Whether a heat engine is reversible or irreversible is a question we can answer by considering the change in entropy. The second law of thermodynamics tells us that $\text{Δ}{S}_{\text{universe}}\ge 0$, with equality only in the case of a reversible process. Let’s apply this idea to a heat engine.

A heat engine consists of the system itself plus the hot and cold reservoirs. During one complete cycle of a heat engine, the total entropy changes by

where $\text{Δ}{S}_{\text{H}}$ and $\text{Δ}{S}_{\text{C}}$ are the entropy changes of the reservoirs. Entropy is a state variable that depends on only the state of the gas. Regardless of what happens during a cycle, a heat engine returns to the same state and thus to the same entropy at the end of each cycle. Thus $\text{Δ}{S}_{\text{system}}=0$ over the course of a complete cycle.

The reservoirs, by definition, are so large that their temperatures don’t change. The removal of heat energy from the hot reservoir and the addition of heat energy to the cold reservoir are both isothermal processes, and we found in 《 SECTION 20.9 that

$\text{Δ}{S}_{\text{H}}$ is negative because $Q_{\text{H}}$ is the quantity of heat withdrawn from the hot reservoir. Thus the total entropy change over one cycle is

All real heat engines have $\text{Δ}{S}_{\text{universe}}>0$, which makes them irreversible; that is, running the cycle backward would not restore the reservoirs to their initial conditions. An entropy change greater than zero means that $Q_{\text{C}}/T_{\text{C}}>Q_{\text{H}}/T_{\text{H}}$. Making the engine more efficient by increasing $W_{\text{out}}$ while reducing $Q_{\text{C}}$ reduces the entropy change, but $\text{Δ}{S}_{\text{universe}}$ can’t go below zero. The efficiency that reduces $\text{Δ}{S}_{\text{universe}}$ to zero is the maximum possible efficiency, and it occurs for a perfectly reversible engine—a Carnot engine.

$\text{Δ}{S}_{\text{universe}}$ is zero if $Q_{\text{C}}/T_{\text{C}}=Q_{\text{H}}/T_{\text{H}}$ or, equivalently,

The efficiency of any heat engine was found to be $\eta =1-Q_{\text{C}}/Q_{\text{H}}$, so the efficiency of a perfectly reversible engine—the maximum efficiency allowed by the second law—is

This is exactly the efficiency we found for a Carnot engine (Equation 21.27), so the Carnot efficiency is the maximum efficiency allowed by the second law of thermodynamics. It is the efficiency of a perfectly reversible engine for which $\text{Δ}{S}_{\text{universe}}=0$. We can summarize our conclusions:

As Example 21.5 showed, real engines usually fall well short of the Carnot limit.

We also found, in informal statement #6 of the second law, that no refrigerator can exceed the coefficient of performance of a perfectly reversible refrigerator. We’ll leave the proof as a homework problem, but an entropy analysis very similar to what we just did for a heat engine shows that the coefficient of performance of a Carnot refrigerator is

Thus we can state:

Statements #7 and #8 of the second law are a major result of this chapter, one with profound implications. The efficiency limit of a heat engine is set by the temperatures of the hot and cold reservoirs. High efficiency requires $T_{\text{C}}\text{/}{T}_{\text{H}}\ll 1$ and thus $T_{\text{H}}\gg T_{\text{C}}$. However, practical realities often prevent $T_{\text{H}}$ from being significantly larger than $T_{\text{C}}$, in which case the engine cannot possibly have a large efficiency. This limit on the efficiency of heat engines is a consequence of the second law of thermodynamics.

A limit on the efficiency of heat engines was not expected. We are used to thinking in terms of energy conservation, so it comes as no surprise that we cannot make an engine with $\eta >1$. But the limits arising from the second law were not anticipated, nor are they obvious. Nonetheless, they are a very real fact of life and a very real constraint on any practical device. No one has ever invented a machine that exceeds the second-law limits, and we have seen that the maximum efficiency for realistic engines is surprisingly low.

CHAPTER 21 CHALLENGE EXAMPLE Calculating efficiency

A heat engine using a monatomic ideal gas goes through the following closed cycle:

• Isochoric heating until the pressure is doubled.

• Isothermal expansion until the pressure is restored to its initial value.

• Isobaric compression until the volume is restored to its initial value.

What is the thermal efficiency of this heat engine? What would be the thermal efficiency of a Carnot engine operating between the highest and lowest temperatures reached by this engine?

MODEL The cycle consists of three familiar processes; we’ll need to analyze each. The amount of work and heat will depend on the quantity of gas, which we don’t know, but efficiency is a work-to-heat ratio that is independent of the amount of gas.

VISUALIZE FIGURE 21.21 shows the cycle. The initial pressure, volume, and temperature are p, V, and T. The isochoric process increases the pressure to 2p and, because the ratio $p\text{/}T$ is constant in an isochoric process, increases the temperature to 2T. The isothermal expansion is along the 2T isotherm. The product $pV$ is constant in an isothermal process, so the volume doubles to 2V as the pressure returns to p.

FIGURE 21.21 The pV cycle of the heat engine.

FIGURE 21.21 Full Alternative Text

SOLVE We know, symbolically, the state variables at each corner of the $pV$ diagram. That is sufficient for calculating $W_{\text{s}}$, Q, and $\text{Δ}{E}_{\text{th}}$.

Process $1\to 2$: An isochoric process has $W_{\text{s}}=0$ and

$$Q=\text{Δ}{E}_{\text{th}}=n{C}_{\text{V}}\text{Δ}T=\frac{3}{2}nRT$$

where we used $C_{\text{V}}=\frac{3}{2}R$ for a monatomic gas and $\text{Δ}T=2T-T=T$.

Process $2\to 3$: An isothermal process has $\text{Δ}{E}_{\text{th}}=0$ and

$$Q=W_{\text{s}}=nR\left(2T\right)\text{ln}\left({\displaystyle \frac{2V}{V}}\right)=(2\text{ln}2)nRT$$

Here we used the Table 21.1 result for the work done in an isothermal process.

Process $3\to 1$: The work done by the gas is the area under the curve, which is negative because $\text{Δ}V=V-2V=-V$ in the compression:

$$W_{\text{s}}=\text{area}=p\text{Δ}V=-pV=-nRT$$

We used the ideal-gas law in the last step to express the result in terms of n and T. The heat transfer is also negative because $\text{Δ}T=T-2T=-T:$

$$Q=n{C}_{\text{P}}\text{Δ}T=-\frac{5}{2}nRT$$

where we used $C_{\text{P}}=\frac{5}{2}R$ for a monatomic gas. Based on the first law, $\text{Δ}{E}_{\text{th}}=Q-W_{\text{s}}=-\frac{3}{2}nRT$.

Summing over the three processes, we see that $(\text{Δ}{E}_{\text{th}}){}_{\text{net}}=0$, as expected, and

$$W_{\text{out}}=(2\text{ln}2-1)nRT$$

Heat energy is supplied to the gas $(Q>0)$ in processes $1\to 2$ and $2\to 3$, so

$$Q_{\text{H}}=(2\text{ln}2+\frac{3}{2})nRT$$

Thus the thermal efficiency of this heat engine is

$$\eta ={\displaystyle \frac{W_{\text{out}}}{Q_{\text{H}}}}={\displaystyle \frac{(2\text{ln}2-1)nRT}{(2\text{ln}2+\frac{3}{2})nRT}}=0.134=13.4\text{\%}$$

A Carnot engine would be able to operate between a high temperature $T_{\text{H}}=2T$ and a low temperature $T_{\text{C}}=T$. Its efficiency would be

$${\eta }_{\text{Carnot}}=1-{\displaystyle \frac{T_{\text{C}}}{T_{\text{H}}}}=1-{\displaystyle \frac{T}{2T}}=0.500=50.0\text{\%}$$

REVIEW As we anticipated, the thermal efficiency depends on the shape of the $pV$ cycle but not on the quantity of gas or even on the values of p, V, or T. The heat engine’s 13.4% efficiency is considerably less than the 50% maximum possible efficiency set by the second law of thermodynamics.