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10.4Heat engines
It is possible (and easy) to transfer mechanical energy to thermal energy: just think of a car when the brakes are applied. Some of the carâs kinetic energy is transferred to thermal energy in the brake pads (the temperature of the brake pads increases). All of the mechanical energy can be transferred to thermal energy in this way. Is the reverse possible? In other words, can we transfer thermal energy to mechanical energy with 100% efficiency? Thermodynamics says that this is not possible, as we will see.
A heat engine is a device that transfers thermal energy to mechanical or electrical energy, which can then be used to do mechanical work.
A schematic example is shown in Figure 10.19.

Figure 10.19: In a heat engine, energy flowing from the hot to the cold reservoir may be used to perform mechanical work.

A very simple attempt at a âheat engineâ can be illustrated by the pressureâvolume diagram of an isothermal expansion, Figure 10.20.

Figure 10.20: a A gas expanding isothermally transferring thermal energy which can be used to do mechanical work. b The process is reversed, bringing the system back to its original state.

The gas is in contact with a hot reservoir (a source of thermal energy). The temperature of the source is ever so slightly higher than that of the gas and so thermal energy gets transferred to the gas that expands doing work. This is what we want: the transfer of thermal energy can be used to do mechanical work. But this is a useless engine! An engine needs to be used over and over again to keep producing work continuously. If the engine is a water pump that needs to pump water out of flooded house, we want it to work over and over again until all the water is out. So the engine must work in a cycle: we must somehow bring the state of the gas at B in Figure 10.20a back to the original state at A so we can repeat the process. One way of doing this is to reverse the expansion of that in Figure 10.20a, leading to Figure 10.20b. But this is useless too: we would have to do the same work compressing the gas as the gas did for us as it expanded. The net work in this âcycleâ would be zero! Figure 10.21 shows a better version of a useful cycle.

Figure 10.21: A cycle on a PâV diagram for a heat engine.

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A hot reservoir (a source of thermal energy) at temperature TH transfers heat into the engine. The hot reservoir could be a fire created by burning a fuel. Think of the engine as a gas in a cylinder with a piston. From A to B, the gas expands pushing the piston out, and this can be exploited to do mechanical work by, for example, raising a load, Figure 10.22. (The âloadâ could be the mass of water that we want to pump out of the flooded house.) At B the load is removed. From B to C the gas is allowed to cool down to its original temperature at constant volume. A âpinâ can lock the piston in place, keeping the volume constant. From B to C, heat is removed from the gas and is rejected into a sink of temperature TC. The sink could be the surroundings of the engine. An external force now compresses the gas isothermally from C back to the original state at A. Heat again leaves the gas and goes into the sink at temperature TC. The external force did less work than the gas did upon expanding, so we achieved a net work out of this engine. The net work done is the heat in, QH, minus the heat out, QC. Figure 10.22 shows the process just described in pictures.

Figure 10.22: The hot reservoir provides heat to the gas which expands, raising a load. The load is removed, and the gas is allowed to cool at constant volume. Heat is released to the surroundings. An external force compresses the gas back to its original state. The process can then repeat so that the transfer of thermal energy can be used to do mechanical work.

The efficiency of a heat engine is:
Î·=Â useful workÂ Â input energyÂ 
Here the input energy is QH and so,
Î·=WQH=QHâQCQH=1âQCQH.
CHECK YOURSELF 7
A heat engine has efficiency 0.40. The heat entering the engine from the hot reservoir is 500 J.
How much heat is rejected?
WORKED EXAMPLE 10.11
A heat engine has efficiency 0.20. The heat rejected into the cold reservoir is 800 J. What is the useful work done by the engine?
Answer
From 0.20=1â800QH we find QH = 1000 J. Hence W = 200 J.
A refrigerator, Figure 10.23, is another type heat engine in which the performance of mechanical work extracts heat from the cold reservoir and deposits it in the hot reservoir. (The food that is to be kept cool is the cold reservoir.)

Figure 10.23: In a refrigerator mechanical work is used to extract heat from a cold reservoir and deposit it in the hot reservoir.

Note that it impossible to operate this refrigerator without the performance of work! If that were possible we would have a situation where heat is transferred from a cold to a warmer body, and as we have seen this would violate the second law of thermodynamics.
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The Carnot cycle
In 1824 the French engineer Sadi Carnot investigated the question of how the transfer of thermal energy can be used to do mechanical work in the most efficient way. He asked if there was a limit to how efficient an engine can be. Carnot showed that there is.
Figure 10.24 shows the Carnot cycle that he investigated. It consists of two isothermals and two adiabatic curves.

Figure 10.24: The Carnot cycle consists of two isothermals and two adiabatics.

The engine starts its cycle at 1. The gas is compressed isothermally until 2. During this stage heat QC leaves the gas and work is done on the gas. From 2 to 3 it is compressed adiabatically; work is again done on the gas. From 3 to 4 the gas expands isothermally receiving heat QH from a hot reservoir; work is done by the gas. From 4 to 1 the gas expands adiabatically; work is done by the gas. The net work done by the engine is therefore
W=QHâQC
The temperature along the isothermal 3 â 4 is TH and along the lower isothermal 1 â 2 is TC. The total change in entropy of the engine is
ÎS=ÎS12+ÎS23+ÎS34+ÎS41
But the stages 2 â 3 and 4 â 1 are adiabatics: Q = 0 and so there is no change in entropy during the adiabatic stages. Along 3 â 4, ÎS34=QHTH and along 1 â 2, ÎS12=QCTC. Because the process starts and ends at A and because entropy is state function, the total entropy change is zero:
0=QHTHâQCTC which implies that QCQH=TCTH.
As we saw earlier the efficiency is Î·=1âQCQH.
But for the Carnot engine QCQH=TCTH so we find the Carnot efficiency to be:
Î·C=1âTCTH
Carnotâs engine must have an efficiency less than 100%. It would be 100% in the impossible cases of either TC = 0 or TH = â.
The importance of the Carnot engine is that no engine is more efficient than the Carnot engine operating between the same temperatures. It can be shown that this follows from the second law of thermodynamics.
(Despite this, the Carnot engine is of no practical use since it consists of isothermal processes, which are very slow to complete, and adiabatic processes, which must be very fast.)
CHECK YOURSELF 8
A heat engine removes 100 J from the hot reservoir at 600 K and rejects 30 J of energy into the cold reservoir at 400 K. Is this possible?
THEORY OF KNOWLEDGE
The argument presented in the extension material is almost like a geometrical proof argument or a logic argument. This is another way of seeing that the second law is a very special law of physics!
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EXTENSION
To see why no engine can have a higher efficiency than the Carnot efficiency, consider a hypothetical such engine connected to a refrigerator running on a Carnot cycle, Figure 10.25.

Figure 10.25: A hypothetical high efficiency engine connected to a Carnot refrigerator.

The work output (W) of the heat engine is work input (W) in the Carnot refrigerator. If the Carnot refrigerator ran as a heat engine its efficiency would be WQâ²H. Now if we assume that the engine on the left has greater efficiency that the Carnot efficiency then
WQH>WQâ²H
This implies that Qâ²H > QH. This means that Qâ²H â QH > 0. Since W = QH â QC = Qâ²H â Qâ²C it means that Qâ²C â QC = Qâ²H â QH = Q. The combination has the effect of transferring heat Q > 0 from the cold to the hot reservoir without performance of work. This violates the second law. So the assumption of an engine with better efficiency than the Carnot efficiency must be false.
NATURE OF SCIENCE
The second law of thermodynamics stands out as a special law of physics. A law such as the conservation of energy places restrictions on what can and cannot happen: A 1.0 kg ball with total energy 10 J cannot jump over a barrier that is higher than 1.0 m; therefore such an event is completely forbidden (in classical physics). The second law says that certain processes do not happen not because they really are impossible but because they are so highly unlikely so as to be impossible for all practical purposes. It is quite amazing that Carnot was able to reach his conclusions about the limitations on the efficiencies on heat engines before the concept of entropy was discovered and before the first law of thermodynamics. The connection of the second law with probability and the number of microstates that realise a particular macroscopic state make this law a truly special law of physics.However, many physicists are not happy with the probabilistic interpretation of entropy and the second law. Recent work appears to have interpreted the second law not in terms of what is likely and unlikely but in terms of deeper quantum mechanical principles such as quantum entanglement. See the article Physicists Rewrite the Fundamental Law That Leads to Disorder, in Quanta magazine, May 27, 2022.
Links
The idea of the âheat deathâ of the universe goes back to Kelvin. If we could consider the universe as a closed system to which the laws of thermodynamics could be applied, then as the universe gets older and older we are driven to a state of ever increasing entropy and disorder. As we approach thermodynamic equilibrium temperature differences disappear. With no energy differences there can be no transfer of energy and no work done. The universe would be an uninteresting and boring place in which nothing happens. Estimates give us a time of about 10100 years until we get there, which is a very long time. Not everyone agrees with this reasoning, arguing that it is by no means clear if the laws of thermodynamics can be applied to the universe as a whole or whether the concept of entropy as a whole has any meaning when it is applied to the whole universe. For a heretical view of how this eternal boredom can be avoided, see the article Time without an end, by Freeman Dyson in Rev. Mod. Phys. 51, 447.
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TEST YOUR UNDERSTANDING
33Figure 10.26 shows the cycles of three engines.

Figure 10.26: For question 33.

Rank the engines in order of increasing efficiency.
34Explain whether you should invest money in a revolutionary new design of a heat engine that claims to operate between temperatures of 300 K and 500 K with an efficiency of 0.42.
35Figure 10.27 shows three heat engines. Which are possible and which are impossible?

Figure 10.27: For question 35.

36Figure 10.28 shows the thermodynamic cycle of a heat engine working with an ideal gas. A quantity of thermal energy is provided to the gas.
aIdentify the part of the cycle where this energy is absorbed.
bCalculate the efficiency of this engine.

Figure 10.28: For question 36.

37aState the second law of thermodynamics.
bExplain, by reference to a Carnot engine, why a heat engine operating in a cycle must reject energy into the surroundings.
38The graph (Figure 10.29) shows an isothermal expansion, followed by an isovolumetric change and finally an adiabatic compression back to the initial state of an ideal gas. The work done by the gas during the isothermal expansion is 900 J, and the temperature of the isothermal curve is 300 K.

Figure 10.29: For question 38.

What is the entropy change of the gas during:
athe isothermal expansion?
bthe isovolumetric change?
39Figure 10.30 shows a cycle ABCDA of an ideal gas. Curves AB and CD are adiabatic.

Figure 10.30: For question 39.

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aState what is meant by an adiabatic curve.
bState the name of the processes BC and DA.
cDetermine along which stage heat is given to the gas.
dAlong DA the pressure drops from 4.0 Ã 106 Pa to 1.4 Ã 106 Pa. The volume along DA is 8.6 Ã 10â3 m3. The efficiency of the engine is 0.36. Calculate:
ithe heat taken out of the gas
iithe heat given to the gas
iiithe area of the loop ABCDA.
40Figure 10.31 shows the isothermal compression of an ideal gas from state 1 of volume V1 to state 2 of volume V2.

Figure 10.31: For question 40.

aOn the axes in the figure draw a curve to show the adiabatic compression of the same gas from state 1 to a state of volume V2.
bExplain in which case the work done on the gas is the largest.
cThe temperature of the gas in state 1 is 300 K. The work done on the gas during the isothermal compression is 25 kJ. Determine the change in entropy of the gas during the isothermal compression.
41A heat engine has an efficiency of 0.25. In one cycle, 300 J of work is performed.
aWhat is the heat input to the engine?
bHow much heat is expelled to the cold reservoir?
42A Carnot engine has an efficiency of 0.30. The engine receives 900 J of energy from a hot reservoir at 600 K.
aHow much heat is expelled to the cold reservoir?
bWhat is the temperature of the cold reservoir?
43An isothermal compression is said to be a reversible process.
aWhat is the change in the entropy of the universe during a reversible isothermal compression?
bHow is the answer to part a reconciled with the fact that the entropy of the gas decreases?
cSuggest why for this reversible process the temperature of the surroundings must be infinitesimally lower than that of the gas.
dHence deduce why the isothermal compression must be conducted very slowly.
