=\dfrac{Q}{m\,\Delta T} and measured in \mathrm{J\,kg^{-1}\,K^{-1}}.

(2) Calculate the amount of heat required to raise the temperature of 25 kg of water by 50^\circC.

Given

m = 25\,\mathrm{kg} mass of water
\Delta T = 50\,\mathrm{K} rise in temperature (same as 50^\circ\mathrm{C})
c = 4200\,\mathrm{J\,kg^{-1}\,K^{-1}} specific heat of water, as used in the book

To find

heat required Q

Use the formula from the list printed on the paper.
Q=mc\,\Delta T
Substitute the values.
Q=25\times4200\times50\,\mathrm{J}
Evaluate.
Q=5.25\times10^{6}\,\mathrm{J}=5.25\,\mathrm{MJ}

Q=5.25\times10^{6}\,\mathrm{J}=5.25\,\mathrm{MJ}

Book answer. 5.25 MJ

Note. The book uses c=4200\,\mathrm{J\,kg^{-1}\,K^{-1}} for water. The paper does not print the value of c for water; with 4180 the answer would be 5.23\,\mathrm{MJ}.

FBISE SSC-II 2026 Annual I, Set M, Q.2 (vi) OR3 marks

How convection in seawater supports marine life?

Solution

Convection in seawater is caused by differences in density due to temperature and salinity. Warm water is less dense and rises; cold or salty water is denser and sinks. These convection currents support marine life in three ways.

Distribution of food. Where winds blow along a coast, surface water is pushed away and colder water rises (upwelling). This water is rich in nutrients, which phytoplankton uses with sunlight to grow. Phytoplankton feeds zooplankton, fish and larger animals such as whales.
Distribution of oxygen. Oxygen is dissolved mainly in the top water, which touches the air. Convection carries this water down into the deep ocean (downwelling), so deep-sea animals get oxygen.
Distribution of heat. Convection spreads heat through the ocean and keeps its temperature stable, which suits the habitats of marine life.

FBISE SSC-II 2026 Annual I, Set M, Q.32+1+2 = 5 marks

Explain green house effect. Is it good or bad? Discuss the consequences of greenhouse effect as well.

Solution

(1) Explain greenhouse effect.

The Earth absorbs the high-energy radiation from the Sun and emits low-energy infrared radiation back towards space. Greenhouse gases in the atmosphere (water vapour, carbon dioxide, methane and others) absorb this infrared radiation and re-emit it back to the Earth. Because of this the Earth warms. This process, by which heat is trapped near the Earth's surface by greenhouse gases, is the greenhouse effect.

The greenhouse gases absorb infrared radiation from the Earth and send part of it back, so heat is trapped near the surface.

(2) Is it good or bad?

The natural greenhouse effect is good: it is necessary for life. Without an atmosphere the Earth would emit all the radiation it absorbs and its night temperature would fall to about -180^\circ\mathrm{C}, like the Moon, and life could not exist.

It becomes bad when human activities (burning fossil fuels, vehicles, industry and agriculture, deforestation) add more greenhouse gases. More heat is trapped and the Earth warms further. This is global warming.

(3) Discuss the consequences of greenhouse effect as well.

The extra greenhouse effect raises the average temperature of the atmosphere and oceans (global warming). Its consequences are:

Flooding. Glaciers melt fast and sea levels rise, which floods coasts and erodes land.
Hurricanes. Warmer oceans give hurricanes more energy, so they are stronger, with faster winds and heavier rain.
Heavier rainfall and droughts. More evaporation puts more moisture in the air, giving more intense rain where it already rains heavily, and dries the soil and plants elsewhere.
Wildfires. Drier trees and vegetation give more fuel for fires, which spread over large areas and pollute the air.
Winter storms and heat waves. More moisture in the air gives heavier snowfall, and the weather patterns and seasons change, affecting ecosystems and human health.

Exam focus

Define specific heat capacity, give c=\dfrac{Q}{m\,\Delta T} and solve Q=mc\,\Delta T numericals (heat for a given rise, e.g. 25 kg of water through 50^\circ\mathrm{C}). (short)
Q.2 (i) OR in all three 2026 sets: B (2 kg at 420, 40 to 45 degrees), D (definition and expression) and M (definition and 25 kg of water, the same as numerical problem 2).
Go to answer
Explain how heat is conducted in solids (lattice vibrations) and through free electrons in metals, with a diagram. (long)
Q.3 in Set B (2+2+1 marks) and Set D (2+3 marks); also comprehensive question 2.
Go to answer
Compare heat transfer in metals and non-metals (free electrons, rate, conductors and insulators). (long)
Q.3 (b), 3 marks, in Set D 2026. Use a table.
Go to answer
Explain the greenhouse effect, say whether it is good or bad, and discuss its consequences (global warming, flooding, hurricanes, droughts). (long)
Q.3 in Set M 2026 (2+1+2 marks); also comprehensive question 5 and MCQ 9.
Go to answer
Why do coastal areas have a moderate climate? Explain with specific heat, and with sea and land breezes. (short)
Q.2 (vi) OR in Set B (moderate coastal temperatures) and Set D (sea and land breezes); Set M asked convection in seawater.
Go to answer
How does convection in seawater support marine life (food, oxygen and heat)? (short)
Q.2 (vi) OR in Set M 2026.
Go to answer
Choose the mode of heat transfer in a situation (warm air rising, a metal pan on a stove) and name measures of thermal insulation in buildings. (mcq)
MCQ 5 of all three 2026 sets: convection (B), conduction (D), thermal insulation (M); also MCQs 2, 3 and 6 of the book.
Go to answer
Describe the three modes of heat transfer and the uses of water's large specific heat (radiator coolant, body temperature, cooking). (long)
Not a separate 2026 question, but it underlies MCQ 5 and the Q.2 and Q.3 questions of all three sets; comprehensive questions 1 and 3 of the book and the unit SLOs.
Go to answer