Thermal expansion and change of state: Physics unit 11 notes (FBISE class 10)

11.1Kinetic theory of matter

Matter exists in three fundamental states: solid, liquid and gas. All matter is made of particles (atoms and molecules). In class 9 you studied matter with the help of the kinetic molecular theory of matter, and this unit begins by revising its main ideas.

The particles of matter attract each other. The strength of this attractive force decides how big the gaps between the particles are: the stronger the attraction, the smaller the distance between the particles, and the weaker the attraction, the larger the distance. These gaps are the main reason that matter is divided into three states.

Figure 11.1: Effect of temperature on the distance between particles of matter. Particles are packed in a regular pattern in a solid, close but free to move in a liquid and far apart in a gas. Heating (coloured arrow) increases the gaps.

Solids

In a solid the particles are close to each other, so a solid has a high density. A strong attractive force keeps the particles closely packed. The particles can only vibrate about their mean positions. When a solid absorbs heat, the vibrational kinetic energy and the amplitude of vibration of its particles increase. The particles push each other away and the gap between them grows. This is how solids expand on heating.

Liquids

A liquid has an intermediate density (smaller than a solid and higher than a gas). The attractive force between its particles is stronger than in a gas but weaker than in a solid. For this reason the distance between particles in a liquid is greater than in a solid. Liquid particles keep changing their positions, which is why a liquid can flow. When a liquid is heated, the attractive force between its particles decreases and the particles start moving more.

Gases

In a gas the attractive force between the particles is negligibly weak, so the distance between particles is large compared with the size of the particles. The particles are in constant random motion and keep colliding with each other and with the walls of the container. These collisions with the walls are what produce the pressure of a gas. Gases are much less dense than liquids and solids. When a gas is heated, the kinetic energy of its molecules increases, so they move faster and the gas expands quickly.

The three states of matter in terms of particles (Figure 11.1).
Property	Solid	Liquid	Gas
Attractive force between particles	Strong	Weaker than in a solid	Negligibly weak
Gaps between particles	Very small	Larger than in a solid	Very large
Motion of particles	Vibrate about mean positions	Keep changing positions and flow	Constant random motion, colliding with walls
Density	High	Intermediate	Very low
Effect of heating	Vibrations and amplitude increase, solid expands	Attraction decreases, particles move more	Molecules move faster, gas expands quickly

11.1.1Changing states of matterOften examined

The state of a substance can change when it gains or loses heat. Figure 11.2 of the book shows ice (solid) changing into water (liquid) and then into water vapour (gas) as its temperature increases. The attractive forces among the particles usually decrease as the temperature increases.

Temperature is not the only thing that affects the state of matter. Pressure also affects the changes of phase: the particles of liquids and gases come close to each other when the pressure on them is increased, and move apart when it is decreased.

A. Changes between liquids and solids

Freezing: The change of a liquid into a solid. A liquid placed where its surroundings are at a low temperature loses heat to the surroundings. Its particles come close to each other and become solid at a temperature called the freezing point. Water changes into ice at $0^{\circ} C$ , its freezing point.
Melting: The change of a solid into a liquid. A solid placed where its surroundings are at a high temperature absorbs heat from them. When it reaches its melting point, its particles overcome the attractive forces that bind them and start flowing. Ice placed in warm surroundings melts at $0^{\circ} C$ .

B. Changes between liquids and gases

Vaporisation: The change of a liquid into a gas (vapour) upon boiling. On continuous heating the particles gain so much kinetic energy that they overcome the attractive forces between them and escape from the liquid. It happens at a certain temperature of the liquid called its boiling point (Figure 11.3(a)).
Evaporation: The change of a liquid into vapour at any temperature. It is discussed in detail later in the unit (sections 11.3 to 11.3.5).
Condensation: The change of a gas (vapour) into a liquid. It happens in vapours when the temperature and pressure are low.

Warm air can hold more moisture than cool air. When warm, moist air cools down, the water vapour condenses because the cooler air has a lower capacity to hold water vapour. When you take a cold water bottle out of the fridge, droplets form on its outside (Figure 11.3(b)). The cold bottle cools the air around it, and the water vapour in that air condenses into liquid water on the bottle surface. Other examples are dew drops on grass, a foggy car windscreen and foggy bathroom mirrors.

C. Changes between solids and gases

Sublimation: The change of a solid directly into a gas. The commonest example is dry ice (solid carbon dioxide) turning into smoke-like carbon dioxide gas (Figure 11.3(c)). Dry ice is a solid at an extremely low temperature ( $- 78.5^{\circ} C$ ). At room temperature and pressure its molecules overcome all the attractive forces and change directly into gas without passing through the liquid phase.
Deposition: The change of a gas directly into a solid, without passing through the liquid phase. It is the reverse of sublimation. Frost is an example: on a cold night, water vapour in the air meets surfaces such as leaves or window panes that are much cooler than the air. The vapour loses heat to these cold surfaces and turns directly into ice crystals, the frost of Figure 11.3(d).

Figure 11.4: Change of state of matter and the process names. Coloured arrows mark the changes in which the substance takes in heat; the other three give heat out.

The seven changes of state (Figures 11.3 and 11.4). The heat column follows the book: melting and boiling need heat, freezing and condensation release it.
Change	From to	Heat	Example from the book
Freezing	Liquid to solid	Given out	Water to ice at $0^{\circ} C$
Melting	Solid to liquid	Taken in	Ice melting at $0^{\circ} C$
Vaporisation (boiling)	Liquid to gas, at the boiling point	Taken in	Boiling water changing into vapour
Evaporation	Liquid to gas, at any temperature	Taken in	Wet clothes drying
Condensation	Gas to liquid	Given out	Droplets on a cold bottle, dew, fog on a mirror
Sublimation	Solid to gas	Taken in	Dry ice into carbon dioxide gas
Deposition	Gas to solid	Given out	Frost on leaves

11.2Thermal expansion

Thermal expansion means "increase in the size of matter on heating". On heating, the particles of matter gain energy and start to vibrate or move around faster. They hit each other and increase the gap between them, so matter expands.

Why a solid expands. When heated, the particles vibrate with a larger amplitude (the arrows) and push each other apart, so the whole row becomes longer by

Δ L

(coloured).

How much a body expands depends on three things: the nature of its material, the increase in temperature, and its original size (the amount of matter in it before heating). For the same increase in temperature, solids expand the least and gases expand the most.

Table 11.1: Difference in relative expansion for solids, liquids and gases.
State	Expansion
Solids	Expand slightly, because the low-energy molecules cannot overcome the intermolecular forces of attraction holding them together.
Liquids	Expand more than solids, because the molecules have enough energy to partly overcome the intermolecular forces of attraction holding them together.
Gases	Expand significantly, because the molecules have enough energy to overcome the intermolecular forces of attraction holding them together completely.

There are three types of thermal expansion in solids: linear thermal expansion (one dimension), surface area thermal expansion (two dimensions) and volumetric thermal expansion (three dimensions). The book discusses linear and volumetric expansion.

11.2.1 ALinear thermal expansion of solidsOften examined

Linear thermal expansion is the change in the length of a solid on heating. A change in one dimension of a solid object on heating is called linear thermal expansion of the solid. It depends on three factors.

Factors on which linear expansion depends

Increase in temperature of the rod. When a solid rod is heated, its atoms gain kinetic energy and vibrate with greater amplitude about their mean positions. The bigger vibrations push the atoms slightly further apart, so the rod becomes longer. If the temperature keeps rising, the energy and the amplitude increase further. So the expansion in length is directly proportional to the increase in temperature.
Original length of the rod. The increase in length is directly proportional to the original length. A longer rod has more atoms and each atom adds to the total expansion, so a longer rod expands more than a shorter rod of the same material heated through the same temperature.
Nature of the material of the rod. Different materials expand differently on heating. Thermal expansion depends on the atomic structure and on the strength of the attractive force between the atoms. For example, an aluminium rod expands more than a steel rod of the same length for the same rise in temperature.

Consider a metal rod of original length $L_{0}$ at initial temperature $T_{0}$ (Figure 11.5). It is heated to a final temperature $T$ and its final length becomes $L$ . The increase in length is $Δ L = L - L_{0}$ for the increase in temperature $Δ T = T - T_{0}$ .

Figure 11.5: Linear thermal expansion of a rod. A rod of original length

L_{0}

at temperature

T_{0}

is heated to

T = T_{0} + Δ T

and its length becomes

L

, an increase of

Δ L

(coloured).

From the three factors, the change in length is directly proportional to the original length $L_{0}$ and to the change in temperature $Δ T$ . Mathematically

Δ L \propto L_{0} Δ T

Δ L = α L_{0} Δ T

11.1

Here $α$ is the constant of proportionality. It is called the coefficient of linear thermal expansion. Rearranging equation 11.1 gives

α = \frac{Δ L}{L _{0} Δ T} = \frac{Δ L / L _{0}}{Δ T} = \frac{fractional change in length}{change in temperature}

11.2

Coefficient of linear thermal expansion: The fractional change in length per kelvin change in temperature. It tells how much the length of a rod changes, compared with its original length, when its temperature rises by $1 K$ .

The coefficient of linear expansion is an intrinsic property of every material, so it varies from one material to another. Its SI unit is $K^{- 1}$ (per kelvin). Table 11.2 gives some values.

Table 11.2: Coefficients of linear thermal expansion.
Material	Value of $α$ ( $K^{- 1}$ )
Steel	$1.2 \times 1 0^{- 5}$
Quartz	$0.04 \times 1 0^{- 5}$
Invar	$0.09 \times 1 0^{- 5}$
Glass	$0.4 \times 1 0^{- 5}$
Copper	$1.7 \times 1 0^{- 5}$
Brass	$2 \times 1 0^{- 5}$
Aluminium	$2.4 \times 1 0^{- 5}$

A solid also contracts on cooling, and the same relations can be used for the contraction of solids on cooling. The final length is $L = L_{0} + Δ L$ when the rod is heated.

Example 11.1: change in length of a steel bridge

An old steel bridge has an original length of

1.5 km

(

1500 m

). The average daily temperature change in the region is

15^{\circ} C

(from midnight to noon). Steel has a coefficient of linear expansion of

12 \times 1 0^{- 6} K^{- 1}

. How much will the bridge length change due to this daily temperature swing?

Given

$L_{0} = 1.5 km = 1500 m$
$α = 12 \times 1 0^{- 6} K^{- 1}$
$Δ T = 15^{\circ} C = 15 K$

To find

change in length $Δ L$

A temperature change of $15^{\circ} C$ is the same as a change of $15 K$ , so $Δ T$ can go straight into the formula. Write the formula for linear expansion.
$Δ L = α L_{0} Δ T$
Substitute the values.
$Δ L = (12 \times 1 0^{- 6} K^{- 1}) (1500 m) (15 K)$
Multiply: $12 \times 1 0^{- 6} \times 1500 = 0.018$ , and $0.018 \times 15 = 0.27$ .
$Δ L = 0.27 m$
So the bridge becomes about $27 cm$ longer between midnight and noon. This is why bridges are built with expansion joints.

Δ L = 0.27 m

Worked example: which rod grows more, the longer or the one with the larger coefficient?

A

2.0 m

aluminium rod and a

3.0 m

steel rod are both heated from

20^{\circ} C

to

70^{\circ} C

. A student says "the steel rod is longer, so it must expand more." Use Table 11.2 to check the claim.

Given

$L_{Al} = 2.0 m$
$L_{steel} = 3.0 m$
$α_{Al} = 2.4 \times 1 0^{- 5} K^{- 1}$
$α_{steel} = 1.2 \times 1 0^{- 5} K^{- 1}$
$Δ T = 70 - 20 = 50^{\circ} C = 50 K$

To find

$Δ L$ of each rod, and whether the student is right

Aluminium rod: use $Δ L = α L_{0} Δ T$ .
$Δ L_{Al} = (2.4 \times 1 0^{- 5}) (2.0) (50) = 2.4 \times 1 0^{- 3} m = 2.4 mm$
Steel rod: same formula with its own $α$ and length.
$Δ L_{steel} = (1.2 \times 1 0^{- 5}) (3.0) (50) = 1.8 \times 1 0^{- 3} m = 1.8 mm$
Compare the two. The shorter aluminium rod expands more, because its coefficient is twice as large and this beats the smaller length.
$\frac{Δ L _{Al}}{Δ L _{steel}} = \frac{( 2.4 ) ( 2.0 )}{( 1.2 ) ( 3.0 )} = \frac{4.8}{3.6} = 1.33$
The student was wrong. Length is only one of the three factors; the nature of the material (the value of $α$ ) and the temperature rise matter as well.

Increase in length against rise in temperature

10

20

30

40

50

60

1

2

3

O

Δ T (K)

Δ L (mm)

2.4

1.8

aluminium, 2.0 m

steel, 3.0 m

Both graphs are straight lines through the origin, because

Δ L \propto Δ T

. The aluminium line is steeper (slope

α L_{0}

is larger), so it is above the steel line at every temperature rise.

Δ L_{Al} = 2.4 mm (more), Δ L_{steel} = 1.8 mm

11.2.1 BVolumetric thermal expansion of solidsOften examined

Volumetric thermal expansion is the change in volume of a solid on heating. A change in all three dimensions of a solid object on heating is called volumetric thermal expansion of solids. It depends on three factors, just like linear expansion.

Increase in temperature of the object. When a three-dimensional object (such as a cube) is heated, the kinetic energy of its particles increases, their vibrations become more vigorous and they spread further apart. The particles of the object occupy more volume. So the expansion in volume is directly proportional to the increase in temperature.
Original volume of the object. For the same reason as in linear expansion, the increase in volume of the solid object is directly proportional to its original volume.
Nature of the material of the object. Different materials expand differently on heating, so the increase in volume is different for different materials.

Let the original volume of a solid cube be $V_{0}$ at temperature $T_{0}$ (Figure 11.6). Its final volume becomes $V$ when its final temperature becomes $T = T_{0} + Δ T$ . The increase in its volume is $Δ V = V - V_{0}$ for the increase in temperature $Δ T = T - T_{0}$ .

Figure 11.6: Volumetric thermal expansion of a solid cube. The cube of volume

V_{0}

at temperature

T_{0}

becomes a larger cube of volume

V = V_{0} + Δ V

at the higher temperature

T

.

From the three factors, the increase in volume of the object is

Δ V = β V_{0} Δ T

11.3

Here $β$ is the constant of proportionality, called the coefficient of volume expansion. From equation 11.3,

β = \frac{Δ V}{V _{0} Δ T} = \frac{Δ V / V _{0}}{Δ T} = \frac{fractional change in volume}{change in temperature}

11.4

Coefficient of volume thermal expansion of solids: The fractional change in volume per kelvin change in temperature. It tells how much the volume of the solid object changes, compared with its original volume, when its temperature rises by $1 K$ .

Its value is different for different materials and its SI unit is $K^{- 1}$ . A solid expands on heating and contracts on cooling, and the same relations can be used for the contraction of solids on cooling. Table 11.3 gives some values of $β$ .

Table 11.3: Coefficients of volumetric thermal expansion.
Material	Value of $β$ ( $K^{- 1}$ )
Methanol	$113 \times 1 0^{- 5}$
Glycerin	$49 \times 1 0^{- 5}$
Mercury	$18 \times 1 0^{- 5}$
Turpentine	$90 \times 1 0^{- 5}$
Acetone	$132 \times 1 0^{- 5}$

11.2.1 CExperimental demonstration of thermal expansion of solids

This experiment shows that a heated metal rod expands with great force. A metal bolt that is held tightly against a heated metal rod snaps (Figure 11.7).

Apparatus

A metal rod (for example steel), a metal bolt and nut (for example brass or aluminium), a Bunsen burner, a stand and clamps.

Procedure

Place the metal rod horizontally with a bolt tightly fastened to it.
Gradually heat the rod using a heat source.
As the rod heats up, it expands because of thermal expansion.
The expansion makes the rod exert pressure on the bolt.
If the expansion is large enough, the bolt eventually snaps.

Figure 11.7: Experimental demonstration of thermal expansion in solids. The heated rod expands lengthways (coloured arrow) and presses on the bolt until the bolt snaps.

Conclusion

Different metals expand at different rates when heated. The rod expands and exerts pressure on the bolt. If this pressure exceeds the strength of the bolt, it snaps. The experiment shows that thermal expansion has to be considered in engineering applications, to prevent damage to structures.

11.2.1 DRelation between the coefficients of linear and volume expansionOften examined

The coefficient of linear expansion and the coefficient of volume expansion of a solid are related by

β = 3 α

11.5

Example 11.2: volume increase of hot air balloon fabric

You are designing a hot air balloon. The balloon fabric has a volume of

500 m^{3}

at room temperature (

20^{\circ} C

). You need to know how much the volume will increase when hot air at

80^{\circ} C

fills the balloon. The fabric has a coefficient of volumetric expansion of

3.5 \times 1 0^{- 5} K^{- 1}

. Calculate the volume increase of the balloon fabric when filled with hot air.

Given

$V_{0} = 500 m^{3}$
$β = 3.5 \times 1 0^{- 5} K^{- 1}$
$Δ T = 80 - 20 = 60^{\circ} C = 60 K$

To find

increase in volume $Δ V$

Use the formula for volume expansion.
$Δ V = β V_{0} Δ T$
Substitute the values.
$Δ V = (3.5 \times 1 0^{- 5} K^{- 1}) (500 m^{3}) (60 K)$
Multiply: $3.5 \times 1 0^{- 5} \times 500 = 0.0175$ and $0.0175 \times 60 = 1.05$ .
$Δ V = 1.05 m^{3}$
After the temperature rises to $80^{\circ} C$ , the volume of the balloon increases by $1.05 m^{3}$ .

Δ V = 1.05 m^{3}

Worked example: from

α

to

β

for a copper cube

A solid copper cube of side

5.0 cm

is heated through

80 K

. Using

α

for copper from Table 11.2, find the increase in volume. Then show what error a student makes who uses

α

in place of

β

.

Given

$L_{0} = 5.0 cm$
$α = 1.7 \times 1 0^{- 5} K^{- 1}$
$Δ T = 80 K$

To find

$β$ , $V_{0}$ and $Δ V$

The coefficient for volume is three times the coefficient for length.
$β = 3 α = 3 (1.7 \times 1 0^{- 5}) = 5.1 \times 1 0^{- 5} K^{- 1}$
The original volume of the cube.
$V_{0} = L_{0}^{3} = (5.0)^{3} = 125 c m^{3}$
Use $Δ V = β V_{0} Δ T$ .
$Δ V = (5.1 \times 1 0^{- 5}) (125) (80) = 0.51 c m^{3}$
Check by another route. Each side grows to $5.0 (1 + 1.7 \times 1 0^{- 5} \times 80) = 5.0068 cm$ , so the new volume is about $(5.0068)^{3} = 125.51 c m^{3}$ , an increase of $0.51 c m^{3}$ . The two methods agree.
The common mistake: using $α$ in place of $β$ gives $(1.7 \times 1 0^{- 5}) (125) (80) = 0.17 c m^{3}$ , only one third of the correct value. A volume expands in three dimensions, so $β$ must be used.

Increase in volume of the copper cube

20

40

60

80

100

0.1

0.2

0.3

0.4

0.5

0.6

0.7

O

Δ T (K)

Δ V (c m^{3})

0.51

0.17

using β

using α (wrong)

The correct line (coloured) has slope

β V_{0} = 6.4 \times 1 0^{- 3} c m^{3} K^{- 1}

. The line made with

α

has one third of that slope and under-estimates the expansion.

Δ V = 0.51 c m^{3}

11.2.2Thermal expansion in liquidsOften examined

Liquids have intermediate attractive forces between their particles (molecules), and their particles can move relatively freely compared with the particles of a solid. When a liquid is heated, the kinetic energy of its particles increases, they push each other apart and occupy more space. This is the reason for the expansion of liquids on heating. The thermal expansion of a liquid is greater than that of a solid because of the weak attractive forces between its particles.

A liquid cannot be heated by itself for this purpose. It has to be heated in a container, so both the change in volume of the liquid and the change in volume of the container during heating have to be considered. There are two kinds of thermal volume expansion for a liquid: real volume expansion and apparent volume expansion.

A. Real volume expansion

When heat is applied directly to a liquid, causing it to expand, the increase in its volume is called real volume expansion. Generally this expansion is similar to the volumetric thermal expansion of solids, and the same formula can be used. The coefficient of real volume expansion of a liquid, \gamma_r, is defined as the fractional increase in real volume of the liquid for a 1\,\mathrm{K} rise in temperature.

\gamma_r=\beta=\dfrac{\Delta V}{V_0\,\Delta T}

11.6

B. Apparent volume expansion

When a liquid is heated in a container, the increase in its volume from the initial to the final position is called the apparent expansion of the liquid. It is the expansion of the liquid that is apparently observed without considering the expansion of the container or vessel. The coefficient of apparent expansion of a liquid, \gamma_a, is defined as the fractional increase in apparent volume of the liquid for a 1\,\mathrm{K} rise in temperature. Its SI unit is also per kelvin (\mathrm{K^{-1}}).

\gamma_a=\dfrac{\Delta V_{\text{apparent}}}{V_0\,\Delta T}

11.7

Real volume expansion: The thermal expansion in volume of a liquid when it is heated directly.
Apparent volume expansion: The thermal expansion in volume of a liquid in a container without considering the expansion of its container.

Experiment: real and apparent expansion

Take a coloured liquid in a long-necked glass flask (Figure 11.8). Note the initial level of the liquid before heating as V_1. Now start heating the flask from its bottom. The liquid level first falls to V_2 and then rises to V_3. Why does this happen?

Figure 11.8: Real and apparent expansion of a liquid. On heating, the level first falls from V_1 to V_2 (the flask expands) and then rises to V_3. Real expansion = apparent expansion + expansion of the flask. The sizes are exaggerated.

The heat reaches the glass flask first, so the flask expands first and its volume increases. As a result the liquid level falls in the flask to the level V_2. After this, heat is conducted to the liquid, and the liquid begins to rise above V_2 on getting hot. At a certain temperature it reaches the level V_3. The increase in the volume of the liquid from V_1 to V_3 is called the apparent expansion of the liquid. But heat is actually transferred to the liquid when it is at the level V_2. So the rise in level from V_2 to V_3 is the real expansion of the liquid.

The actual expansion of the liquid is greater than its apparent expansion. The fall in level from V_1 to V_2 represents the expansion of the glass flask. Hence

\text{Real expansion of the liquid}=\text{Apparent expansion of the liquid}+\text{Expansion of the flask}

V_2V_3=V_1V_3+V_1V_2

11.8

Here V_aV_b means the length of level change from V_a to V_b in Figure 11.8.

The coefficient of real volume expansion \gamma_r is always greater than the coefficient of apparent volume expansion \gamma_a, by an amount equal to the coefficient of volume expansion of the container, \gamma_g. Thus

\gamma_r=\gamma_a+\gamma_g

11.9

Different liquids have different coefficients of volume expansion.

Science tidbits

All liquids expand on heating and contract on cooling except water. When water cools from 4\,^\circ\mathrm{C} to 0\,^\circ\mathrm{C}, it actually expands. This strange property is called the anomalous expansion of water. When water freezes at 0\,^\circ\mathrm{C}, the hydrogen bonds become more stable and make the water molecules arrange themselves into a crystalline lattice structure. This structure causes the expansion of water on freezing. At 4\,^\circ\mathrm{C} water has its highest density. That is why ice at 0\,^\circ\mathrm{C} floats on liquid water, a property crucial for life in aquatic ecosystems. As water in lakes and rivers freezes, the ice floats on top, insulating the lower layers of water and preventing them from freezing completely. This allows life to survive harsh winters.

Example 11.3: glycerine in a laboratory flask

A laboratory flask is filled with 250\,\mathrm{ml} of glycerine at a temperature of 20\,^\circ\mathrm{C}. Calculate the volume of glycerine at 70\,^\circ\mathrm{C}. The coefficient of volumetric thermal expansion for glycerine is 500\times10^{-6}\,\mathrm{K^{-1}}.

Given

V_0 = 250\,\mathrm{ml}=250\times10^{-6}\,\mathrm{m^3}
\gamma_r = 500\times10^{-6}\,\mathrm{K^{-1}}
\Delta T = 70-20=50\,^\circ\mathrm{C}=50\,\mathrm{K}

To find

increase in volume \Delta V, then the volume at 70\,^\circ\mathrm{C}

The glycerine itself is heated, so the real expansion of the liquid is needed.
\Delta V=\gamma_rV_0\Delta T
Substitute the values.
\Delta V=(500\times10^{-6}\,\mathrm{K^{-1}})(250\times10^{-6}\,\mathrm{m^3})(50\,\mathrm{K})
Multiply: 500\times250\times50=6\,250\,000, and the powers of ten give 10^{-12}.
\Delta V=6.25\times10^{-6}\,\mathrm{m^3}=6.25\,\mathrm{ml}
The volume of glycerine at 70\,^\circ\mathrm{C} is the original volume plus the increase.
V=V_0+\Delta V=250+6.25=256.25\,\mathrm{ml}

\Delta V=6.25\,\mathrm{ml},\qquad V=256.25\,\mathrm{ml}

Worked example: mercury in a glass flask

A glass flask holds 200\,\mathrm{cm^3} of mercury, filled to the top at 20\,^\circ\mathrm{C}. It is heated to 70\,^\circ\mathrm{C}. Take \beta for mercury from Table 11.3 and \alpha for glass from Table 11.2. Find (a) the real expansion of the mercury, (b) the expansion of the flask, (c) the apparent expansion, that is, the volume of mercury that overflows.

Given

V_0 = 200\,\mathrm{cm^3}
\gamma_r = 18\times10^{-5}\,\mathrm{K^{-1}}\ (\text{mercury})
\alpha_{\text{glass}} = 0.4\times10^{-5}\,\mathrm{K^{-1}}
\Delta T = 50\,\mathrm{K}

To find

real expansion, expansion of flask, apparent expansion

Real expansion of the mercury, from \Delta V=\gamma_rV_0\Delta T.
\Delta V_{\text{real}}=(18\times10^{-5})(200)(50)=1.8\,\mathrm{cm^3}
The flask is a solid, so its coefficient of volume expansion is three times its \alpha (equation 11.5).
\gamma_g=3\alpha=3(0.4\times10^{-5})=1.2\times10^{-5}\,\mathrm{K^{-1}}
The inside volume of the flask expands as if it were solid glass.
\Delta V_{\text{flask}}=(1.2\times10^{-5})(200)(50)=0.12\,\mathrm{cm^3}
Real expansion = apparent expansion + expansion of the flask (equation 11.8), so
\Delta V_{\text{apparent}}=1.8-0.12=1.68\,\mathrm{cm^3}
Check with coefficients (equation 11.9). \gamma_a=\gamma_r-\gamma_g=18\times10^{-5}-1.2\times10^{-5}=16.8\times10^{-5}\,\mathrm{K^{-1}}, and (16.8\times10^{-5})(200)(50)=1.68\,\mathrm{cm^3}. The two methods agree.
The mercury expanded by 1.8\,\mathrm{cm^3}, but the flask made 0.12\,\mathrm{cm^3} of extra room for it, so only 1.68\,\mathrm{cm^3} spills. A student who forgets the expansion of the flask would say 1.8\,\mathrm{cm^3} overflows.

\Delta V_{\text{real}}=1.8\,\mathrm{cm^3},\ \Delta V_{\text{flask}}=0.12\,\mathrm{cm^3},\ \Delta V_{\text{apparent}}=1.68\,\mathrm{cm^3}

11.2.3Applications and consequences of thermal expansion in real lifeOften examined

There are many applications of thermal expansion in daily life. Some of them are discussed below.

A. Thermometers (mercury and alcohol thermometers)

The expansion of liquids (mercury or alcohol) is used to measure temperature in mercury and alcohol thermometers. As the temperature increases, the liquid expands and rises in the calibrated capillary tube. The temperature is read from the scale.

B. Bimetallic strips (thermostats and switches)

A bimetallic strip consists of two thin strips of different metals, such as brass and iron, joined together (Figure 11.9). On heating the strip, one metal strip expands more than the other. This unequal expansion causes the strip to bend. Bimetallic strips are used as switches and thermostats. For example, in an electric iron a bimetallic thermostat switch controls the temperature of the heating coil.

Figure 11.9: Bimetallic strip. (a) The straight strip at room temperature. (b) On heating, the brass layer expands more than the iron layer, so the strip bends towards the contact. Heat is shown in colour.

Worked example: which way does a bimetallic strip bend when it is cooled?

A bimetallic strip of brass (on top) joined to iron (below) is straight at room temperature. In a cold room it is cooled well below room temperature. The brass layer expands more than the iron layer when heated. (a) Does the strip stay straight, bend up or bend down? (b) Explain. (c) Draw the strip.

A solid contracts on cooling, and the same relations as for expansion apply. The change in length \Delta L=\alpha L_0\Delta T is then negative, because \Delta T is negative.
The brass layer has the larger coefficient of expansion (it expands more on heating), so for the same fall in temperature it also contracts more. After cooling, the brass layer is shorter than the iron layer.
The two layers are joined along their whole length, so they cannot slip past each other. The only way the strip can fit a shorter top layer and a longer bottom layer is to curve with the shorter brass layer on the inside of the bend. The tip of the strip therefore moves up.
On heating it was the other way round: the brass layer became longer, so it was on the outside of the curve and the strip bent down towards the contact of Figure 11.9. A bimetallic strip bends in both cases, in opposite directions, so it can respond to a rise or to a fall of temperature.

A bimetallic strip that is cooled below room temperature. The brass layer contracts more than the iron layer, so the strip bends the other way (the tip moves up). The bending is exaggerated.

The strip bends upwards, with the brass layer on the inside of the curve.

C. Expansion joints

The purpose of expansion joints is to accommodate the expansion and contraction of materials resulting from changes in temperature. This prevents structural damage in bridges, railways, pipelines and buildings (Figure 11.10(a) and (b)).

D. Opening the cap of a bottle

To open the cap of a bottle that is tight, immerse it in hot water for a minute or so. The metal cap expands and becomes loose. It is now easy to turn it open (Figure 11.10(c)).

E. Power lines

Overhead transmission lines are given a certain amount of sag so that they can contract in winter without snapping (Figure 11.11).

Figure 11.11: Sagging in power lines. The conductor is hung with a sag (coloured) so that it can contract in cold weather without snapping.

F. Designing instruments and machinery

Different materials expand differently. Understanding thermal expansion is crucial for designing reliable and safe infrastructure, instruments and machinery.

G. Consequences of thermal expansion

If thermal expansion is ignored during construction, or in designing machines and instruments, different problems appear. Some of them are:

Temperature changes can cause materials like concrete and asphalt to expand and contract. Without proper joints or gaps, this can lead to cracks and structural damage.
Overheated power lines might sag excessively, causing short circuits or power outages.
In hot weather, railway tracks can expand and buckle if there are no expansion gaps.
Pipes carrying fluids can burst if the liquid inside expands due to heating and there is no room for expansion, causing leaks and significant damage.
Instruments with parts that expand when heated may give wrong readings if they are not properly calibrated.
Metals get weaker when they expand and contract many times, which can damage items like aeroplanes, machines and buildings.

11.3Evaporation and its cause

Have you ever wondered how wet clothes dry on a clothesline, or why puddles disappear on a hot summer day? This happens by a process called evaporation.

Evaporation: The process by which a liquid changes into vapours (gas) at any temperature. It occurs at the surface of the liquid.

Cause of evaporation

The molecules of a liquid have different kinetic energies, because they collide with each other and with the walls of the container. The high-energy molecules moving near the surface of the liquid escape from the surface in the form of vapour (Figure 11.12). This is how evaporation takes place.

Figure 11.12: Evaporation. Fast-moving molecules near the surface escape from the liquid (coloured arrows) as vapour.

11.3.2Evaporation causes coolingOften examined

During evaporation, the fast-moving molecules escape from the surface of the liquid. The molecules that have lower kinetic energies are left behind. This lowers the average kinetic energy of the liquid molecules and hence the temperature of the liquid.

Why the liquid cools

A sketch (not to scale) of the kinetic energies in a liquid. Molecules in the shaded high-energy part escape from the surface. The molecules left behind have a lower average kinetic energy, so the temperature of the liquid falls.

The temperature of a substance is related to the average kinetic energy of its molecules. A decrease in temperature causes the sensation of cooling. For example, when a person sweats, the evaporation of moisture from the skin cools the body. Similarly, when a liquid evaporates from the surface of a solid, it lowers the temperature of that solid. This principle is used in refrigerators and air conditioning systems.

11.3.3Factors affecting the rate of evaporationOften examined

The rate of evaporation tells how quickly the molecules at the surface of a liquid escape and become a gas. A higher rate of evaporation means that the liquid evaporates faster. The rate of evaporation depends on the following factors.

A. Temperature

The rate of evaporation is directly proportional to temperature. A higher temperature of the liquid increases the kinetic energy of its molecules and so increases the rate of evaporation. Water in sunshine (a high temperature area) evaporates faster than water in shade (a low temperature area).

B. Surface area

A larger surface area lets more molecules escape, which speeds up evaporation. The larger the surface area of a liquid, the greater the number of molecules that have the chance to escape from its surface. This is why wet clothes are stretched out to dry quickly.

C. Humidity

Humidity is the content of water in air. Dry air is less humid (it has a low value of moisture in it) and has the ability to store more moisture. Lower humidity therefore increases the rate of evaporation, because dry air can absorb more water vapour. Room air coolers (Figure 11.13) work most efficiently in the dry air months of June and less efficiently in the humid months of August.

Figure 11.13: Schematic diagram of a room air cooler. Hot air is pulled through the wet pad, where water evaporates and takes heat from the air, and the cool air (coloured) is blown out by the fan.

D. Wind speed

Increased wind speed blows away the moist air near the surface, which speeds up the rate of evaporation. This is why wet clothes dry more quickly on a windy day than on a calm day.

The four factors that affect the rate of evaporation.
Factor	Change	Effect on the rate of evaporation	Reason
Temperature	Higher	Increases	Molecules have more kinetic energy
Surface area	Larger	Increases	More molecules at the surface can escape
Humidity	Lower (drier air)	Increases	Dry air can take in more water vapour
Wind speed	Higher	Increases	Moist air near the surface is blown away

11.3.4Difference between evaporation and boilingOften examined

Both boiling and evaporation are the change of a liquid into vapour, but there are some differences between the two processes, as shown in Table 11.4.

Table 11.4: Difference between evaporation and boiling.
Property	Evaporation	Boiling
Occurs	At the surface of the liquid	Throughout the liquid
Temperature	At any temperature	Only at the boiling point of the liquid
Energy source	Uses surrounding energy	Uses a continuous external heat source
Rate or speed	Generally slower and depends on many factors	A fast process (after reaching the boiling point)
Process completion time	Takes longer to complete	A quick process compared with evaporation
Effect of pressure	Less affected by external pressure	Highly affected by external pressure
Formation of bubbles	No bubble formation	Bubbles form throughout the liquid and rise from the surface
Dependence on surface area	Depends on surface area	Does not depend on surface area
Effect on temperature of surroundings	Causes cooling in the surroundings	Increases the temperature of the surroundings

11.3.5Applications of evaporationOften examined

A. Refrigeration using evaporation (without CFCs)

Cooling in a fridge or refrigerator is produced by the evaporation of a refrigerant (a liquefied gas), which absorbs heat and creates a cooling effect.

Refrigerant: A material used to produce cooling in a refrigerator. It absorbs and releases heat by cycling between the liquid and the gaseous state.

Freons, or chlorofluorocarbons (CFCs), were the type of refrigerant gas used in fridges. However, it was found that CFCs deplete the ozone in the upper atmosphere, which increases the amount of UV rays from the sun, and all living things are harmed by ultraviolet radiation. So the use of CFCs has been prohibited. CFCs are replaced by environment-friendly substances such as water, ammonia and hydrocarbons (propane, butane) and so on.

The refrigerator contains a coil (called the evaporation coil) filled with refrigerant at low pressure. When the warm air inside the fridge comes into contact with the evaporator coil, the refrigerant absorbs heat and changes from liquid to gas, which cools the inside of the fridge. This gas is then compressed, releasing heat outside, and is condensed back into a liquid by the condenser. This cycle continues to maintain a cool temperature inside the fridge.

Figure 11.14: The refrigeration cycle. The refrigerant takes in heat from inside the fridge as it evaporates in the evaporator, is compressed, and gives out heat outside as it condenses in the condenser (heat flow in colour).

B. Food preservation

Many food items are preserved by removing the water content through evaporation. This concentrates the flavours and prevents spoilage. Examples include dried fruits and jerky (dried meat), as shown in Figure 11.15(a).

C. Production of salt

Sea water is stored in large pools. As the water evaporates, the salt is left behind (Figure 11.15(b)).

D. Air conditioning

Air conditioners use evaporation to cool air. They work by passing warm air over a coil that is filled with a cool liquid. As the liquid evaporates, it takes heat from the air, cooling it down.

E. Fever control

When someone has a high fever, a wet cloth is applied to the forehead. The reason is that heat from the head is removed when the water evaporates. As a result, the temperature of the patient's brain stays within acceptable bounds and no brain injury occurs.

11.4Latent heatOften examined

Imagine you are boiling a pot of water. The temperature of the water increases gradually as you turn up the heat. But then something interesting happens: the water reaches 100\,^\circ\mathrm{C} and, even when you apply more heat, the temperature does not change. Where is all that energy going?

The answer lies in a concept called latent heat. It is the "hidden" heat absorbed or released by a substance during a phase change (such as melting, freezing or solidification, boiling or condensation) without a change in temperature.

Latent heat: The amount of heat energy absorbed or lost by a substance during its phase change without change in its temperature.

To understand this idea, look at the graph of temperature against the heat added uniformly to ice at -20\,^\circ\mathrm{C} until it has become steam above its boiling point (Figure 11.16). The graph has five portions.

Heating ice at -20\,^\circ\mathrm{C} until it is steam

Figure 11.16: Phase change of ice to vapour (the sketch is not to scale). The two flat parts, B to C and D to E (coloured), are the changes of state: heat is added but the temperature does not change.

The five portions of Figure 11.16.
Portion	What happens	Temperature	Where the heat goes
A to B	Heat is given to ice at -20\,^\circ\mathrm{C}	Rises to 0\,^\circ\mathrm{C}; it is still ice	Raises the temperature of the ice
B to C	Ice changes to water at its melting point 0\,^\circ\mathrm{C}	Constant	Changes the state from solid to liquid: latent heat of fusion (L_f). It continues until all the ice has become water.
C to D	Added heat raises the temperature of the water	Rises uniformly to the boiling point	Raises the temperature of the water
D to E	Water changes to vapour at its boiling point 100\,^\circ\mathrm{C}	Constant	Changes the state from liquid to gas: latent heat of vaporisation (L_v)
E to F	Added heat raises the temperature of the vapour (steam)	Rises	Raises the temperature of the steam

The temperature of a substance is related to the average kinetic energy of its particles. Whenever added heat increases the average kinetic energy of the particles, the temperature rises. But when the added heat is used to change the state of the substance, the temperature does not change. Here the heat is used to weaken the attractive forces between the particles instead of increasing their average kinetic energy. This happens at the melting point and at the boiling point of the substance.

During melting, heat is absorbed by the substance to change its state from solid to liquid at its melting point without a change in temperature. In the same way, the same amount of heat is released by the same substance at its freezing point during solidification (liquid to solid). The same happens for the boiling of a liquid and the condensation of its vapour: the heat energy absorbed during boiling is equal to the heat energy released during condensation. In all these processes, the heat absorbed or released by the substance is used to accommodate the attractive forces between its particles.

11.4.1Latent heat of fusionOften examined

Latent heat of fusion: The amount of heat energy required to convert a given mass of a substance from the solid state to the liquid state without change in its temperature. During freezing, a liquid releases the same amount of heat to change into a solid.

If \Delta Q is the amount of heat energy needed to melt a mass m of a solid into liquid (or to freeze liquid into solid), then mathematically

\Delta Q=mL_f

11.10

Here L_f is the specific latent heat of fusion of the substance. It is defined as the amount of heat energy required to convert unit mass (one kilogram) of the solid at its melting point to liquid (or of liquid into solid) without any change in temperature.

L_f=\dfrac{\Delta Q}{m}

11.11

The SI unit of specific latent heat of fusion is joule per kilogram (\mathrm{J\,kg^{-1}}). Different substances have different specific latent heats of fusion. The latent heat of fusion of ice is 3.33\times10^{5}\,\mathrm{J\,kg^{-1}}. Table 11.5 gives some values.

Table 11.5: Latent heats of fusion and vaporisation of selected materials.
Material	Melting point (^\circ\mathrm{C})	Latent heat of fusion (\mathrm{J\,kg^{-1}})	Boiling point (^\circ\mathrm{C})	Latent heat of vaporisation (\mathrm{J\,kg^{-1}})
Helium	-269.65	5.23\times10^{3}	-269	2.09\times10^{4}
Oxygen	-218.79	1.38\times10^{4}	-183	2.13\times10^{5}
Water	0	3.33\times10^{5}	100	2.26\times10^{6}
Mercury	-39	1.13\times10^{4}	357	2.93\times10^{5}
Sulphur	119	3.81\times10^{4}	445	3.26\times10^{5}
Lead	327	2.45\times10^{4}	1750	8.70\times10^{5}
Aluminium	660	3.97\times10^{5}	2450	1.14\times10^{7}
Copper	1083	1.34\times10^{5}	1187	5.06\times10^{6}
Silver	961	8.82\times10^{4}	2193	2.33\times10^{6}
Gold	1063	6.44\times10^{4}	2660	1.58\times10^{6}

Latent heat of fusion compared with specific latent heat of fusion.
	Latent heat of fusion	Specific latent heat of fusion
Meaning	The total heat \Delta Q needed to melt the given mass of a solid at its melting point	The heat needed to melt one kilogram of the solid at its melting point
Formula	\Delta Q=mL_f	L_f=\Delta Q/m
Unit	Joule (\mathrm{J})	Joule per kilogram (\mathrm{J\,kg^{-1}})
Depends on mass?	Yes, it is proportional to m	No, it is a property of the material

Worked example: ice at -10\,^\circ\mathrm{C} to water at 30\,^\circ\mathrm{C}

How much heat is needed to change 0.20\,\mathrm{kg} of ice at -10\,^\circ\mathrm{C} into water at 30\,^\circ\mathrm{C}? Take c_{\text{ice}}=2100\,\mathrm{J\,kg^{-1}K^{-1}}, c_{\text{water}}=4200\,\mathrm{J\,kg^{-1}K^{-1}} and L_f=3.33\times10^{5}\,\mathrm{J\,kg^{-1}}. Draw the graph of temperature against heat added.

Given

m = 0.20\,\mathrm{kg}
c_{\text{ice}} = 2100\,\mathrm{J\,kg^{-1}K^{-1}}
c_{\text{water}} = 4200\,\mathrm{J\,kg^{-1}K^{-1}}
L_f = 3.33\times10^{5}\,\mathrm{J\,kg^{-1}}

To find

total heat \Delta Q

The process has three steps because the temperature stops changing during melting. Step 1: warm the ice from -10\,^\circ\mathrm{C} to 0\,^\circ\mathrm{C} (\Delta T=10\,\mathrm{K}).
\Delta Q_1=mc_{\text{ice}}\Delta T=(0.20)(2100)(10)=4200\,\mathrm{J}
Step 2: melt all the ice at 0\,^\circ\mathrm{C}. The temperature does not change, so use the latent heat.
\Delta Q_2=mL_f=(0.20)(3.33\times10^{5})=66\,600\,\mathrm{J}
Step 3: warm the water from 0\,^\circ\mathrm{C} to 30\,^\circ\mathrm{C} (\Delta T=30\,\mathrm{K}).
\Delta Q_3=mc_{\text{water}}\Delta T=(0.20)(4200)(30)=25\,200\,\mathrm{J}
Add the three amounts.
\Delta Q=4200+66\,600+25\,200=96\,000\,\mathrm{J}=9.6\times10^{4}\,\mathrm{J}
The graph (below) has a sloping part, a long flat part and another sloping part. A student who leaves out the melting step gets 4200+25\,200=29\,400\,\mathrm{J}, less than a third of the correct heat.

Heat needed to turn ice at -10\,^\circ\mathrm{C} into water at 30\,^\circ\mathrm{C}

The melting step (coloured) is by far the longest part of the graph: it needs more heat than the two warming steps together.

\Delta Q=9.6\times10^{4}\,\mathrm{J}\ (96\,\mathrm{kJ})

11.4.2Latent heat of vaporisationOften examined

Latent heat of vaporisation: The amount of heat energy required to convert a given mass of a substance from the liquid state to the gaseous state without any change in its temperature. During condensation the same amount of heat is released by the gas as it changes into a liquid.

If \Delta Q is the amount of heat energy needed to vaporise a mass m of a liquid into gas (or to condense gas into liquid), then mathematically

\Delta Q=mL_v

11.12

Here L_v is the specific latent heat of vaporisation of the substance. It is defined as the amount of heat energy required to convert unit mass (one kilogram) of the liquid at its boiling point to gas (or of gas into liquid) without any change in temperature.

L_v=\dfrac{\Delta Q}{m}

11.13

The SI unit of specific latent heat of vaporisation is joule per kilogram (\mathrm{J\,kg^{-1}}). Different substances have different specific latent heats of vaporisation. The latent heat of vaporisation of water is 2.26\times10^{6}\,\mathrm{J\,kg^{-1}} (see Table 11.5).

Latent heat of vaporisation compared with specific latent heat of vaporisation.
	Latent heat of vaporisation	Specific latent heat of vaporisation
Meaning	The total heat \Delta Q needed to vaporise the given mass of a liquid at its boiling point	The heat needed to vaporise one kilogram of the liquid at its boiling point
Formula	\Delta Q=mL_v	L_v=\Delta Q/m
Unit	Joule (\mathrm{J})	Joule per kilogram (\mathrm{J\,kg^{-1}})

11.4.3Experiment to find the latent heat of fusion of iceOften examined

The latent heat of fusion of ice can be found by heating ice at a steady rate and noting how long each stage takes.

Procedure

Place a beaker over a stand as shown in Figure 11.17.
Add small pieces of ice to the beaker.
To find out the temperature of the ice, hang a thermometer inside the beaker.
Set a burner below the beaker.
Start measuring the time with a stopwatch.
The ice will begin to melt.
Now there is a mixture of ice and water, but the temperature will not rise above 0\,^\circ\mathrm{C} until all the ice melts.
Note the time that the ice takes to melt completely into water at 0\,^\circ\mathrm{C}.
When all the ice has changed to water, its temperature will begin to rise. Continue heating the water at 0\,^\circ\mathrm{C} in the beaker.
Note the time that the water in the beaker takes to reach its boiling point of 100\,^\circ\mathrm{C} from 0\,^\circ\mathrm{C}.
Draw a temperature-time graph such as the one shown in Figure 11.18.

Figure 11.17: Experimental arrangement to find the latent heat of fusion of ice.

Temperature against time for ice and water

Figure 11.18: Temperature-time graph for ice and water, heated at a uniform rate. Melting lasts from t_1 to t_2 (coloured), and the water then warms from 0\,^\circ\mathrm{C} to 100\,^\circ\mathrm{C} from t_2 to t_3.

Let the mass of the ice be m. Reading the times from the graph: the time taken by the ice to melt completely at 0\,^\circ\mathrm{C} (the time for fusion) is t_F=t_2-t_1=3.7 minutes, and the time taken by the water to rise from 0\,^\circ\mathrm{C} to 100\,^\circ\mathrm{C} is t_R=t_3-t_2=4.6 minutes.

Derivation Latent heat of fusion of ice from the temperature-time graph

Aim. Find the specific latent heat of fusion L_f of ice from the times read off the graph of Figure 11.18.

Assumptions.

Heat is supplied by the burner at a uniform rate, so the heat absorbed is proportional to the time.
The specific heat of water is c=4200\,\mathrm{J\,kg^{-1}K^{-1}}.

Read the times from the graph. The time for the ice to melt completely at 0\,^\circ\mathrm{C} is the time of fusion.
t_F=t_2-t_1=3.7\,\mathrm{min}
The time taken by the water to rise from 0\,^\circ\mathrm{C} to 100\,^\circ\mathrm{C}.
t_R=t_3-t_2=4.6\,\mathrm{min}
Take the mass of ice as m. The temperature of the water rises by \Delta T=100\,^\circ\mathrm{C}=100\,\mathrm{K}, and the specific heat of water is c=4200\,\mathrm{J\,kg^{-1}K^{-1}}. The heat absorbed by the water in warming is
\Delta Q=mc\Delta T=m(4200)(100)=m\,(4.2\times10^{5}\,\mathrm{J\,kg^{-1}})
This heat is absorbed during the time t_R, and the heat is supplied at a uniform rate. So the rate of absorption of heat (the heat absorbed in unit time) is
\text{rate}=\dfrac{\Delta Q}{t_R}
The rate is the same during fusion and during warming. So the heat absorbed in the time t_F of fusion is the rate multiplied by that time. Call this equation (1).
\Delta Q_F=\dfrac{\Delta Q}{t_R}\times t_F=\Delta Q\times\dfrac{t_F}{t_R}\qquad(1)
The same heat, by the definition of latent heat of fusion, is equation (2).
\Delta Q_F=m\times L_f\qquad(2)
Compare (1) and (2).
m\,L_f=\Delta Q\times\dfrac{t_F}{t_R}
Put in the values. The mass m cancels from both sides.
L_f=(4.2\times10^{5}\,\mathrm{J\,kg^{-1}})\times\dfrac{3.7\,\mathrm{min}}{4.6\,\mathrm{min}}=3.38\times10^{5}\,\mathrm{J\,kg^{-1}}

ResultL_f=\dfrac{c\,\Delta T\;t_F}{t_R}\approx3.4\times10^{5}\,\mathrm{J\,kg^{-1}}

The experiment gives about 3.4\times10^{5}\,\mathrm{J\,kg^{-1}} for the latent heat of fusion of ice, close to the accepted value of 3.33\times10^{5}\,\mathrm{J\,kg^{-1}} (Table 11.5). The method works because the time spent at 0\,^\circ\mathrm{C} measures the heat used, when the heat is supplied at a steady rate.

11.4.4Experiment to find the latent heat of vaporisation of waterOften examined

To find the latent heat of vaporisation of water, keep heating the beaker that contains boiling water, as shown in Figure 11.19, until all the water changes into steam. Note the time taken by the water to change completely into steam at its boiling point of 100\,^\circ\mathrm{C}, and extend the temperature-time graph (Figure 11.20).

Figure 11.19: Experimental arrangement to find the latent heat of vaporisation of water.

Temperature against time for water, up to steam

Figure 11.20: Temperature-time graph for water. The long flat part at 100\,^\circ\mathrm{C} (coloured), from t_3 to t_4, is the time t_v in which all the water changes into steam. It is much longer than the melting stage.

Derivation Latent heat of vaporisation of water from the temperature-time graph

Aim. Find the specific latent heat of vaporisation L_v of water by extending the heating of Figure 11.18 until all the water has turned to steam.

Keep heating the beaker of boiling water until all the water has changed into steam at its boiling point of 100\,^\circ\mathrm{C}, and note the time. Extend the graph (Figure 11.20). The time taken by the water at 100\,^\circ\mathrm{C} to change into steam is the time of vaporisation.
t_v=t_4-t_3=24.4\,\mathrm{min}
From the experiment of section 11.4.3 the rate of absorption of heat by the water is known (heat absorbed in warming divided by the time t_R=4.6\,\mathrm{min}).
\text{rate}=\dfrac{\Delta Q}{t_R}
The heat absorbed during vaporisation is this rate multiplied by the time of vaporisation. Call this equation (1).
\Delta Q_v=\dfrac{\Delta Q}{t_R}\times t_v=\Delta Q\times\dfrac{t_v}{t_R}\qquad(1)
By the definition of latent heat of vaporisation, equation (2).
\Delta Q_v=m\times L_v\qquad(2)
Compare (1) and (2), with \Delta Q=m\,(4.2\times10^{5}\,\mathrm{J\,kg^{-1}}) from the warming of the water.
m\,L_v=m\,(4.2\times10^{5}\,\mathrm{J\,kg^{-1}})\times\dfrac{24.4\,\mathrm{min}}{4.6\,\mathrm{min}}
Simplify. The mass m cancels.
L_v=2.23\times10^{6}\,\mathrm{J\,kg^{-1}}

ResultL_v=\dfrac{c\,\Delta T\;t_v}{t_R}\approx2.23\times10^{6}\,\mathrm{J\,kg^{-1}}

The experiment gives 2.23\times10^{6}\,\mathrm{J\,kg^{-1}} while the accepted value is 2.26\times10^{6}\,\mathrm{J\,kg^{-1}} (Table 11.5), so the method is reliable.

General procedure for latent heat of fusion and vaporisation

From these two experiments the following general steps give the latent heat of fusion and of vaporisation of any substance.

Take the solid at its melting point in a beaker and heat it at a uniform rate.
Note the following times taken by the substance: (a) the time in which it completely melts, without any change of temperature, at its melting point (fusion process); (b) the time in which its temperature rises from the melting point to the boiling point; (c) the time in which it changes its state from liquid to gas, without change of temperature, at its boiling point (boiling process).
Calculate the heat absorbed by the substance during the time its temperature rises from the melting point to the boiling point.
Calculate the rate of absorption of heat by dividing the heat calculated in step 3 by the time in which its temperature rises: \text{rate of absorption of heat}=\Delta Q/t_R.
Latent heat of fusion: multiply this rate of absorption of heat by the time in which the substance completely melts without any change in its temperature at the melting point.
Latent heat of vaporisation: multiply this rate of absorption of heat by the time in which the substance changes from liquid to gas without change in temperature at its boiling point.

Example 11.4: heat to melt ice-cream

Find the amount of heat required for melting an ice-cream of mass 0.5\,\mathrm{kg} that is at -20\,^\circ\mathrm{C}. The latent heat of fusion of ice-cream is L_f=3.347\times10^{5}\,\mathrm{J\,kg^{-1}}, its melting point is -10\,^\circ\mathrm{C}, and its specific heat capacity is c=2.1\times10^{3}\,\mathrm{J\,kg^{-1}\,^\circ C^{-1}}.

Given

m = 0.5\,\mathrm{kg}
T_0 = -20\,^\circ\mathrm{C}\ (\text{initial})
T = -10\,^\circ\mathrm{C}\ (\text{melting point})
c = 2.1\times10^{3}\,\mathrm{J\,kg^{-1}K^{-1}}
L_f = 3.347\times10^{5}\,\mathrm{J\,kg^{-1}}

To find

total heat \Delta Q

Change in temperature to reach the melting point.
\Delta T=T-T_0=-10-(-20)=10\,^\circ\mathrm{C}=10\,\mathrm{K}
First heat is needed to raise the temperature of the ice-cream from -20\,^\circ\mathrm{C} to its melting point -10\,^\circ\mathrm{C}.
\Delta Q_1=mc\Delta T=(0.5)(2.1\times10^{3})(10)=1.05\times10^{4}\,\mathrm{J}
Now the heat needed to melt it at its melting point.
\Delta Q_2=mL_f=(0.5)(3.347\times10^{5})=16.735\times10^{4}\,\mathrm{J}
Total heat required.
\Delta Q=\Delta Q_1+\Delta Q_2=1.05\times10^{4}+16.735\times10^{4}=17.785\times10^{4}\,\mathrm{J}=177.85\,\mathrm{kJ}
So a total of 177.85\,\mathrm{kJ} of heat is required to melt 0.5\,\mathrm{kg} of ice-cream that starts at -20\,^\circ\mathrm{C}.

\Delta Q=1.785\times10^{5}\,\mathrm{J}\ (177.85\,\mathrm{kJ})

Worked example: how long should the boiling plateau be?

In the experiment, the burner supplies heat at a uniform rate and the ice took t_F=3.7\,\mathrm{min} to melt. Use Table 11.5 to predict how long the same mass of water at 100\,^\circ\mathrm{C} would need to turn completely into steam, and compare with 24.4\,\mathrm{min} from the experiment. Why is the second flat part so much longer than the first?

Given

t_F = 3.7\,\mathrm{min}
L_f = 3.33\times10^{5}\,\mathrm{J\,kg^{-1}}
L_v = 2.26\times10^{6}\,\mathrm{J\,kg^{-1}}

To find

predicted time t_v

The heat is absorbed at a constant rate, so the heat absorbed is proportional to the time, \Delta Q=(\text{rate})\times t. For the same mass m, the heats needed are mL_f and mL_v, so the times are in the ratio of the latent heats.
\dfrac{t_v}{t_F}=\dfrac{mL_v}{mL_f}=\dfrac{L_v}{L_f}
Work out the ratio.
\dfrac{L_v}{L_f}=\dfrac{2.26\times10^{6}}{3.33\times10^{5}}=6.79
Predicted time.
t_v=6.79\times3.7\,\mathrm{min}=25.1\,\mathrm{min}
Check using the rate found from the warming of the water: \text{rate}=c\Delta T/t_R=(4.2\times10^{5})/4.6=9.13\times10^{4}\,\mathrm{J\,kg^{-1}\,min^{-1}}. Then t_v=L_v/\text{rate}=(2.26\times10^{6})/(9.13\times10^{4})=24.8\,\mathrm{min}. The two predictions, 25.1 and 24.8\,\mathrm{min}, are close to the measured 24.4\,\mathrm{min}.
The second flat part is about seven times longer because each kilogram needs about seven times as much heat to turn into steam as to melt: the particles have to be pulled completely apart, not just loosened. The length of a flat part therefore shows the size of the latent heat.

t_v\approx25\,\mathrm{min}\ (\text{measured }24.4\,\mathrm{min})

11.5Pressure exerted by gas particles

The molecules of a gas are in a state of continuous random motion in a container. They collide with one another and also with the walls of the container (Figure 11.21). Pressure is the measure of the force exerted by the gas particles on a unit area of the walls of the container, perpendicular to the wall.

Figure 11.21: Random motion of gas particles in a container. The collisions with the walls (coloured arrows) push on the walls, and these pushes together are the pressure of the gas.

P=\dfrac{F}{A}=\dfrac{\text{force exerted by the particles due to collisions}}{\text{area of the walls of the container}}

The pressure of a gas in a container changes with a change in its temperature, its volume and the amount of substance in it (the number of particles). The three effects are discussed one by one.

A. Effect of temperature

If the temperature of the gas increases, the particles move faster. Faster particles hit the walls with a greater force and more frequently, so the pressure of the gas in the container increases. The pressure of the gas decreases if its temperature decreases. So the pressure of a gas in a container is directly proportional to its temperature, provided the volume remains constant.

B. Effect of volume

The pressure of a gas is inversely proportional to its volume if its temperature is kept constant (Boyle's law). If the volume of the container is reduced while the number of particles is kept constant, the particles have less space to move around. They hit the walls more often, so the pressure increases. Conversely, if the volume is increased, the particles have more space and collide with the walls less frequently, so the pressure decreases (Figure 11.22).

C. Effect of the amount of gas (number of particles)

Increasing the number of particles of the gas in a fixed volume increases the number of collisions of the gas molecules with the walls of the container, and hence it increases the pressure. Decreasing the number of particles in the container decreases the pressure.

Figure 11.22: Effect of temperature and volume on the pressure of a gas. (a) A rise in temperature makes the particles faster (longer coloured arrows), so the collisions are harder and more frequent. (b) Pushing the piston in (coloured arrow) reduces the volume, so the same particles collide with the walls more frequently. Either change raises the pressure.

How the pressure of a gas changes, explained with particles.
Change made	What the particles do	Effect on pressure	Condition
Temperature raised	Move faster, so they hit the walls harder and more often	Increases (proportional to the temperature)	Volume constant
Volume reduced	Have less space, so they hit the walls more often	Increases (inversely proportional to the volume, Boyle's law)	Temperature constant
More particles added	More particles hit the walls	Increases	Volume constant
Temperature lowered, volume increased or particles removed	The opposite of the changes above	Decreases	As above

11.6SuperconductivityOften examined

When a wire or a device such as a bulb is connected to a cell, current flows through it. Current flow means that the free electrons in the wire move through the gaps between the atoms of the conductor. But they collide with the atoms of the conductor. This causes a hindrance to their flow, which is called the resistance of the conductor. When it is related to the material, it is called the specific resistance or resistivity of the material of the conductor.

The atoms of the conductor are in vibrational motion. With a decrease in temperature, their amplitude of vibration decreases. This increases the size of the gap between the atoms that the electrons can move through. Therefore the collisions of the flowing electrons with the atoms of the conductor decrease, and hence the resistivity of the material also decreases.

If the temperature keeps on decreasing, then at a certain temperature (the critical temperature) the resistivity of the material becomes zero, and the material is then called a superconductor.

Superconductor: A material through which current flows without resistance.
Critical temperature: The temperature at which and below which the resistance and resistivity of a substance become zero. It is also called the superconducting transition temperature.

Resistivity of a superconductor and of a normal metal

Figure 11.23: Superconductors. The resistivity of a normal metal falls as the temperature falls but stays above zero (ink curve). A superconductor (coloured) follows a similar curve until the critical temperature T_c, where its resistivity drops suddenly to zero and stays zero below T_c. The curves are sketches, not to scale.

Superconductivity was discovered by the Dutch physicist Heike Kamerlingh Onnes in 1911. While studying the properties of materials at very low temperatures, he observed that the electrical resistance of mercury becomes zero when it is cooled to about 4.2\,\mathrm{K} (-268.95\,^\circ\mathrm{C}). He received the Nobel Prize in Physics in 1913.

Table 11.6: Critical temperatures of some substances.
Substance	Critical temperature
Mercury (Hg)	4.2\,\mathrm{K}
Aluminium (Al)	1.18\,\mathrm{K}
Tin (Sn)	3.72\,\mathrm{K}
Lead (Pb)	7.2\,\mathrm{K}
Yttrium barium copper oxide (YBCO)	92\,\mathrm{K}
Bismuth strontium calcium copper oxide (BSCCO)	108\,\mathrm{K}

Applications of superconductors

Superconductors are used in powerful electromagnets, particle accelerators, magnetic levitation (Maglev) trains (Figure 11.24(a)), small but powerful electric motors, fast computer chips, and potentially more efficient power transmission lines. Figure 11.24(b) shows magnetic levitation by a superconductor. Researchers continue to explore ways to achieve superconductivity at higher temperatures, which would make it more practical for everyday use.

Powerful electromagnets
Particle accelerators
Magnetic levitation (Maglev) trains
Small but powerful electric motors
Fast computer chips
Potentially more efficient power transmission lines

Formula sheet

Linear thermal expansion

\Delta L=\alpha L_0\,\Delta T

\Delta L is the increase in length, L_0 the original length, \alpha the coefficient of linear thermal expansion and \Delta T the rise in temperature (equation 11.1)

Units: \Delta L and L_0 in \mathrm{m}; \alpha in \mathrm{K^{-1}}; \Delta T in \mathrm{K}

Coefficient of linear thermal expansion

\alpha=\dfrac{\Delta L}{L_0\,\Delta T}

Fractional change in length per kelvin change in temperature (equation 11.2)

Unit: \mathrm{K^{-1}}

Volume thermal expansion

\Delta V=\beta V_0\,\Delta T

\Delta V is the increase in volume, V_0 the original volume, \beta the coefficient of volume expansion and \Delta T the rise in temperature (equation 11.3)

Units: \Delta V and V_0 in \mathrm{m^3}; \beta in \mathrm{K^{-1}}; \Delta T in \mathrm{K}

Coefficient of volume thermal expansion

\beta=\dfrac{\Delta V}{V_0\,\Delta T}

Fractional change in volume per kelvin change in temperature (equation 11.4)

Unit: \mathrm{K^{-1}}

Relation between the coefficients

\beta=3\alpha

\beta is the coefficient of volume expansion and \alpha the coefficient of linear expansion of a solid (equation 11.5)

Unit: \mathrm{K^{-1}}

Coefficient of real volume expansion of a liquid

\gamma_r=\beta=\dfrac{\Delta V}{V_0\,\Delta T}

\Delta V is the real increase in the volume of the liquid (equation 11.6)

Unit: \mathrm{K^{-1}}

Coefficient of apparent volume expansion of a liquid

\gamma_a=\dfrac{\Delta V_{\text{apparent}}}{V_0\,\Delta T}

\Delta V_{\text{apparent}} is the expansion seen without counting the expansion of the container (equation 11.7)

Unit: \mathrm{K^{-1}}

Real and apparent expansion of a liquid

\gamma_r=\gamma_a+\gamma_g

\gamma_g is the coefficient of volume expansion of the container; equivalently, real expansion = apparent expansion + expansion of the flask (equations 11.8 and 11.9)

Unit: \mathrm{K^{-1}}

Latent heat of fusion

\Delta Q=mL_f

\Delta Q is the heat needed to melt (or released on freezing) a mass m at its melting point; L_f is the specific latent heat of fusion, L_f=\Delta Q/m (equations 11.10 and 11.11)

Units: \Delta Q in \mathrm{J}; m in \mathrm{kg}; L_f in \mathrm{J\,kg^{-1}}

Latent heat of vaporisation

\Delta Q=mL_v

\Delta Q is the heat needed to vaporise (or released on condensation) a mass m at its boiling point; L_v is the specific latent heat of vaporisation, L_v=\Delta Q/m (equations 11.12 and 11.13)

Units: \Delta Q in \mathrm{J}; m in \mathrm{kg}; L_v in \mathrm{J\,kg^{-1}}

Rate of absorption of heat (experiment)

\text{rate}=\dfrac{\Delta Q}{t_R},\qquad L_f=\dfrac{c\,\Delta T\,t_F}{t_R},\qquad L_v=\dfrac{c\,\Delta T\,t_v}{t_R}

\Delta Q=mc\Delta T is the heat that warms the water from 0\,^\circ\mathrm{C} to 100\,^\circ\mathrm{C} in time t_R; t_F and t_v are the times for fusion and vaporisation at a uniform rate of heating

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

Pressure

P=\dfrac{F}{A}

F is the force exerted perpendicularly on an area A (for a gas, the force of the particle collisions on the walls)

Unit: \mathrm{Pa}=\mathrm{N\,m^{-2}}

Key terms

Freezing: The phenomenon in which a liquid changes into a solid.
Melting: The phenomenon in which a solid changes into a liquid.
Vaporisation: The phenomenon in which a liquid is converted into gas (vapours) upon boiling.
Evaporation: The phenomenon of conversion of a liquid into vapours at any temperature.
Condensation: The phenomenon in which gases (vapours) change into liquids.
Sublimation: The phenomenon in which a solid changes directly into a gas.
Deposition: The phenomenon in which a gas changes directly into a solid.
Thermal expansion: The increase in the size of matter on heating.
Linear thermal expansion: The increase in the length of a substance on heating.
Volume thermal expansion: The increase in the volume of a substance on heating.
Coefficient of linear thermal expansion: The fractional change in length per kelvin change in temperature.
Coefficient of volume thermal expansion: The fractional change in volume per kelvin change in temperature.
Real volume expansion: The thermal expansion in volume of a liquid when it is heated directly.
Apparent volume expansion: The thermal expansion in volume of a liquid in a container without considering the expansion of its container.
Bimetallic strip: Two thin strips of different metals joined together, which bend on heating.
Humidity: The content of water in air.
Refrigerant: A material used to produce cooling in a refrigerator, which absorbs and releases heat by cycling between the liquid and gaseous state.
Latent heat: The amount of heat energy absorbed or lost by a substance during its phase change without change in its temperature.
Latent heat of fusion: The amount of heat energy required to convert a given mass of a substance from the solid state to the liquid state without change in its temperature.
Latent heat of vaporisation: The amount of heat energy required to convert a given mass of a substance from the liquid state to the gaseous state without any change in its temperature.
Pressure: The force on a unit area of an object, acting perpendicularly.
Pressure of a gas: The measure of the force exerted by the gas particles on a unit area of the walls of the container, perpendicular to the walls.
Superconductor: A material through which current flows without resistance.
Critical temperature: The temperature at which and below which the resistance and resistivity of a substance become zero.

Multiple choice questionsp. 70–72

1

During a hot summer day, a metal bridge might expands slightly. This expansion is caused by

Athe bridge rusting and weakening.
Bthe metal atoms in the bridge vibrating more intensely.Answer
Cthe weight of cars driving over the bridge.
Da decrease in the air pressure around the bridge.

Why

Thermal expansion happens because the particles of a solid gain kinetic energy when heated. They vibrate with a larger amplitude and push their neighbours further away, so the bridge lengthens (option B).

Rusting (A), the weight of cars (C) and a fall in air pressure (D) do not make the atoms move further apart, so they are not causes of thermal expansion.

Note. The printed question reads "might expands"; the grammar is corrected here.

2

Fog forms on a cold window pane in the morning. Which change of state is occurring?

AMelting
BBoiling
CCondensationAnswer
DDeposition

Why

Fog is tiny droplets of liquid water. Water vapour in the air loses heat to the cold pane and changes from gas to liquid. A gas changing into a liquid is condensation (option C).

Melting (A) is solid to liquid and boiling (B) is liquid to gas, so neither applies. Deposition (D) is gas to solid, which gives frost rather than droplets.

3

Which process involves the change of state from a gas to a solid without passing through the liquid phase?

ADepositionAnswer
BMelting
CFreezing
DSublimation

Why

Deposition is the change of a gas directly into a solid, with no liquid stage (option A). Frost on leaves is an example (book Figure 11.3 d).

Melting (B) and freezing (C) both involve the liquid phase, and sublimation (D) is the reverse order, solid directly to gas.

4

Boiling point of water is:

A212\,^\circ\mathrm{C}
B212\,^\circ\mathrm{F}Answer
C100\,\mathrm{K}
D373\,^\circ\mathrm{C}

Why

Water boils at 100\,^\circ\mathrm{C} at standard atmospheric pressure. Converting to Fahrenheit, F=\tfrac{9}{5}\times100+32=212, so the boiling point is 212\,^\circ\mathrm{F} (option B).

A has the Fahrenheit number with the Celsius unit. C gives the Celsius number in kelvin, but the kelvin value is 373\,\mathrm{K}. D is the kelvin value with the Celsius unit, which is wrong.

5

\mathrm{J\,kg^{-1}\,K^{-1}} is the unit of:

Aspecific heat capacityAnswer
Bheat capacity
Clatent heat of fusion
Dheat energy

Why

Specific heat capacity is c=\dfrac{\Delta Q}{m\,\Delta T}, so its unit is joule divided by (kilogram times kelvin), which is \mathrm{J\,kg^{-1}\,K^{-1}} (option A).

Heat capacity (B) is \mathrm{J\,K^{-1}} because it has no mass term. Latent heat of fusion (C) is \mathrm{J\,kg^{-1}} and heat energy (D) is \mathrm{J}.

6

Evaporation takes place from

AsurfaceAnswer
Bbottom
Ccenter
Dany location

Why

Evaporation is the change of a liquid into vapour at any temperature, and it happens only at the surface of the liquid, where the fastest molecules can escape (option A).

B and C are inside the liquid, where molecules are surrounded by others and cannot escape easily. D is wrong because the book places evaporation at the surface, and a change inside the whole liquid is boiling.

7

Relation between linear and volume expansion of solids is

A\beta=\alpha/3
B\beta=1.5\,\alpha
C\alpha=\beta/3Answer
D\alpha=3\,\beta

Why

For a solid, \beta=3\alpha (book equation 11.5), so \alpha=\dfrac{\beta}{3} (option C).

A has the factor inverted. B uses 1.5 instead of 3, and D reverses the relation so that \alpha would be three times \beta.

8

Heat added to a substance, at its melting point, is used to:

Aincrease K.E. of particles
Bdecrease K.E. of particles
Cincrease the attraction between particles
Ddecrease the attraction between particlesAnswer

Why

At the melting point the temperature does not rise, so the heat does not increase the average kinetic energy of the particles. This rules out A and B.

The heat is used to weaken the attractive forces between the particles so that they can move apart and form a liquid (book, section 11.4). The attraction decreases (option D), and C is the opposite.

9

336 J/g is latent heat of fusion of a material. How much heat is required to melt 10 g of material at its melting point?

A336 J
B3360 JAnswer
C33600 J
D3.36\times10^{5} J

Why

Heat for a change of state is \Delta Q=mL_f=(10\,\mathrm{g})(336\,\mathrm{J\,g^{-1}})=3360\,\mathrm{J} (option B). The same answer comes from 0.010\,\mathrm{kg}\times3.36\times10^{5}\,\mathrm{J\,kg^{-1}}.

A is the latent heat per gram without multiplying by the mass. C is ten times too large, and D is the heat for 1 kg, not 10 g.

10

Which of the following factors increases the rate of evaporation?

ADecrease in temperature
BIncrease in humidity
CIncrease in wind speedAnswer
DDecrease in surface area

Why

Wind blows away the moist air just above the liquid, so more water vapour can leave the surface and evaporation speeds up (option C).

Lower temperature (A), higher humidity (B) and smaller surface area (D) all slow evaporation, because the molecules have less energy, the air already holds water vapour, or fewer molecules can escape.

11

What is value of \alpha for a solid if its \beta is 9\times10^{-7}\,\mathrm{K^{-1}}?

A3\times10^{-7}\,\mathrm{K^{-1}}Answer
B4.5\times10^{-7}\,\mathrm{K^{-1}}
C9\times10^{-7}\,\mathrm{K^{-1}}
D27\times10^{-7}\,\mathrm{K^{-1}}

Why

For a solid \beta=3\alpha, so \alpha=\dfrac{\beta}{3}=\dfrac{9\times10^{-7}}{3}=3\times10^{-7}\,\mathrm{K^{-1}} (option A).

B comes from dividing by 2 (which would need \beta=2\alpha). C is \beta itself, and D is 3\beta, the inverted relation.

12

The process that involves the latent heat of vaporization is

AMelting of ice
BFreezing of water
CEvaporation
DCondensation of steamAnswer

Why

Latent heat of vaporisation is the heat absorbed when a liquid turns into vapour at its boiling point, and the same amount is released when the vapour condenses back to liquid. Condensation of steam at 100\,^\circ\mathrm{C} therefore releases L_v=2.26\times10^{6}\,\mathrm{J\,kg^{-1}} per kilogram (option D).

Melting (A) and freezing (B) involve the latent heat of fusion, not vaporisation.

Note. No key is printed. Option C is printed "Eveporation" and is read as evaporation. Evaporation is also a liquid to vapour change that takes in heat, but the book's latent heat of vaporisation is the heat at the boiling point, so D is the best answer.

13

The sum and difference of the coefficient of real and apparent expansion of a liquid are in the ratio 2:1. The ratio of the coefficient of real expansion and apparent expansion must be

A1:1
B2:1
C2:3
D3:1Answer

Why

Let the real coefficient be \gamma_r and the apparent coefficient be \gamma_a. The condition gives \dfrac{\gamma_r+\gamma_a}{\gamma_r-\gamma_a}=\dfrac{2}{1}, so \gamma_r+\gamma_a=2\gamma_r-2\gamma_a, which simplifies to \gamma_r=3\gamma_a (option D).

Check with \gamma_r=\gamma_a+\gamma_g (book equation 11.9): \gamma_g=2\gamma_a gives the sum to difference ratio (2\gamma_a+\gamma_g):\gamma_g=2:1. A (1:1) gives zero difference, B (2:1) would give 3:1 instead, and C (2:3) gives a negative difference, which is impossible because the real coefficient is the larger one.

14

Latent heat refers to the energy absorbed or released by a substance during a change of state, but with no change in temperature. What does "latent" mean in this context?

AConstant
BVisible
CHiddenAnswer
DTransparent

Why

"Latent" means hidden. The heat enters or leaves the substance, but it does not show up as a temperature change, because it is used to change the state (option C).

Constant (A), visible (B) and transparent (D) do not describe the meaning of latent heat.

Constructed response questionsp. 72

1

Look at these graphs for heating the ice. The energy (E)-time (t) graph shows that energy is supplied to the ice at a constant rate to change it into liquid and finally into steam, while the temperature (T)-time (t) graph shows the change of temperature of the ice and water during the different stages.

Solution

(a) There are two horizontal parts of this graph, where the temperature is not changing but energy is being supplied to the substance at a constant rate. Explain why the temperature does not change at these two stages.

The two horizontal parts are the melting stage (ice at 0\,^\circ\mathrm{C} changing into water) and the boiling stage (water at 100\,^\circ\mathrm{C} changing into steam). In both, the heat supplied is used as latent heat.

Temperature measures the average kinetic energy of the particles. During a change of state the energy weakens the attractive forces between the particles, so it is used to pull them apart rather than to make them move faster. The average kinetic energy stays the same, so the temperature stays constant until the whole substance has changed state.

(b) Explain why the temperature and energy trends are not the same for the same time intervals.

The heater supplies energy at a constant rate P, so the energy supplied grows in a straight line with time: E=Pt. This is the straight E-t graph.

The temperature does not follow a straight line, because the energy is used in two different ways. While a substance is warming within one state, the energy raises its temperature, and the rise depends on its mass and specific heat capacity. During melting and boiling the energy changes the state, so the temperature stays flat.

Each state has its own specific heat capacity. Ice (2100\,\mathrm{J\,kg^{-1}\,K^{-1}}) and steam (2000\,\mathrm{J\,kg^{-1}\,K^{-1}}) warm about twice as fast as water (4200\,\mathrm{J\,kg^{-1}\,K^{-1}}) for the same energy. The steepness of the T-t graph therefore changes from stage to stage, while the E-t graph stays straight.

Temperature against time (schematic)

The temperature rises in steps: flat at the melting and boiling points, with steeper rises for ice and steam than for water. The time intervals are not to scale.

Energy against time (schematic)

Energy supplied at a constant rate gives a straight line.

2

A heater supplies energy at a constant rate to a 100 g sample of a substance. The variation with time of the temperature of the substance is shown in the graph. The substance is perfectly insulated from its surroundings. The power of the heater is 250 W. From the graph, answer the following questions.

Temperature against time for the substance

The graph from the question, redrawn from the printed figure. Values are read from the graph, so they are approximate.

Solution

(a) What is the melting and boiling point of the substance?

The temperature stops rising at two levels. The lower flat part, at about -50\,^\circ\mathrm{C}, is the melting stage. The upper flat part, just above the 30\,^\circ\mathrm{C} mark, is the boiling stage.

Reading the graph, the melting point is -50\,^\circ\mathrm{C} and the boiling point is about 33\,^\circ\mathrm{C}.

T_{\mathrm{melt}}=-50\,^\circ\mathrm{C},\quad T_{\mathrm{boil}}\approx33\,^\circ\mathrm{C}

(b) How long does it take to melt completely, and how long does it take to vaporise from the liquid state?

Melting is the flat part at -50\,^\circ\mathrm{C}, which runs from about 1 min to about 2.5 min. Vaporisation is the flat part at the boiling point, from about 7.5 min to about 21.5 min.

Time to melt completely.
t_{\mathrm{melt}}=2.5-1\approx1.5\,\mathrm{min}\;(=90\,\mathrm{s})
Time to vaporise from liquid.
t_{\mathrm{vap}}=21.5-7.5\approx14\,\mathrm{min}\;(=840\,\mathrm{s})

t_{\mathrm{melt}}\approx1.5\,\mathrm{min},\quad t_{\mathrm{vap}}\approx14\,\mathrm{min}

(c) How much does the temperature increase in its liquid phase (the sloping part of the graph between melting and boiling)?

In the liquid phase the temperature rises from the melting point to the boiling point. The rise is the difference of the two levels.

Temperature increase in the liquid phase.
\Delta T=T_{\mathrm{boil}}-T_{\mathrm{melt}}=33-(-50)=83

\Delta T=83\,^\circ\mathrm{C}\quad(=83\,\mathrm{K})

(d) Calculate the latent heat of vaporisation of the substance.

The heater supplies energy at 250\,\mathrm{W} for the whole vaporisation time, and all of it is used as latent heat. The mass is 100\,\mathrm{g}=0.1\,\mathrm{kg}.

Energy used to vaporise the substance, from P=E/t.
Q=Pt=250\times(14\times60)=250\times840=2.1\times10^{5}\,\mathrm{J}
Use \Delta Q=mL_v to find the specific latent heat.
L_v=\dfrac{\Delta Q}{m}=\dfrac{2.1\times10^{5}}{0.1}=2.1\times10^{6}\,\mathrm{J\,kg^{-1}}

L_v=2.1\times10^{6}\,\mathrm{J\,kg^{-1}}

Short response questionsp. 73

1

Why do dew-drops form on leaves and grass in a spring morning?

Solution

Dew forms when water vapour in the air condenses. On a clear night the leaves and grass lose heat to the sky and cool down faster than the air around them. The air in contact with them cools, and it can no longer hold all the water vapour it contains (warm air holds more moisture than cool air).

The water vapour then changes from gas to liquid, which is condensation, and it forms droplets on the leaves and grass.

2

What is the effect of pressure and temperature variation on sublimation and deposition?

Solution

Sublimation (solid to gas) is faster at a higher temperature, because the particles of the solid gain more energy and escape more easily. It is also helped by a lower pressure: the book's food-preservation example (Point to Ponder) places frozen food in a vacuum chamber, so the frozen water sublimes directly into the very low pressure space.

Deposition (gas to solid) is the reverse. It is helped by a low temperature, as on a surface colder than the air, which is when frost forms on leaves. A higher pressure of the vapour brings the molecules closer together, which helps them settle into a solid.

In short: sublimation is favoured by high temperature and low pressure, and deposition by low temperature and higher vapour pressure.

3

Why do vapours form on the handle and cylinder of a fire extinguisher when it is discharged?

Solution

The gas inside the extinguisher is under high pressure. When it is discharged, it expands rapidly as it leaves the cylinder, and an expanding gas becomes colder. The handle and the cylinder are cooled by this gas.

Water vapour in the surrounding air touches these cold surfaces and loses heat, so it condenses into tiny liquid droplets. These droplets look like a white vapour on the handle and cylinder.

4

How does sweating help to cool our body down during exercise?

Solution

Sweat is water on the skin. As it evaporates, the molecules with the most kinetic energy escape from the liquid surface, so the average kinetic energy of the remaining liquid falls. Evaporation takes its energy from the surroundings, which here is the skin and the body.

The body therefore loses heat to the evaporating sweat, so its temperature is kept down. This is the cooling caused by evaporation (book, section 11.3.2).

5

Consider a piece of metal X initially at a temperature of 100\,^\circ\mathrm{C}. It is placed on a heater, which provides heat at a constant rate, until it reaches a final temperature of 400\,^\circ\mathrm{C}. The metal has a melting point of 250\,^\circ\mathrm{C}.

Solution

(i) Draw a temperature-time graph illustrating the changes in temperature of metal X as it is heated.

The metal is solid while its temperature rises from 100\,^\circ\mathrm{C} to 250\,^\circ\mathrm{C}, so the graph rises. At 250\,^\circ\mathrm{C} it melts, and the temperature stays flat while the heat changes it into liquid. After melting, the liquid is heated from 250\,^\circ\mathrm{C} to 400\,^\circ\mathrm{C}, so the graph rises again.

Temperature-time graph for metal X

Two sloping parts with a flat part at the melting point, 250\,^\circ\mathrm{C}. The length of the flat part depends on the mass and latent heat of fusion, so the time intervals are not to scale.

(ii) Metal Y has a higher latent heat of fusion than metal X. Both metals have identical melting points and heat capacities, and metal Y is heated at the same energy supply rate as metal X. Describe how the temperature-time graph for metal Y will differ from that of metal X.

The two graphs have the same sloping parts before and after melting, because the heat capacities, the masses and the energy supply rate are the same, so each state warms at the same rate. Both start at 100\,^\circ\mathrm{C} and both melt at 250\,^\circ\mathrm{C}.

The flat part at 250\,^\circ\mathrm{C} is longer for metal Y. Melting needs \Delta Q=mL_f, and the heat supplied at a fixed rate P is P\,t. A larger latent heat needs more heat, so it takes a longer time, t=mL_f/P, to melt the same mass. Metal Y therefore stays at its melting point for longer and reaches 400\,^\circ\mathrm{C} later than metal X.

Temperature-time graphs for metals X and Y

Metal X (solid line) and metal Y (dashed line). The flat part of Y is longer because its latent heat of fusion is greater. The time intervals are not to scale.

Note. The time intervals are not to scale: the question gives no mass, latent heat or power, so the length of the flat part is schematic.

6

How does the pressure in a car tyre change during a long drive on a hot day?

Solution

Friction between the tyre and the road, and the repeated flexing of the rubber, do work on the tyre and the air inside it. This energy becomes internal energy, so the temperature of the air in the tyre rises.

The tyre is almost rigid, so the volume of the air stays nearly constant. At constant volume the pressure is proportional to the absolute temperature, \dfrac{P_1}{T_1}=\dfrac{P_2}{T_2}, so the pressure rises. The temperature must be taken in kelvin.

7

How does understanding thermal expansion help prevent cracks in sidewalks during hot weather?

Solution

Concrete and asphalt expand when they are heated. If the slabs were laid tightly against each other, the expansion would have no room, the pressure would build up, and the slabs would crack or buckle.

Engineers leave small gaps, called expansion joints, between the slabs (book, section 11.2.3). Each slab can then expand into the gap and contract again when it cools, so the stress does not build up.

8

Imagine you are stuck in the snow with your car. Which would be more effective for melting the snow trapped underneath, a pot of hot water or a high-powered heat lamp?

Solution

Melting the snow needs a large amount of heat, the latent heat of fusion of ice, L_f=3.3\times10^{5}\,\mathrm{J\,kg^{-1}}. This heat is used at 0\,^\circ\mathrm{C}, where the temperature does not rise.

A pot of hot water has only a limited store of heat. It can give up heat only as it cools, and it must be carried and poured onto the snow. Its heat is used up within a short time, and the water it makes adds to the wet snow, which can refreeze. The heat lamp supplies a steady, high rate of energy for as long as it is switched on, and it keeps melting the snow without needing to be refilled.

So the high-powered heat lamp is the more effective choice, because it keeps delivering heat at a high constant rate. The hot water is limited by the amount of heat it holds.

9

How does the evaporation of water from a plant's leaves help to transport water and nutrients throughout the plant?

Solution

Water evaporates from the surface of the leaves into the air. This removes water from the leaves, so the water content there falls and more water is drawn up from the roots through the stem to replace it.

The water moving up the plant carries the dissolved nutrients from the soil with it. The water that leaves the leaves as vapour is replaced continuously, so the water and nutrients keep moving through the whole plant.

10

Why does adding ice to a drink cool it down more effectively than adding cold water?

Solution

Ice at 0\,^\circ\mathrm{C} has to absorb the latent heat of fusion before it can melt, and this heat is taken from the drink without any change in the ice's temperature. Cold water, which is above 0\,^\circ\mathrm{C}, can only cool the drink by warming up by its own specific heat capacity.

Heat absorbed by 1 g of ice melting at 0\,^\circ\mathrm{C}.
Q_{\mathrm{ice}}=mL_f=(0.001)(3.3\times10^{5})=330\,\mathrm{J}
Temperature drop of 1 g of water that releases the same heat.
\Delta T=\dfrac{Q}{mc}=\dfrac{330}{(0.001)(4200)}\approx79\,\mathrm{K}

So one gram of melting ice takes in as much heat as one gram of water cooling by about 79\,^\circ\mathrm{C}. The ice therefore cools the drink much more effectively than the same mass of cold water.

11

What are the challenges associated with maintaining the extremely low temperatures required for superconductivity to occur?

Solution

A material shows superconductivity only when it is cooled below its critical temperature. Table 11.6 gives these temperatures from 1.18\,\mathrm{K} for aluminium to 108\,\mathrm{K} for BSCCO, and most are a few kelvin. The temperature has to stay below this value all the time, not only at the start.

Keeping a material this cold needs cooling equipment that is expensive to run and needs good insulation, because any heat leaking in from the surroundings can raise the temperature above the critical value and destroy the superconducting state.

The materials with higher critical temperatures are still very cold, so researchers continue to look for superconductors that work at higher temperatures, which would make them more practical for everyday use (book, section 11.6).

Long response questionsp. 73–74

1

Define latent heat. Differentiate between latent heat and specific latent heat of fusion. Write their formulae.

Solution

Latent heat

Latent heat is the amount of heat energy absorbed or released by a substance during a change of state, without any change in its temperature. The heat is used to weaken the attractive forces between the particles, not to increase their kinetic energy, so the temperature stays the same.

Melting of ice, freezing of water, boiling of water and condensation of steam all involve latent heat.

Latent heat of fusion

The latent heat of fusion is the heat energy needed to change a given mass of a solid into a liquid at its melting point, with no change in temperature. A liquid gives out the same heat when it freezes.

Specific latent heat of fusion

The specific latent heat of fusion L_f is the heat needed to change unit mass (1 kg) of a solid at its melting point into liquid, with no change in temperature.

L_f=\dfrac{\Delta Q}{m}

\Delta Q=mL_f

Here \Delta Q is the heat energy, m the mass and L_f the specific latent heat of fusion. Its SI unit is \mathrm{J\,kg^{-1}}. For ice, L_f=3.33\times10^{5}\,\mathrm{J\,kg^{-1}} (Table 11.5).

Differences

Point	Latent heat of fusion	Specific latent heat of fusion
Meaning	Heat needed to melt a given mass at its melting point	Heat needed to melt 1 kg at its melting point
Formula	\Delta Q=mL_f	L_f=\dfrac{\Delta Q}{m}
SI unit	joule (J)	joule per kilogram (\mathrm{J\,kg^{-1}})
Depends on	the mass melted and the substance	only the substance\; each substance has its own value

Latent heat on a heating curve

The graph shows the heating of 1 kg of ice from -20\,^\circ\mathrm{C} (the same idea as book Figure 11.16). The flat parts are the latent heats: 330\,\mathrm{kJ} to melt the ice and 2260\,\mathrm{kJ} to boil the water. The sloping parts are the heating of each state, which uses its specific heat capacity.

Heating 1 kg of ice from -20\,^\circ\mathrm{C} to water at its boiling point

Heat supplied to 1 kg of ice. The flat parts are the latent heats: melting at 0\,^\circ\mathrm{C} and boiling at 100\,^\circ\mathrm{C}.

2

Differentiate between latent heat of vaporisation and specific latent heat of vaporisation. Give their formulae and units.

Solution

Latent heat of vaporisation

The latent heat of vaporisation is the heat energy needed to change a given mass of a liquid into vapour (gas) at its boiling point, with no change in temperature. The same heat is released when the vapour condenses back into liquid.

Specific latent heat of vaporisation

The specific latent heat of vaporisation L_v is the heat needed to change unit mass (1 kg) of a liquid at its boiling point into gas, with no change in temperature.

\Delta Q=mL_v

L_v=\dfrac{\Delta Q}{m}

Here \Delta Q is the heat energy, m the mass and L_v the specific latent heat of vaporisation. The SI unit of L_v is \mathrm{J\,kg^{-1}}. For water, L_v=2.26\times10^{6}\,\mathrm{J\,kg^{-1}}.

Differences

Point	Latent heat of vaporisation	Specific latent heat of vaporisation
Meaning	Heat needed to vaporise a given mass at its boiling point	Heat needed to vaporise 1 kg at its boiling point
Formula	\Delta Q=mL_v	L_v=\dfrac{\Delta Q}{m}
SI unit	joule (J)	joule per kilogram (\mathrm{J\,kg^{-1}})
Example	heat to turn 2 kg of water into steam at 100 ^\circC: 2\times2.26\times10^{6}=4.52\times10^{6}\,\mathrm{J}	L_v=2.26\times10^{6}\,\mathrm{J\,kg^{-1}} for water

The specific value depends only on the substance, while the total heat also depends on the mass.

3

What is evaporation? Differentiate between boiling and evaporation. On what factors evaporation depends?

Solution

Evaporation

Evaporation is the process in which a liquid changes into vapour at any temperature. It takes place only at the surface of the liquid. The molecules near the surface with the highest kinetic energy escape. As they leave, the average kinetic energy of the liquid falls, so the liquid and its surroundings cool.

Difference between boiling and evaporation

Both change a liquid into vapour, but they differ in the ways shown in the table (book Table 11.4).

Property	Evaporation	Boiling
Where it occurs	at the surface of the liquid	throughout the liquid
Temperature	at any temperature	only at the boiling point
Energy source	uses energy from the surroundings	needs a continuous external heat source
Rate	generally slower\; depends on many factors	fast once the boiling point is reached
Completion time	takes longer to complete	quick
Effect of pressure	less affected by external pressure	highly affected by external pressure
Bubbles	no bubbles form	bubbles form throughout the liquid and rise to the surface
Surface area	depends on the surface area	does not depend on the surface area
Effect on surroundings	causes cooling	increases the temperature of the surroundings

Factors on which evaporation depends

Temperature. Evaporation is faster at a higher temperature, because more molecules have enough kinetic energy to escape.
Surface area. A larger surface lets more molecules escape at once, which is why wet clothes are spread out to dry.
Humidity. Dry air can absorb more water vapour, so evaporation is faster in dry air. In humid air the air already holds much vapour, so evaporation is slower.
Air movement (wind speed). Moving air blows away the moist air above the surface, so more molecules can escape and evaporation speeds up.

4

What are linear and volume thermal expansion of solids? Discuss qualitatively the factors upon which these thermal expansions depend.

Solution

Thermal expansion of solids

Thermal expansion is the increase in the size of matter on heating. In a solid, the increase in one dimension is linear thermal expansion, and the increase in volume is volume thermal expansion.

Linear thermal expansion

A rod of original length L_0 at temperature T_0 becomes L when its temperature rises to T. Its increase in length is \Delta L=L-L_0 for a temperature change \Delta T=T-T_0. Experiment shows that

\Delta L=\alpha L_0\,\Delta T

where \alpha is the coefficient of linear thermal expansion. It is the fractional change in length per kelvin change in temperature, \alpha=\dfrac{\Delta L}{L_0\,\Delta T}, and its SI unit is \mathrm{K^{-1}}.

Linear expansion of a rod. The rod is drawn with its original length above and its heated length below. The extra length ΔL is drawn larger than it is in reality.

Volume thermal expansion

For a solid of original volume V_0 heated through \Delta T, the increase in volume is

\Delta V=\beta V_0\,\Delta T

where \beta is the coefficient of volume thermal expansion (fractional change in volume per kelvin, unit \mathrm{K^{-1}}). For a solid, the two coefficients are related by \beta=3\alpha, as derived below.

Derivation Relation between \beta and \alpha

Aim. Show that \beta=3\alpha for a solid cube.

Assumptions.

The cube has side L and volume V=L^{3}.
The change is small, so the squares and cubes of \Delta L are negligible.

Write the volume after heating, when the side increases from L to L+\Delta L.
V+\Delta V=(L+\Delta L)^{3}
Expand the cube and drop the terms with \Delta L^{2} and \Delta L^{3}, which are very small.
V+\Delta V\approx L^{3}+3L^{2}\Delta L
Subtract the original volume V=L^{3} to get the increase in volume.
\Delta V=3L^{2}\,\Delta L
Replace \Delta L by \alpha L\,\Delta T from linear expansion.
\Delta V=3L^{2}\,\alpha L\,\Delta T=3\alpha L^{3}\Delta T
Replace L^{3} by the original volume V_0 and compare with \Delta V=\beta V_0\,\Delta T.
\Delta V=3\alpha V_0\,\Delta T\;\Rightarrow\;\beta=3\alpha

Result\beta=3\alpha

So the coefficient of volume expansion is three times the coefficient of linear expansion. For steel, \alpha=1.2\times10^{-5}\,\mathrm{K^{-1}} gives \beta=3.6\times10^{-5}\,\mathrm{K^{-1}}.

Factors on which thermal expansion depends

Increase in temperature. The expansion is directly proportional to the rise in temperature: the atoms vibrate more and move further apart as the temperature rises.
Original size. The expansion is directly proportional to the original length or volume, because a longer rod has more atoms, and each one adds to the expansion.
Nature of the material. Each material has its own coefficient, which depends on the strength of the attractive forces between its atoms. Aluminium expands more than steel for the same rise in temperature.

Book Table 11.2 (coefficients of linear thermal expansion)
Material	Coefficient of linear expansion \alpha (\mathrm{K^{-1}})
Steel	1.2\times10^{-5}
Quartz	0.04\times10^{-5}
Invar	0.09\times10^{-5}
Glass	0.4\times10^{-5}
Copper	1.7\times10^{-5}
Brass	2\times10^{-5}
Aluminium	2.4\times10^{-5}

5

What are uses of thermal expansion in our daily life? Give examples.

Solution

Thermal expansion is used in many everyday devices and structures, and designers must also allow for it. The main examples given in the book are:

Expansion joints in bridges and roads. Gaps or sliding joints let the structure expand in summer and contract in winter without cracking or buckling (book Figure 11.10 a).
Gaps between railway tracks. The gaps leave room for the steel rails to expand. Without them, the rails would buckle in hot weather (Figure 11.10 b).
Overhead power lines. The conductors are given a sag, so that they can contract in winter without snapping (Figure 11.11).
Opening a tight bottle cap. Dipping the bottle in hot water for a minute expands the metal cap more than the glass, so the cap loosens and can be turned open (Figure 11.10 c).
Mercury and alcohol thermometers. The liquid expands along a calibrated capillary tube as the temperature rises, and the level of the liquid is read as a temperature.
Bimetallic strips. Two metals with different expansions are joined together to make switches and thermostats (see question 7).
Eiffel Tower. It is made of iron and grows up to 15 cm taller in summer. Engineers allow for this in its design to keep it safe.

Thermal expansion also causes problems if it is ignored: concrete and asphalt crack, pipes carrying liquid can burst when the liquid expands and there is no room for it, and instruments with parts that expand unevenly give wrong readings.

6FBISE SSC-II 2026 Annual I, Set M, Q.5

Explain thermal expansion of liquids in detail.

Solution

Expansion of liquids

Liquids expand when heated. Their particles gain kinetic energy and push each other further apart. The attractive forces between liquid particles are weaker than in solids, so liquids expand more than solids for the same rise in temperature. A liquid cannot be heated on its own, so it is heated in a container. The volume change must therefore be split into two types: real expansion and apparent expansion.

Real volume expansion

The real volume expansion is the increase in the volume of the liquid when heat is applied to it directly. Its coefficient is \gamma_r, which is the fractional increase in real volume for a rise of 1 K:

\gamma_r=\dfrac{\Delta V}{V_0\,\Delta T}

Apparent volume expansion

The apparent volume expansion is the expansion of the liquid as it appears in the container, without considering the expansion of the container. Its coefficient is \gamma_a:

\gamma_a=\dfrac{\Delta V_{\mathrm{apparent}}}{V_0\,\Delta T}

Experiment with a flask

A coloured liquid is kept in a flask with a long narrow neck (book Figure 11.8). When the flask is heated from below, the liquid level first falls from V_1 to V_2, because the glass flask expands first and its volume grows. The heat then reaches the liquid, which expands and rises from V_2 to V_3.

Liquid in a flask with a narrow neck. The levels V1, V2 and V3 are marked. The real expansion is the rise from V2 to V3, the apparent expansion is the rise from V1 to V3, and the flask expansion is the fall from V1 to V2.

The rise of the liquid seen against the glass is the apparent expansion (V1 to V3). The real expansion is the rise from V2 to V3, measured from the level after the flask has grown. The three quantities are related by

\Delta V_r=\Delta V_a+\Delta V_g

where \Delta V_g is the expansion of the flask. Dividing by V_0\,\Delta T gives the relation between the coefficients (book equation 11.9):

\gamma_r=\gamma_a+\gamma_g

So the real coefficient is always greater than the apparent coefficient, by the coefficient of the container. Different liquids have different coefficients, as Table 11.3 shows: for example \beta=113\times10^{-5}\,\mathrm{K^{-1}} for methanol and 18\times10^{-5}\,\mathrm{K^{-1}} for mercury.

Water is an exception

Almost all liquids expand on heating and contract on cooling. Water does not behave in this way between 4\,^\circ\mathrm{C} and 0\,^\circ\mathrm{C}: as it cools from 4\,^\circ\mathrm{C} to 0\,^\circ\mathrm{C} it expands, so ice floats on liquid water (book Science Tidbits).

7

Discuss the application of bimetallic strips in temperature control devices like thermostats. Explain the principle of thermal expansion that allows the bimetallic strip to function, and illustrate how the differing expansion rates of the two metals are utilized in practical.

Solution

What a bimetallic strip is

A bimetallic strip is two thin strips of different metals, such as brass and iron, joined together along their length (book Figure 11.9). The metals are chosen so that they expand by different amounts for the same rise in temperature.

Principle

When the strip is heated, both metals expand, but the metal with the larger coefficient of linear expansion lengthens more. The two layers therefore have different lengths and the strip bends. It curves towards the metal with the smaller coefficient, which is the one that expands less.

Bimetallic strip. When cool it is straight (left). When hot the brass layer, which expands more, is on the outside and the strip bends towards the iron side (right). The dashed line marks the straight strip when cool.

Worked illustration with numbers

Take a brass and a steel strip, each 10\,\mathrm{cm} long, heated through 80\,\mathrm{K}. Table 11.2 gives \alpha_{\mathrm{brass}}=2\times10^{-5}\,\mathrm{K^{-1}} and \alpha_{\mathrm{steel}}=1.2\times10^{-5}\,\mathrm{K^{-1}}. (Iron is not in the table, so steel is used for the numbers.)

Increase in length of the brass layer, from \Delta L=\alpha L_0\,\Delta T.
\Delta L_{\mathrm{brass}}=(2\times10^{-5})(0.10)(80)=1.6\times10^{-4}\,\mathrm{m}
Increase in length of the steel layer.
\Delta L_{\mathrm{steel}}=(1.2\times10^{-5})(0.10)(80)=9.6\times10^{-5}\,\mathrm{m}
Difference in lengths, which is what makes the strip bend.
\Delta L_{\mathrm{brass}}-\Delta L_{\mathrm{steel}}=6.4\times10^{-5}\,\mathrm{m}=0.064\,\mathrm{mm}

A small difference of this kind is enough to bend a strip noticeably, which is why the difference in expansion rates is used in practice.

Use in temperature control

In an electric iron, a bimetallic thermostat switch controls the heating coil (book, section 11.2.3). When the iron becomes too hot, the strip bends far enough to open the contact and cut off the current. As the iron cools, the strip straightens and closes the contact again, so the temperature stays within a set range. Bimetallic strips are also used in other switches.

8

What is superconductivity? Explain the phenomenon in detail. What is their scope in the future world?

Solution

What superconductivity is

Superconductivity is the property of certain materials whose electrical resistance, and resistivity, becomes zero when they are cooled below a particular temperature. A current set up in a superconductor keeps flowing without any resistance, so there is no loss of energy as heat. Superconductors are the materials through which current flows without resistance (book, section 11.6).

Critical temperature

The temperature at which, and below which, the resistance and resistivity of a substance become zero is its critical temperature T_c, also called its superconducting transition temperature.

Above T_c the material behaves as an ordinary conductor. Its resistivity falls as the temperature falls, because the atoms vibrate less, the gaps between them grow and the electrons collide with them less often (book Figure 11.23). At T_c the resistivity drops suddenly to zero.

Resistivity against temperature

Schematic version of book Figure 11.23. The superconductor's resistivity drops to zero at T_c, while a normal metal keeps a small resistivity. The values are schematic, and T_c=4.2\,\mathrm{K} is the value for mercury.

Critical temperatures of some substances

Book Table 11.6
Substance	Critical temperature
Mercury (Hg)	4.2 K
Aluminium (Al)	1.18 K
Tin (Sn)	3.72 K
Lead (Pb)	7.2 K
Yttrium barium copper oxide (YBCO)	92 K
Bismuth strontium calcium copper oxide (BSCCO)	108 K

Discovery

The Dutch physicist Heike Kamerlingh Onnes discovered superconductivity in 1911. He found that the electrical resistance of mercury becomes zero when it is cooled to about 4.2\,\mathrm{K}, which is -268.95\,^\circ\mathrm{C}. He was awarded the Nobel Prize in Physics in 1913.

Scope in the future

Superconductors are used in powerful electromagnets, particle accelerators, magnetic levitation (Maglev) trains, small but powerful electric motors, fast computer chips and possibly more efficient power transmission lines (book Figure 11.24).

The main limitation is the very low temperature needed. Researchers continue to look for ways to achieve superconductivity at higher temperatures, which would make it far more practical for everyday use.

Numerical response questionsp. 74

1

Consider a steel bar of length 1.5\,\mathrm{m} at 10\,^\circ\mathrm{C}. It is heated to raise its temperature to 100\,^\circ\mathrm{C}. Calculate (a) the increase in length and (b) the final length at 100\,^\circ\mathrm{C}. The coefficient of linear thermal expansion of steel is 1.2\times10^{-5}\,\mathrm{K^{-1}}.

Solution

Given

L_0 = 1.5\,\mathrm{m} original length of the bar
T_0 = 10\,^\circ\mathrm{C}
T = 100\,^\circ\mathrm{C}
\alpha = 1.2\times10^{-5}\,\mathrm{K^{-1}} coefficient of linear expansion of steel

To find

(a) \Delta L
(b) L

The change in temperature is \Delta T=100-10=90\,\mathrm{K}, and this is the same as a change of 90\,^\circ\mathrm{C}.

(a) Increase in length of the bar.

Use the linear expansion formula.
\Delta L=\alpha L_0\,\Delta T
Substitute the values.
\Delta L=(1.2\times10^{-5})(1.5)(90)=1.62\times10^{-3}\,\mathrm{m}
Convert to millimetres.
\Delta L=1.62\,\mathrm{mm}

\Delta L=1.62\,\mathrm{mm}

(b) Final length of the bar at 100\,^\circ\mathrm{C}.

The final length is the original length plus the increase.
L=L_0+\Delta L=1.5+0.00162=1.50162\,\mathrm{m}
Round to the precision of the data.
L\approx1.5\,\mathrm{m}

L\approx1.5\,\mathrm{m}

Book answer. (a) 1.62 mm, (b) 1.5 m

2

A solid cube of side length 10\,\mathrm{cm} at 25\,^\circ\mathrm{C} is heated. What will be the increase in its volume at 125\,^\circ\mathrm{C} if its coefficient of linear thermal expansion is 9\times10^{-4}\,\mathrm{K^{-1}}?

Solution

Given

L = 10\,\mathrm{cm} side of the cube
V_0 = L^{3}=1000\,\mathrm{cm^{3}} original volume
\Delta T = 125-25=100\,\mathrm{K}
\alpha = 9\times10^{-4}\,\mathrm{K^{-1}} as printed in the question

To find

\Delta V

For a solid \beta=3\alpha, so find the coefficient of volume expansion.
\beta=3\alpha=2.7\times10^{-3}\,\mathrm{K^{-1}}
Use \Delta V=\beta V_0\,\Delta T.
\Delta V=(2.7\times10^{-3})(1000)(100)
Evaluate.
\Delta V=270\,\mathrm{cm^{3}}

\Delta V=2.7\times10^{2}\,\mathrm{cm^{3}}

Book answer. 2.7 cm^3

Note. With the printed coefficient 9\times10^{-4}\,\mathrm{K^{-1}} the answer is 270\,\mathrm{cm^{3}}. The book's 2.7\,\mathrm{cm^{3}} follows only if \alpha=9\times10^{-6}\,\mathrm{K^{-1}}. That value fits Table 11.2, where every coefficient is of the order of 10^{-5}\,\mathrm{K^{-1}}, so the exponent in the question is probably misprinted. The working above uses the printed value.

3

A steel railroad track segment is 10\,\mathrm{m} long at a cool morning temperature of 15\,^\circ\mathrm{C}. The coefficient of linear expansion for steel is 1.2\times10^{-5} per ^\circ\mathrm{C}. Later in the day, the temperature rises to 30\,^\circ\mathrm{C}. How much will the steel track segment expand in length?

Solution

Given

L_0 = 10\,\mathrm{m}
\Delta T = 30-15=15\,\mathrm{K}
\alpha = 1.2\times10^{-5}\,\mathrm{K^{-1}} read as 10^{-5} per ^\circC, the printed minus sign is lost

To find

\Delta L

Use the linear expansion formula.
\Delta L=\alpha L_0\,\Delta T
Substitute the values.
\Delta L=(1.2\times10^{-5})(10)(15)=1.8\times10^{-3}\,\mathrm{m}
Convert to millimetres.
\Delta L=1.8\,\mathrm{mm}

\Delta L=1.8\,\mathrm{mm}

Book answer. 1.8 mm

4

A 2-litre (2000\,\mathrm{cm^{3}}) glass bottle is filled completely with orange juice at a room temperature of 20\,^\circ\mathrm{C}. The coefficient of volumetric expansion for orange juice is 5\times10^{-5} per ^\circ\mathrm{C}. If the bottle is left in a hot car where the temperature reaches 40\,^\circ\mathrm{C}, what volume of orange juice will overflow? Expansion of the glass bottle is negligible.

Solution

Given

V_0 = 2000\,\mathrm{cm^{3}} volume of juice in the full bottle
\Delta T = 40-20=20\,\mathrm{K}
\beta = 5\times10^{-5}\,\mathrm{K^{-1}} read as 10^{-5} per ^\circC\; the printed minus sign is lost

To find

volume that overflows

The bottle does not expand, so the overflow is the real expansion of the juice.
Use \Delta V=\beta V_0\,\Delta T.
\Delta V=(5\times10^{-5})(2000)(20)
Evaluate.
\Delta V=2.0\,\mathrm{cm^{3}}

\Delta V=2.0\,\mathrm{cm^{3}}

Book answer. 2 cm^3

5

How much heat is required to change 15 kg of ice into water at its melting point?

Solution

Given

m = 15\,\mathrm{kg} mass of ice
L_f = 3.3\times10^{5}\,\mathrm{J\,kg^{-1}} latent heat of fusion of ice

To find

\Delta Q

Heat needed at the melting point, with no change in temperature.
\Delta Q=mL_f
Substitute the values.
\Delta Q=(15)(3.3\times10^{5})=4.95\times10^{6}\,\mathrm{J}

\Delta Q=4.95\times10^{6}\,\mathrm{J}

Book answer. 4.95 \times 10^6 J

Note. With the Table 11.5 value 3.33\times10^{5}\,\mathrm{J\,kg^{-1}} the answer is 4.995\times10^{6}\,\mathrm{J}, which rounds to 5.0\times10^{6}\,\mathrm{J}. The book's 4.95\times10^{6}\,\mathrm{J} uses 3.3\times10^{5}\,\mathrm{J\,kg^{-1}}.

6

How much heat is required to change 7 kg of water into steam at its boiling point?

Solution

Given

m = 7\,\mathrm{kg} mass of water
L_v = 2.26\times10^{6}\,\mathrm{J\,kg^{-1}} latent heat of vaporisation of water

To find

\Delta Q

Heat needed at the boiling point, with no change in temperature.
\Delta Q=mL_v
Substitute the values.
\Delta Q=(7)(2.26\times10^{6})=1.582\times10^{7}\,\mathrm{J}

\Delta Q=1.58\times10^{7}\,\mathrm{J}

Book answer. 1.58 \times 10^7 J

7

4 kg of ice has a temperature of -20\,^\circ\mathrm{C}. It is heated to convert it into steam, and its final temperature is 120\,^\circ\mathrm{C}. Calculate the total amount of heat energy involved in this conversion of ice into steam. (Specific heat of ice =2100\,\mathrm{J\,kg^{-1}\,K^{-1}}, specific heat of water =4200\,\mathrm{J\,kg^{-1}\,K^{-1}}, specific heat of steam =2000\,\mathrm{J\,kg^{-1}\,K^{-1}}, latent heat of fusion of ice =3.3\times10^{5}\,\mathrm{J\,kg^{-1}} and latent heat of vaporisation of water =2.26\times10^{6}\,\mathrm{J\,kg^{-1}}.)

Solution

Given

m = 4\,\mathrm{kg}
T_i = -20\,^\circ\mathrm{C} ice
T_f = 120\,^\circ\mathrm{C} steam
c_{\mathrm{ice}} = 2100\,\mathrm{J\,kg^{-1}\,K^{-1}}
c_{\mathrm{w}} = 4200\,\mathrm{J\,kg^{-1}\,K^{-1}}
c_{\mathrm{s}} = 2000\,\mathrm{J\,kg^{-1}\,K^{-1}}
L_f = 3.3\times10^{5}\,\mathrm{J\,kg^{-1}}
L_v = 2.26\times10^{6}\,\mathrm{J\,kg^{-1}}

To find

total heat Q

The process has five stages: heating the ice to 0\,^\circ\mathrm{C}, melting it, heating the water to 100\,^\circ\mathrm{C}, boiling it, and heating the steam to 120\,^\circ\mathrm{C}. The heat for each stage is added up.

Stage 1: heat the ice from -20\,^\circ\mathrm{C} to 0\,^\circ\mathrm{C}.
Q_1=mc_{\mathrm{ice}}\Delta T=4\times2100\times20=1.68\times10^{5}\,\mathrm{J}
Stage 2: melt the ice at 0\,^\circ\mathrm{C}.
Q_2=mL_f=4\times3.3\times10^{5}=1.32\times10^{6}\,\mathrm{J}
Stage 3: heat the water from 0\,^\circ\mathrm{C} to 100\,^\circ\mathrm{C}.
Q_3=mc_{\mathrm{w}}\Delta T=4\times4200\times100=1.68\times10^{6}\,\mathrm{J}
Stage 4: boil the water at 100\,^\circ\mathrm{C}.
Q_4=mL_v=4\times2.26\times10^{6}=9.04\times10^{6}\,\mathrm{J}
Stage 5: heat the steam from 100\,^\circ\mathrm{C} to 120\,^\circ\mathrm{C}.
Q_5=mc_{\mathrm{s}}\Delta T=4\times2000\times20=1.6\times10^{5}\,\mathrm{J}
Add the five stages.
\begin{aligned}Q&=Q_1+Q_2+Q_3+Q_4+Q_5\\ &=(0.168+1.32+1.68\\ &\quad+9.04+0.16)\times10^{6}\\ &=12.368\times10^{6}\,\mathrm{J}\end{aligned}

Q\approx1.237\times10^{7}\,\mathrm{J}

Book answer. 1.236 \times 10^7 J

Note. The exact sum is 1.2368\times10^{7}\,\mathrm{J}, which rounds to 1.237\times10^{7}\,\mathrm{J}. The book's 1.236\times10^{7}\,\mathrm{J} looks truncated rather than rounded.

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 81 mark

Linear expansion in solids refers to the change in:

AMass only
BArea only
CVolume only
DLength onlyAnswer

Why

Linear thermal expansion is the change in one dimension, the length, of a solid on heating, given by \Delta L=\alpha L_0\,\Delta T (option D).

The mass does not change on heating (A). A change in area is area expansion (B), and a change in volume is volume expansion (C), which are different from linear expansion.

FBISE SSC-II 2026 Annual I, Set B, Q.2 (xi)1x3 = 3 marks

State three differences between boiling and evaporation.

Solution

Both boiling and evaporation change a liquid into vapour, but they differ as follows (book Table 11.4).

Property	Evaporation	Boiling
Where it occurs	Only at the surface of the liquid	Throughout the liquid
Temperature	At any temperature	Only at the boiling point
Bubbles	No bubbles form	Bubbles form throughout the liquid and rise

Other differences that can be written instead: evaporation is slow and boiling is fast; evaporation uses energy from the surroundings and causes cooling, while boiling needs a continuous external source of heat; evaporation depends on the surface area, boiling does not.

FBISE SSC-II 2026 Annual I, Set B, Q.52+3 = 5 marks

a. What is meant by linear expansion of solids? b. Why are gaps left between the railway tracks? Explain.

Solution

(a) What is meant by linear expansion of solids? (2 marks)

Linear thermal expansion is the change in the length (one dimension) of a solid when it is heated. When the atoms are heated they gain kinetic energy, vibrate with a larger amplitude about their mean positions and push each other slightly further apart, so the solid becomes longer.

For a rod of original length L_0 heated through \Delta T, the increase in length is directly proportional to L_0 and to \Delta T:

\Delta L=\alpha L_0\,\Delta T\qquad\Rightarrow\qquad \alpha=\dfrac{\Delta L}{L_0\,\Delta T}

where \alpha is the coefficient of linear thermal expansion: the fractional change in length per kelvin change in temperature. Its SI unit is \mathrm{K^{-1}} and it depends on the material (steel 1.2\times10^{-5}\,\mathrm{K^{-1}}, aluminium 2.4\times10^{-5}\,\mathrm{K^{-1}}).

(b) Why are gaps left between the railway tracks? Explain. (3 marks)

Reason. Metals expand when heated. A steel rail that is laid tight on a cool day would have nowhere to expand on a hot day, so the rails would push against each other with a very large force and the track would bend or buckle (and could derail a train). A small gap between the ends of consecutive rails gives room for the expansion, as in expansion joints (book section 11.2.3).

Rails with a gap between them. (a) On a cool day the gap is wide. (b) On a hot day each rail has expanded into the gap, which is now narrow (coloured), so the rails do not press on each other. The change is exaggerated.

How much room is needed. The size of the gap depends on the expansion \Delta L=\alpha L_0\,\Delta T. As an illustration, a steel rail 10\,\mathrm{m} long (\alpha=1.2\times10^{-5}\,\mathrm{K^{-1}}) that is heated through 40\,\mathrm{K} grows by

\Delta L=1.2\times10^{-5}\times10\times40=4.8\times10^{-3}\,\mathrm{m}=4.8\,\mathrm{mm}

which is small, but a force large enough to buckle the track acts if this growth is stopped. On a cold day the rails contract and the gaps become wider, so the rails do not snap or pull apart the fastenings.

FBISE SSC-II 2026 Annual I, Set D, MCQ 81 mark

Which of the following expands the most when heated?

AAll expand equally
BSolids
CLiquids
DGasesAnswer

Why

In a gas the particles are far apart and the attractive forces between them are almost zero, so a gas expands the most for the same rise in temperature (option D). In a liquid the forces are weaker than in a solid, so a liquid expands more than a solid, but still much less than a gas.

Option C (liquids) is the tempting choice only if gases are left out. The book compares solids and liquids: liquids expand more than solids because of the weaker forces. Gases have the weakest forces of all, so they expand the most. Option A is wrong because the three states do not expand equally.

FBISE SSC-II 2026 Annual I, Set D, Q.2 (xi)1x3 = 3 marks

State three factors that affect the rate of evaporation.

Solution

The rate of evaporation depends on the following factors (any three, with the reason for each).

Temperature. A higher temperature gives the molecules more kinetic energy, so more of them can escape from the surface and evaporation is faster.
Surface area. A larger surface lets more molecules escape at the same time, so evaporation is faster. This is why wet clothes are spread out to dry.
Humidity. In dry air (low humidity) the air can take in more water vapour, so evaporation is faster. In humid air it is slower.
Wind speed. Moving air blows away the moist air near the surface, so more molecules can leave and evaporation is faster.

FBISE SSC-II 2026 Annual I, Set M, MCQ 81 mark

Coefficient of volume expansion, \beta=

A\dfrac{V_o\Delta T}{\Delta V}
B\dfrac{\Delta V}{V_o\Delta T}Answer
C\dfrac{\Delta V\Delta T}{V_o}
D\dfrac{V_o}{\Delta V\Delta T}

Why

From \Delta V=\beta V_0\,\Delta T (book equation 11.3), dividing both sides by V_0\,\Delta T gives

\beta=\dfrac{\Delta V}{V_0\,\Delta T}

which is the fractional change in volume per kelvin (option B). Option A is the inverse of this. C and D put \Delta T in the wrong place, so they do not have the unit \mathrm{K^{-1}}.

FBISE SSC-II 2026 Annual I, Set M, Q.2 (xi)1x3 = 3 marks

State three differences between boiling and evaporation.

Solution

Both boiling and evaporation change a liquid into vapour, but they differ as follows (book Table 11.4).

Property	Evaporation	Boiling
Where it occurs	Only at the surface of the liquid	Throughout the liquid
Temperature	At any temperature	Only at the boiling point
Bubbles	No bubbles form	Bubbles form throughout the liquid and rise

Other differences that can be written instead: evaporation is slow and boiling is fast; evaporation uses energy from the surroundings and causes cooling, while boiling needs a continuous external source of heat; evaporation depends on the surface area, boiling does not.

FBISE SSC-II 2026 Annual I, Set M, Q.55 marks

Explain the thermal expansion of liquids [real and apparent expansion].

Solution

Expansion of liquids

Liquids expand when heated. Their particles gain kinetic energy and push each other further apart. The attractive forces between liquid particles are weaker than in solids, so liquids expand more than solids for the same rise in temperature. A liquid cannot be heated on its own, so it is heated in a container. The volume change must therefore be split into two types: real expansion and apparent expansion.

Real volume expansion

The real volume expansion is the increase in the volume of the liquid when heat is applied to it directly. Its coefficient is \gamma_r, which is the fractional increase in real volume for a rise of 1 K:

\gamma_r=\dfrac{\Delta V}{V_0\,\Delta T}

Apparent volume expansion

The apparent volume expansion is the expansion of the liquid as it appears in the container, without considering the expansion of the container. Its coefficient is \gamma_a:

\gamma_a=\dfrac{\Delta V_{\mathrm{apparent}}}{V_0\,\Delta T}

Experiment with a flask

A coloured liquid is kept in a flask with a long narrow neck (book Figure 11.8). When the flask is heated from below, the liquid level first falls from V_1 to V_2, because the glass flask expands first and its volume grows. The heat then reaches the liquid, which expands and rises from V_2 to V_3.

Liquid in a flask with a narrow neck. The levels V1, V2 and V3 are marked. The real expansion is the rise from V2 to V3, the apparent expansion is the rise from V1 to V3, and the flask expansion is the fall from V1 to V2.

The rise of the liquid seen against the glass is the apparent expansion (V1 to V3). The real expansion is the rise from V2 to V3, measured from the level after the flask has grown. The three quantities are related by

\Delta V_r=\Delta V_a+\Delta V_g

where \Delta V_g is the expansion of the flask. Dividing by V_0\,\Delta T gives the relation between the coefficients (book equation 11.9):

\gamma_r=\gamma_a+\gamma_g

So the real coefficient is always greater than the apparent coefficient, by the coefficient of the container. Different liquids have different coefficients, as Table 11.3 shows: for example \beta=113\times10^{-5}\,\mathrm{K^{-1}} for methanol and 18\times10^{-5}\,\mathrm{K^{-1}} for mercury.

Water is an exception

Almost all liquids expand on heating and contract on cooling. Water does not behave in this way between 4\,^\circ\mathrm{C} and 0\,^\circ\mathrm{C}: as it cools from 4\,^\circ\mathrm{C} to 0\,^\circ\mathrm{C} it expands, so ice floats on liquid water (book Science Tidbits).

Exam focus

Explain the thermal expansion of liquids: real and apparent expansion, the flask experiment (levels V_1, V_2, V_3) and \gamma_r=\gamma_a+\gamma_g. (long)
Set M Q.5 (5 marks) in 2026; the same as comprehensive question 6. The 5-mark long question of this unit was a thermal expansion question in two of the three sets.
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Define linear thermal expansion, write \Delta L=\alpha L_0\,\Delta T, and explain why gaps are left between railway tracks. (long)
Set B Q.5 (2+3 marks) in 2026, built on comprehensive question 4 and the expansion joints in section 11.2.3.
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State three differences between boiling and evaporation (surface or throughout, any temperature or boiling point only, bubbles). (short)
Q.2 (xi) of both Set B and Set M in 2026; also part of comprehensive question 3. Learn Table 11.4.
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State three factors that affect the rate of evaporation and give the reason for each. (short)
Set D Q.2 (xi) in 2026; also part of comprehensive question 3.
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MCQs on the thermal expansion definitions: linear expansion, \beta=\Delta V/(V_0\Delta T), and which state expands most. (mcq)
MCQ 8 in all three 2026 sets (B: linear expansion, D: which state expands most, M: the formula for \beta). Book exercise MCQs 1, 7, 11 and 13 test the same ideas.
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Numericals with \Delta L=\alpha L_0\,\Delta T, \Delta V=\beta V_0\,\Delta T and \beta=3\alpha, including a liquid in a glass flask. (numerical)
No numerical from this unit was set in 2026, but numerical questions 1 to 4 of the book use these formulas and the 2026 theory questions rest on them.
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Latent heat of fusion and of vaporisation: definitions, \Delta Q=mL_f and \Delta Q=mL_v, and the heating curve of ice. (numerical)
Not asked in the 2026 papers, but comprehensive questions 1 and 2, numerical questions 5 to 7 and conceptual questions 1 and 2 cover it, and it is the other half of the unit's SLOs.
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Applications and consequences of thermal expansion: bimetallic strip, thermometer, expansion joints, power lines. (long)
Railway gaps (Set B Q.5 b) are one such application; comprehensive questions 5 and 7 ask for the others.
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