1.5: Heat Engines

Introduction

Heat and work are both forms of transferring energy, and under the right circumstance, one form may be transformed into the other. However, the second law of thermodynamics puts a limitation on this. To go from work to heat is called dissipation and there is no limitation on this at all. In fact it was through dissipation (by friction) that we discovered that heat and work were both forms of energy. There is, however, a limitation on converting heat to work. In this section we will begin by considering the theoretical Carnot cycle which uses only reversible processes. From past analyses we know that reversible processes yield the maximum work possible from a gas expansion. Thus, the efficiency of the Carnot cycle is the upper limit to the efficiency of conversion of heat into work.

The Carnot cycle

The idealized Carnot cycle involves four steps as illustrated in figure \(\PageIndex{1}\).

	\(\Rightarrow\)	\(\Downarrow\)
\(\Uparrow\)	\(\Leftarrow\)

Figure \(\PageIndex{1}\): The four reversible steps in the Carnot cycle. Step 1 starting at high pressure and low volume (A) and expanding isothermally (heat in from the hot reservoir) to a lower pressure and larger volume (B). Step 2 starting at (B) and expanding adiabatically (no thermal contact with either reservoir) to an even lower pressure and larger volume (C). Step 3 starting at (C) and being compressed isothermally (heat out to the cold reservoir) to a higher pressure and lower volume (D). Step 4 starting at (D) being compressed back to the initial pressure and volume (A) adiabatically (no thermal contact with reservoirs). Figures courtesy of BlyumJ, CC BY-SA 4.0 , via Wikimedia Commons.

One way to imagine the heat exchange happening in the process is to imagine that the piston containing the working gas is separated from the cold and hot reservoirs by barriers that can shift from insulating to heat conducting frictionlessly. A realistic physical approximation to this would be movable barriers where one part is insulating foam and another a heat conductor such as copper metal. 

This idealized cycle is shown on a PV diagram in figure \(\PageIndex{2}\).

Figure \(\PageIndex{2}\): PV diagram for the Carnot cycle.

We can analyze the heat total heat and work of this cycle assuming an ideal gas.This cycle forms the 4-stage Carnot cycle heat engine. The cycle consists of:

1. Isothermal expansion at the hot temperature, \(T_h\). As we have shown for an ideal gas an isothermal change leads to zero change in internal energy. Thus: \[\Delta U_1=w_1+q_h=0. \nonumber \]
2. During adiabatic cooling from \(T_h\) to \(T_c\) no heat is exchanged. Thus: \[\Delta U_2=w_2.\nonumber \]
3. Isothermal compression at the cold temperature, \(T_c\): \[\Delta U_3=w_3+q_c=0 \nonumber \]
4. Adiabatic heating from \(T_c\) to \(T_h\): \[\Delta U_4=w_4 \nonumber \]

The total four-step process produces work because \(w_{hot} \gt w_{cold}\), where \(w_{hot} = w_1 + w_2\) and \(w_{cold} = w_3 + w_4\). The work is the integral under the upper curves minus the lower curves, i.e.the surface area in between.

Sadi Carnot

Sadi Carnot was a French engineer at the beginning of the 19th century. He considered a cyclic process involving a cylinder filled with gas. This cycle the Carnot cycle contributed greatly to the development of thermodynamics and the improvement of the steam engine. Carnot demonstrated that the cold temperature on the right is as important as the heat source on the left in defining the possible efficiency of a heat engine

Figure 20.7.2 : Animation of a typical vertical triple-expansion engine. (CC SA-BY 2.5; Emoscopes).

Efficiency

Of course we spend good money on the fuel to start the cycle by heating things up. So how much work do we get for the heat we put in? In other words, we want to know how efficient our heat engine is. The efficiency, \(\eta\) of a heat engine is:

\[\eta=\frac{|w_\text{cycle}|}{q_h}=\frac{q_h+q_c}{q_h}=1+\frac{q_c}{q_h} \nonumber \]

To get the work of the cycle, we can make use of internal energy as a state function. As the path is circular the circular integrals for \(U\) is zero:

\[\begin{align*} \oint{dU} &= \Delta U_{\text{cycle}} \\[4pt] &=\sum{\Delta U_i} \\[4pt] &=w_1+q_h+w_2+w_3+q_c+w_4 \\[4pt] &=0 \end{align*}  \nonumber \]

Rearranging:

\[\begin{align*} q_h + q_c &=-w_1-w_2-w_3-w_4 \\[4pt] &=-w_\text{cycle} \end{align*}  \nonumber \]

An ideal engine would take \(q_h\rightarrow q_c\) with 100% efficiency. The work of the cycle will be equivalent to the heat transfer. For ideal gases:

1. \(w_1=-RT_h\ln{\left(\frac{V_B}{V_A}\right)}=-q_h\)
2. \(dU=\delta w=C_VdT\rightarrow w_2=C_V\left(T_c-T_h\right)\)
3. \(w_3=-RT_c\ln{\left(\frac{V_D}{V_C}\right)}=-q_c\)
4. \(w_4=C_V\left(T_h-T_c\right)\)

Finding an expression for \(w_\text{cycle}\):

\[\begin{split} w_\text{cycle} &= -RT_h\ln{\left(\frac{V_B}{V_A}\right)}+C_V\left(T_c-T_h\right)-RT_c\ln{\left(\frac{V_D}{V_C}\right)}+C_V\left(T_h-T_c\right) \\ &= -RT_h\ln{\left(\frac{V_B}{V_A}\right)}-RT_c\ln{\left(\frac{V_D}{V_C}\right)}= -R(T_h-T_c)\ln{\left(\frac{V_B}{V_A}\right)} \end{split} \nonumber \]

The last simplification step used the fact that (V_B/V_A) = (V_C/V_D) = (V_D/V_C)^-1 because the two adiabats involve changes between the same high and low temperatures plus our gas is ideal making \(P_c V_c^\gamma = P_h V_h^\gamma\).

We have an expression for work, so we can evaluate the efficiency, \(\eta\). The efficiency of the Carnot engine is:

\[\eta=\frac{|w_\text{cycle}|}{q_h}=\frac{R\left(T_h-T_c\right)\ln{\left(\frac{V_B}{V_A}\right)}}{RT_h\ln{\left(\frac{V_B}{V_A}\right)}}=\frac{T_h-T_c}{T_h}=1-\frac{T_c}{T_h} \nonumber \]

Paths (2) and (4) are adiabats, so we can also use entropy, \(S\), to get the same solution:

\[\oint{dS}= \frac{q_h}{T_h}+\frac{q_c}{T_c}=0 \nonumber \]

Therefore:

\[ \dfrac{q_c}{q_h} = -\dfrac{T_c}{T_h} \nonumber \]

And we get that:

\[η= 1+ \dfrac{q_c}{q_h} = 1-\dfrac{T_c}{T_h} \label{eff} \]

As you see we can only get full efficiency if \(T_{cold}\) is 0 K, which is never (i.e., we always waste energy). Another implication is that if \(T_c = T_h\) then no work can be obtained, no matter how much energy is available in the from of heat. Or in other words, if one dissipates work into heat isothermally, none of it can be retrieved. Equation \(\ref{eff}\) is not very forgiving at all. Imagine that you have a heat source of 400 K (a superheated pool of water, e.g. a geyser) and you are dumping the waste heat into a river near room temperature at 300 K. The best efficiency you'll ever get is:

\[η= 1-\dfrac{300}{400}= 25\% \nonumber \]

Sadly, you'd be dumping three quarters of your energy as heat in the river (and that is best case scenario as there are always more losses, e.g. due to friction). 

This is another illustration of the second law of thermodynamics. For a process to move forward entropy must be created in the Universe. In the case of heat engines this means that some of the heat must end up as waste heat in the Universe. Thus heat cannot be converted to work with 100% efficiency.

Heat pumps

So getting our work from heat is hard and always less than 100% successful. The other way around should be easy. After all, we can dissipate work into heat freely even under isothermal conditions!

What happens if we let the heat engine run backwards? Consider reversing all the flows in the above diagram. Obviously we must put in work to make the cycle run in reverse. The heat will now flow from cold to hot, say from your cold garden into your nice and warm apartment. The amount of heat you get in your humble abode will be the sum of all the work (say 100 Joules) you dissipate plus the heat you pumped out of the garden (say 300 Joules). Thus if you are willing to pay for the energy you dissipate (100 Joules), you may well end up with a total of 400 Joules of heat in your apartment! Obviously if it is heat you are after, this is a better deal than just dissipating the work in your apartment (by burning some oil). Then you'd only get 100 J for your precious buck.

Dissipating it as electrical heating is even worse because you would:

1. first burn (a lot more!) oil to generate heat;
2. use this heat to produce electrical work at great expense because a lot of the heats gets dumped at the low temperature side;
3. dissipate the work again in your apartment (without using it to pump any heat out of the garden).

Refrigerators are also heat pumps. They heat the kitchen by pumping heat from its innards to the kitchen. If I keep the door open, however, all it does is dissipate precious electrical work, because the pumped heat will flow back into its innards and spoil the milk.

More realistic cycles

Stirling

Let's consider another type of heat engine, the Stirling engine. The Stirling engine uses a circular reversible path in an ideal PV diagram:

The path consists of four steps:

1. Isothermal expansion at the hot temperature \(T_h\): \[q_1=q_h =-w \nonumber \] and we get work out (i.e., negative work): \[w_h =-RT_h \ln \left( \dfrac{V_B}{V_A} \right) \nonumber \]
2. Isochoric cooling from \(T_h\) to \(T_c\) (with constant \(C_V\)): \[q_2= \dfrac{C_V}{ΔT} \label{q2} \]
3. Isothermal compression at the cold temperature \(T_c\): \[q_3=q_c =-w \nonumber \] and we must put work in (i.e., positive work): \[w_c =-RT_c \ln \left( \dfrac{V_D}{V_C}\right)=RT_c \ln \left( \dfrac{V_B}{V_A} \right) \nonumber \]
4. Isochoric heating from \(T_c\) to \(T_h\) (with constant \(C_V\)): \[q_4= -\dfrac{C_V}{ΔT} \label{q4} \]

Notice that this gray area vanishes if \(T_{h}= T_{c}\). Obviously how cold the cold side is of great importance! The amount of work is also equal to the difference in the heat picked up at high temperature \(q_h\) and dumped at low temperature \(q_{c}\). The isochoric heats cancel. The problem is that \(q_{c}\) is only zero if the cold temperature is 0 K. That means that we can never get all the heat we pick up at high temperatures to come out as work.

Otto

This cycle is quite close to what occurs in an internal combustion automobile engine. The cycle starts with the compressed fuel-air mixture at P_A, V_A and T_A.

1. The fuel is then ignited and assumed to heat at constant volume to T_h.
2. The heated gas then expands adiabatically (isentropicly) to V_B, P_B and T_B.
3. The gas then cools to T_C while V_B is maintained. This last step is a simplification of exhausting of the spent combustion products and replacement by a new fuel-air mixture, which usually requires an additional piston cycle to force out the exhaust and suck in the fresh fuel-air mixture.
4. The fuel-air mixture is adiabatically compressed to the initial P_A, V_A and T_A.

Analysis of the cycle shows that the efficiency is equal to:

\[η=  1 - \left(\frac{V_A}{V_B}\right)^{(C_P - C_V)\div C_V} = 1 - \left(\frac{V_A}{V_B}\right)^{\gamma -1}. \label{ottoeff}\]

This is why high compression ratio engines are desirable. The higher the compression ratio the higher T_h and thus the higher the efficiency. The ratio is limited by the available materials for constructing the engine and the properties of the fuel used. Typical modern high compression engines operate with compression ratios 12:1 (V_B:V_A) or slightly greater.¹ This limits their thermodynamic maximum efficiency. Assume the best case scenario where the working gas is an ideal monatomic gas (no internal degrees of freedom to absorb the energy, so it all ends up a kinetic energy generating pressure). For ideal gases, \(C_P = (C_V + nR)\). For a monatomic ideal gas \(C_V = \frac{3}{2}nR\), so :

\[η=1 - \left(\frac{1}{12}\right)^{\frac{2}{3}} =0.81.\nonumber\]

This looks pretty good. But there are two problems: 1) most of the gases are at least diatomics with ideal values of \(C_V= \frac{5}{2}nR\) or nonlinear polyatomics with values of \(C_V=3nR\); 2) there are additional frictional and mechanical losses. Assuming ideal polyatomic molecules we would expect a maximum efficiency of:

\[η=1 - \left(\frac{1}{12}\right)^{\frac{1}{3}} =0.56.\nonumber\]

Actual automobiles achieve in the 20 - 30% range of conversion of thermal energy to vehicle motion. This compares very poorly with the round trip efficiency of 80+% when storing solar energy in batteries and then using it to drive electric motors in an EV. This is why even using electricity generated by burning fossil fuels in modern electric generation facilities, which can get thermal efficiencies close to 60%,² EV's are more energy efficient than ICE automobiles.

Under ideal operating conditions where the adiabatic expansion in step 2 cools the gases to T_C, so no additional heat is wasted in cooling the gases to T_C in step 3 (a "warmed up" engine should be close), you can that the efficiency depends only on the minimum T_C exhaust gas temperature and the maximum T_h of the gases, just as for the Carnot cycle:

\[η(\text{otto ideal})= 1-\dfrac{T_c}{T_h} \label{otto_ideal}. \]

References

1. "Compression ratio", Wikipedia.org. https://en.wikipedia.org/wiki/Compression_ratio. Accessed Aug. 31, 2025.

2. "Combined cycle power plant", Wikipedia.org. https://en.wikipedia.org/wiki/Combin...le_power_plant. Accessed Aug. 31, 2025.

Contributors

• Jerry LaRue
• Jonathan Gutow, UW Oshkosh