To illustrate how these ideas actually work, FIGURE 21.7 shows a simple engine that converts heat into the work of lifting mass M.

FIGURE 21.7 A simple heat engine transforms heat into work.

The net effect of this multistep process is to convert some of the fuel’s energy into the useful work of lifting the mass. There has been no net change in the gas, which has returned to its initial pressure, volume, and temperature at the end of step (e). We can start the whole process over again and continue lifting masses (doing work) as long as we have fuel.

FIGURE 21.8 shows the heat-engine process on a pV diagram. It is a closed cycle because the gas returns to its initial conditions. No work is done during the isochoric process, and, as you can see from the areas under the curve, the work done by the gas to lift the mass is greater than the work the environment must do on the gas to recompress it. Thus this heat engine, by burning fuel, does net work per cycle: $W_{\text{net}}=W_{\text{lift}}-W_{\text{ext}}=(W_{\text{s}}){}_{1\to 2}+(W_{\text{s}}){}_{3\to 1}$.

FIGURE 21.8 The closed-cycle pV diagram for the heat engine of Figure 21.7.

Notice that the cyclical process of Figure 21.8 involves two cooling processes in which heat is transferred from the gas to the environment. Heat energy is transferred from hotter objects to colder objects, so the system must be connected to a cold reservoir with $T_{\text{C}}<T_{\text{gas}}$ during these two processes. A key to understanding heat engines is that they require both a heat source (burning fuel) and a heat sink (cooling water, the air, or something at a lower temperature than the system).

EXAMPLE 21.1 Analyzing a heat engine I

Analyze the heat engine of FIGURE 21.9 to determine (a) the net work done per cycle, (b) the engine’s thermal efficiency, and (c) the engine’s power output if it runs at 600 rpm. Assume the gas is monatomic.

FIGURE 21.9 The heat engine of Example 21.1.

FIGURE 21.9 Full Alternative Text

MODEL The gas follows a closed cycle consisting of three distinct processes, each of which was studied in Chapters 18 and 19. For each of the three we need to determine the work done and the heat transferred.

SOLVE To begin, we can use the initial conditions at state 1 and the ideal-gas law to determine the number of moles of gas:

$$n={\displaystyle \frac{p_1{V}_1}{R{T}_1}}={\displaystyle \frac{(200\times {10}^3\text{Pa})(2.0\times {10}^{-4}{\text{m}}^3)}{\left(8.31\text{J/mol K}\right)\left(300\text{K}\right)}}=0.0160\text{mol}$$

Process $1\to 2:$ The work done by the gas in the isobaric expansion is

$$(W_{\text{s}}){}_{12}=p\text{Δ}V=(200\times {10}^3\text{Pa})((6.0-2.0)\times {10}^{-4}{\text{m}}^3)=80\text{J}$$

We can use the ideal-gas law at constant pressure to find $T_2=\left(V_2\text{/}{V}_1\right)T_1=3T_1=900\text{K}$. The heat transfer during a constant-pressure process is

$$\begin{array}{ccc}\hfill Q_{12}& \hfill =\hfill & n{C}_{\text{P }}\text{Δ}T\hfill \\ \hfill & \hfill =\hfill & \left(0.0160\text{mol}\right)\left(20.8\text{J/mol K}\right)(900\text{K}-300\text{K})=200\text{J}\hfill \end{array}$$

where we used $C_{\text{P}}=\frac{5}{2}R$ for a monatomic ideal gas.

Process $2\to 3:$ No work is done in an isochoric process, so $(W_{\text{s}}){}_{23}=0$. The temperature drops back to 300 K, so the heat transfer, with $C_{\text{V}}=\frac{3}{2}R$, is

$$\begin{array}{ccc}\hfill Q_{23}& \hfill =\hfill & n{C}_{\text{V}}\text{Δ}T\hfill \\ \hfill & \hfill =\hfill & \left(0.0160\text{mol}\right)\left(12.5\text{J/mol K}\right)(300\text{K}-900\text{K})=-120\text{J}\hfill \end{array}$$

Process $3\to 1:$ The gas returns to its initial state with volume $V_1$. The work done by the gas during an isothermal process is

$$\begin{array}{ccc}\hfill \left(W_{\text{s}}\right){}_{31}& \hfill =\hfill & nRT\text{ ln}\left({\displaystyle \frac{V_1}{V_3}}\right)\hfill \\ \hfill & \hfill =\hfill & (0.0160\text{mol}\left)\left(8.31\text{J/mol K}\right)\right(300\text{ K})\text{ ln}\left({\displaystyle \frac{1}{3}}\right)=-44\text{J}\hfill \end{array}$$

$W_{\text{s}}$ is negative because the environment does work on the gas to compress it. An isothermal process has $\text{Δ}{E}_{\text{th}}=0$ and hence, from the first law,

$$Q_{31}=(W_{\text{s}}){}_{31}=-44\text{J}$$

Q is negative because the gas must be cooled as it is compressed to keep the temperature constant.

1. The net work done by the engine during one cycle is

   $$W_{\text{out}}=(W_{\text{s}}){}_{12}+(W_{\text{s}}){}_{23}+(W_{\text{s}}){}_{31}=36\text{J}$$

   As a consistency check, notice that the net heat transfer is

   $$Q_{\text{net}}=Q_{12}+Q_{23}+Q_{31}=36\text{J}$$

   Equation 21.4 told us that a heat engine must have $W_{\text{out}}=Q_{\text{net}}$, and we see that it does.

2. The efficiency depends not on the net heat transfer but on the heat $Q_{\text{H}}$ transferred into the engine from the flame. Heat enters during process $1\to 2$, where Q is positive, and exits during processes $2\to 3$ and $3\to 1$, where Q is negative. Thus

   $$\begin{array}{l}{Q}_{\text{H}}=Q_{12}=200\text{ J}\\ {\text{Q}}_{\text{C}}=\left|Q_{23}\right|+\left|Q_{31}\right|=164\text{ J}\end{array}$$

   Notice that $Q_{\text{H}}-Q_{\text{C}}=36\text{J}=W_{\text{out}}$. In this heat engine, 200 J of heat from the hot reservoir does 36 J of useful work. Thus the thermal efficiency is

   $$\eta ={\displaystyle \frac{W_{\text{out}}}{Q_{\text{H}}}}={\displaystyle \frac{36\text{J}}{200\text{J}}}=0.18\text{or }18\text{\%}$$

   This heat engine is far from being a perfect engine!

3. An engine running at 600 rpm goes through 10 cycles per second. The power output is the work done per second:

   $$\begin{array}{ccc}\hfill P_{\text{out}}& \hfill =\hfill & (\text{work per cycle})\times (\text{cycles per second})\hfill \\ \hfill & \hfill =\hfill & 360\text{ J}\text{/}\text{s}=360\text{W}\hfill \end{array}$$

REVIEW Although we didn’t need $Q_{\text{net}}$, verifying that $Q_{\text{net}}=W_{\text{out}}$ was a check of self-consistency. Heat-engine analysis requires many calculations and offers many opportunities to get signs wrong. However, there are a sufficient number of self-consistency checks so that you can almost always spot calculational errors if you check for them.

Let’s think about this example a bit more before going on. We’ve said that a heat engine operates between a hot reservoir and a cold reservoir. Figure 21.9 doesn’t explicitly show the reservoirs. Nonetheless, we know that heat is transferred from a hotter object to a colder object. Heat $Q_{\text{H}}$ is transferred into the system during process $1\to 2$ as the gas warms from 300 K to 900 K. For this to be true, the hot-reservoir temperature $T_{\text{H}}$ must be $\ge 900\text{K}$. Likewise, heat $Q_{\text{C}}$ is transferred from the system to the cold reservoir as the temperature drops from 900 K to 300 K in process $2\to 3$. For this to be true, the cold-reservoir temperature $T_{\text{C}}$ must be $\le 300\text{K}$.

So, while we really don’t know what the reservoirs are or their exact temperatures, we can say with certainty that the hot-reservoir temperature $T_{\text{H}}$ must exceed the highest temperature reached by the system and the cold-reservoir temperature $T_{\text{C}}$ must be less than the coldest system temperature.