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7.2Specific heat capacity and change of phase
Calorimetry
KEY POINT
We define the specific heat capacity c of a body to be the energy required to increase the temperature of a unit mass of the body by one kelvin.
So, to increase the temperature of a body of mass m by ÎT degrees the heat Q required is:
Q=Â mcÎT
EXAM TIP
The term âcapacityâ implies that the body somehow contains a certain amount of thermal energy, just as a water bottle contains water. This is incorrect. Heat is thermal energy âin transitâ that moves from one body into another; it is not energy contained in any one body.
Table 7.1 lists the specific heat capacity of some common substances.
Substance
c / J kgâ1 Kâ1
aluminium
900
lead
128
iron
440
copper
385
silver
240
water
4200
ice
2100
ethanol
2430
marble
880
Table 7.1: Specific heat capacities for several substances.
CHECK YOURSELF 3
Two bodies of equal mass are in thermal contact. X has temperature 0 Â°C and Y 100 Â°C. X has smaller specific heat capacity than Y. Will the final temperature of the two bodies after equilibrium is reached be less than, equal to or greater than 50 Â°C?
WORKED EXAMPLE 7.2
A quantity of heat equal to 9800 J is absorbed by a piece of iron of mass 0.50 kg and specific heat capacity 440 J kgâ1 Kâ1.
aCalculate the temperature increase of the iron.
b9800 J of thermal energy was removed from 0.50 kg of water initially at 48 Â°C. The specific heat capacity of water is 4200 J kgâ1 Kâ1. Calculate the final temperature of the water.
Answer
aWe need to use Q = mcÎT. This gives:
9800=0.50Ã440ÃÎT
Solving for the change in temperature gives:
ÎT=98000.50Ã440=44.6â45Â K
(Notice that we do not need to know the initial temperature of the iron to answer this question.)
bWe use Q = mcÎT to get:
9800=0.50Ã4200ÃÎT
Solving for the change in temperature gives:
ÎT=98000.50Ã4200=4.7â5Â K
So the final temperature of the water is:
48â5=43Â°C
(Notice that the temperature changes of the iron and the water are very different. Notice also that it is unnecessary to convert between kelvin and Â°C since the temperature changes are the same on both scales.)
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WORKED EXAMPLE 7.3
A piece of iron of mass 200 g and temperature 300 Â°C is dropped into 1.00 kg of water of temperature 20 Â°C. Predict the final equilibrium temperature of the water.
(Take c for iron as 440 J kgâ1 Kâ1 and for water as 4200 J kgâ1 Kâ1.)
Answer
Let T be the final unknown temperature. The iron will also be at this temperature, so:
amount of thermal energy lost by the iron = mironciron(300 â T)
and
amount of thermal energy gained by the water = mwatercwater(T â 20)
Conservation of energy demands that thermal energy lost = thermal energy gained, so:
mironÂ cironÂ (300âT)=mwaterÂ cwaterÂ (Tâ20)
0.200Ã440Ã(300âT)=1.0Ã4200Ã(Tâ20)
âT=25.7Â°Câ26Â°C
(Note how the large specific heat capacity of water (and the larger mass) results in a small increase in the temperature of the water compared with the huge drop in the temperature of the iron.)
Change of phase
When thermal energy is absorbed by a body or transferred away from a body, the body may not necessarily change its temperature. The body may change phase instead. Changes of phase happen at constant temperature (Figure 7.5) and include:
âmelting: when a solid changes to a liquid (thermal energy is transferred to the solid)
âfreezing: when a liquid changes into a solid (thermal energy is transferred away from the liquid)
âvaporisation (or boiling): when a liquid changes into vapour (thermal energy is transferred to the liquid)
âcondensation: when a vapour changes into a liquid (thermal energy is transferred away from the vapour).
âevaporation: vaporisation taking place at the surface of a liquid at any temperature

Figure 7.5: Hot lava turns into a solid upon contact with water. Thermal energy transfers from the hot lava to the cold water.

The rate of evaporation increases with increasing liquid temperature and increased surface area. It is the fastest molecules that leave the liquid surface and so the average kinetic energy of the remaining liquid molecules decreases; this means the liquid cools.
The energy supplied or removed during a phase change is called latent heat (because it does not result in a temperature change).
KEY POINT
Specific latent heat, L, is the energy required to change the phase of 1 kg of a substance at constant temperature and pressure.
So, the energy required to change the phase of a mass m is Q = mL. If the change is melting or freezing, we call it the specific latent heat of fusion, LF. If the change is vaporisation or condensation then we call it specific latent heat of vaporisation, LV.
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Notice from Table 7.2 that the specific latent heat for vaporisation is greater than that for melting. This is because the increase in separation of the molecules is much larger when going from the liquid to the vapour phase than when going from the solid to the liquid phase. More work is needed to achieve the greater separation, and so more energy is required.
Substance
Specific latent heat of fusion / kJ kgâ1
Melting temperature / Â°C
Specific latent heat of vaporisation / kJ kgâ1
Boiling temperature / Â°C
water
334
0
2260
100
ethanol
109
â114
840
78
aluminium
395
660
10550
2467
lead
23
327
850
1740
copper
205
1078
2600
5190
iron
275
1540
6300
2800
Table 7.2: Specific latent heats of fusion and vaporisation together with the melting and boiling temperatures.
CHECK YOURSELF 4
Why is contact of the skin with steam at 100 Â°C more dangerous than contact with boiling water at 100 Â°C?
WORKED EXAMPLE 7.4
An ice cube of mass 25.0 g and temperature â10.0 Â°C is dropped into water of mass 300.0 g and temperature 20.0 Â°C. Calculate the final temperature.
(Specific heat capacity of ice = 2100 J kgâ1 Kâ1; specific latent heat of fusion of ice = 334 kJ kgâ1, specific heat capacity of water = 4200 J kgâ1 Kâ1.)
Answer
Let the final temperature be T. The water will cool down by transferring thermal energy away.
Using Q = mcÎT, the thermal energy dissipated by the water is:
0.300Ã4200Ã(20âT)
This thermal energy will be absorbed by the ice to:
âincrease its temperature from â10 Â°C to 0 Â°C: the thermal energy required is 25 Ã 10â3 Ã 2100 Ã 10 J
âmelt the ice cube into water at 0 Â°C: the thermal energy required is 25 Ã 10â3 Ã 334 Ã 103 J
âincrease the temperature of the former ice cube from 0 Â°C to the final temperature T: the thermal energy required is 25 Ã 10â3 Ã 4200 Ã T.
Thus:
0.3Ã4200Ã(20âT) = (25Ã10â3Ã2100Ã10) + (25Ã10â3Ã334Ã103) + (25Ã10â3Ã4200ÃT)
Solving for T gives T = 14 Â°C.
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EXAM TIP
You can save yourself time and possible errors if you write this equation, as is, in the equation solver of your graphic display calculator (GDC) and ask the GDC to solve it for you.
WORKED EXAMPLE 7.5
A sample of 120 g of a solid initially at 20 Â°C is heated by a heater of constant power. The specific heat capacity of the solid is 2500 J kgâ1 Kâ1. The temperature of the sample varies with time as shown in Figure 7.6.
Use the graph to determine:
athe power of the heater
bthe melting temperature of the sample
cthe specific latent heat of fusion of the sample
dthe specific heat capacity of the sample in the liquid phase.

Figure 7.6: For worked example 7.5.

Answer
aIt takes 120 s to raise the temperature of the solid sample from 20 Â°C to 48 Â°C.
Using Q = mcÎT, the thermal energy required is:
0.120Ã2500Ã(48â20)=8400Â J
So the power is:
P=Qt=8400120=70Â W
bThe temperature is constant at melting, shown by the flat part of the graph, so the melting temperature is 48 Â°C.
cThe sample is melting from 120 s to 560 s, i.e. for 440 s. The thermal energy supplied during this time is therefore:
Q=Pt=70Ã440=30 800Â J
So the specific latent heat of fusion is:
LF=Qm=30 8000.120=2.6Ã105Â JÂ kgâ1
dThe liquid increases its temperature from 48 Â°C to 56 Â°C in 40 s. In these 40 s the thermal energy provided is:
Q=Pt=70Ã40=2800Â J
Using Q = mcÎT:
0.120ÃcÃ(56â48)=2800Â J
Hence, c=2.9Ã103Â JÂ kgâ1Â Kâ1.
We already mentioned that a phase change occurs at constant temperature. Ice, for example, melts at 0 Â°C. For a mass m of ice at 0 Â°C we need to supply an energy mL which turns the ice into liquid water without a change in temperature. How do we explain this observation? At the melting temperature, changing from a solid to a liquid means that the average distance between the molecules increases. Increasing the separation of the molecules requires work (because there are attractive forces between the molecules that need to be overcome). The thermal energy supplied increases the intermolecular potential energy (recall the graph in Figure 7.3) and not the kinetic energy of the molecules. So the temperature stays the same but the internal energy increases since the potential energy increases. It is an interesting question, of course, to ask what if anything prevents energy from going into kinetic energy during a phase change.
Consider thermal energy provided to melting ice and water at 0 Â°C. Microscopically, some thermal energy will inevitably go to a small quantity of water raising its temperature slightly above zero. This slightly warmer water is surrounded by ice and water at 0 Â°C, and so thermal energy will now be transferred away from this warmer water reducing its temperature back down to 0 Â°C. If this thermal energy is transferred to ice, it will cause the ice to melt. If thermal energy transfers back into water at 0 Â°C, the process repeats. The result is that more and more ice melts and the overall equilibrium temperature stays constant at 0 Â°C.
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The method of mixtures
The electrical method described in worked example 7.5 is one method for measuring specific heat capacity and latent heat. Another method, the method of mixtures, measures the specific heat capacity of a solid as follows. A solid is put in a container of hot water and allowed time to reach a constant temperature. The temperature of the solid is thus that of the water and is recorded. The solid is then transferred into a calorimeter of known specific heat capacity and initial temperature which contains a liquid such as water (Figure 7.7). The calorimeter is insulated. The final temperature of the water is recorded after thermal equilibrium has been reached. Thermal equilibrium means that the temperatures of the bodies involved are the same.

Figure 7.7: The hot metal is placed in the cold water in the calorimeter. The hot metal is removed from the container of boiling water and is quickly placed inside an insulated calorimeter containing cold water.

For example, consider a mass of 0.400 kg of a solid at 80 Â°C that is put in a 100 g copper calorimeter containing 800 g of water at 20 Â°C. The final temperature of the water is measured to be 22 Â°C. From these values, we may deduce the specific heat capacity of the solid as follows.
Using Q = mcÎT, the amount of thermal energy (in joules) lost by the solid is:
0.400ÃcÃ(80â22)=23.2c
The amount of thermal energy gained by the calorimeter (see Table 7.1 for the value of c for copper) and the water is:
0.100Ã385Ã(22â20)îîîîîîîîîîîîîîîîîîîîîîîîîîîîcalorimeterÂ +0.800Ã4200Ã(22â20)îîîîîîîîîîîîîîîîîîîîîîîîîîîîwaterÂ =6797Â J
Equating this to the thermal energy 23.2c lost by the solid, we find that c = 293 J kgâ1 Kâ1.
EXAM TIP
It is likely that some thermal energy (heat) was dissipated to the surrounding air while it was being transferred. This means that the actual temperature of the solid is less than we supposed. The actual specific heat capacity is, therefore, larger than the calculated value.
The same method can be applied to measure the specific latent heat of fusion of ice. To do this, place a quantity of ice at 0 Â°C (the ice must therefore come from a mixture with water at 0 Â°C) into a calorimeter containing water at a few degrees above room temperature. Blot the ice dry before putting it into the calorimeter. The mass of the ice can be determined by weighing the calorimeter at the end of the experiment.
For example, suppose that 25.0 g of ice at 0.00 Â°C is placed in an aluminium calorimeter of mass 250 g containing 300 g of water at 24.0 Â°C. The temperature of the water is measured at regular intervals of time until the temperature reaches its minimum value of 17.0 Â°C. The calorimeter and water lost thermal energy, which the ice gained.
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Thermal energy transferred away from calorimeter and water:
0.250Ã900Ã(24â17)+0.300Ã4200Ã(24â17)=10 395Â J
Thermal energy transferred to the ice:
0.025ÃLF+0.025Ã4200Ã17=0.025ÃLF+1785
Equating the two gives:
1785+0.025ÃLF=10 395âLâ344Â kJÂ kgâ1
NATURE OF SCIENCE
Models change
As already mentioned, heat was once thought to be a fluid (caloric). Conservation of energy was a natural consequence of this model of heat: a body lost a certain amount of fluid and another gained it. Energy was conserved. So the concept of heat as a fluid seemed natural. But there are phenomena that cannot be explained with this simple picture. For one thing, if heat is a fluid it must have mass. So when heat leaves a body, the body must lose mass. This is not observed, so the caloric theory must be wrong. The theory has many other failings and was abandoned in the nineteenth century. A major problem is that it does not take into account the atomic theory of matter. The theory we use now is based on statistical mechanics, which uses probability theory to predict the average behaviour of very large numbers of particles.
Cooling
A hot liquid placed in colder surroundings will cool down by losing thermal energy to the surroundings until its temperature reaches that of the surroundings. Experiments show that the rate of cooling, i.e. the rate at which the temperature decreases, is proportional to the difference in temperature between the liquid and the surroundings: ÎTÎt=k(TâTs) where the constant k is proportional to the area of the surface of the liquid. This implies an exponential decrease of the temperature. The red curve in Figure 7.8 shows a liquid of initial temperature 40 Â°C which approaches the surroundings temperature of 10 Â°C as time goes by. The orange curve shows an identical liquid whose surface area is less than that corresponding to the red curve.

Figure 7.8: The temperature (red and orange curves) decreases exponentially, approaching the temperature of the surroundings (blue curve).

TEST YOUR UNDERSTANDING
8A hot body is brought into contact with a colder body until their temperatures are the same. Assume that no other bodies are nearby.
aDiscuss whether the energy lost by one body is equal to the energy gained by the other.
bDiscuss whether the temperature drop of one body is equal to the temperature rise of the other.
9aA body of mass 0.150 kg has its temperature increased by 5.00 Â°C when 385 J of energy is provided to it. Calculate the bodyâs specific heat capacity.
bAnother body of mass 0.150 kg has its temperature increased by 5.00 K when 385 J of energy is provided to it. Calculate this bodyâs specific heat capacity.
10A calorimeter of mass 90 g and specific heat capacity 420 J kgâ1 Kâ1 contains 310 g of a liquid at 15.0 Â°C. An electric heater rated at 20.0 W warms the liquid to 19.0 Â°C in 3.0 min. Assuming there are no energy losses to the surroundings, estimate the specific heat capacity of the liquid.
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11A calorimeter for which mc = 25 J Kâ1 contains 140 g of a liquid. An immersion heater is used to provide energy at a rate of 40 W for a total time of 4.0 min. The temperature of the liquid increases by 15.8 Â°C. Calculate the specific heat capacity of the liquid. State an assumption made in reaching this result.
12A car of mass 1360 kg descends from a hill of height 86 m at a constant speed. Assuming that all of the gravitational potential energy lost by the car goes into heating the brakes, estimate the rise in the temperature of the brakes. (It takes 16 kJ of energy to increase the temperature of the brake drums by 1 K; ignore any energy losses to the surroundings.)
13A radiator made out of a metal of specific heat capacity 450 J kgâ1 Kâ1 has a mass of 45.0 kg and is filled with 23.0 kg of water of specific heat capacity 4200 J kgâ1 Kâ1.
aDetermine the energy required to raise the temperature of the radiatorâwater system by 1 K.
bIf energy is provided to the radiator at the rate of 450 W, calculate how long it will take for the temperature to increase by 20.0 Â°C.
14How much ice at â10 Â°C must be dropped into an aluminium cup containing 300 g of water at 20 Â°C in order for the temperature of the water to be reduced to 10 Â°C? The cup itself has a mass of 150 g and specific heat capacity 900 J kgâ1 Kâ1. Assume that no energy is lost to the surroundings.
15The surface of a pond of area 20 m2 is covered by ice of uniform thickness 6.0 cm. The temperature of the ice is â5.0 Â°C. Calculate how much energy is required to melt this amount of ice into water at 0 Â°C. (Take the density of ice to be 900 kg mâ3.)
16Radiation from the sun falls on the frozen surface of a pond at a rate of 600 W mâ2. The ice temperature is 0 Â°C.
aCalculate how long it will take to melt a 1.0 cm thick layer of ice. (Take the density of ice to be 900 kg mâ3.)
bSuggest why the actual mass of ice that melts is less than your answer to a.
17aCalculate how much energy is required to warm 1.0 kg of ice initially at â10 Â°C to ice at 0 Â°C.
bCalculate how much energy is needed to melt the ice at 0 Â°C.
cCalculate how much energy is required to further increase the temperature of the water from 0 Â°C to 10 Â°C.
18Ice at 0 Â°C is added to 1.0 kg of water at 20 Â°C, cooling it down to 10 Â°C. Determine how much ice was added.
19A quantity of 100 g of ice at 0 Â°C and 50 g of steam at 100 Â°C are added to a container that has 150 g water at 30 Â°C. Determine the final temperature. Ignore the container itself in your calculations.
20Steam at 100 Â°C is mixed with 200 g of ice at 0 Â°C to produce water at 50 Â°C. What is the mass of steam required?
21Two bodies (same mass and initial temperature) are dropped into two identical containers filled with water. The mass and temperature of the water in both containers is the same. The graphs in Figure 7.9 show the cooling curves of the two bodies.

Figure 7.9: For question 21.

Which body has the higher specific heat capacity?
