Question 20.9: Entropy and the Carnot cycle For the Carnot engine in Exampl......

IDENTIFY and SET UP All four steps in the Carnot cycle (see Fig. 20.13) are reversible, so we can use our expressions for the entropy change \Delta S in a reversible process. We find \Delta S for each step and add them to get \Delta S for the complete cycle.

EXECUTE There is no entropy change during the adiabatic expansion b \rightarrow c or the adiabatic compression d \rightarrow a. During the isothermal expansion a \rightarrow b at T_{\mathrm{H}}=500 \mathrm{~K}, the engine takes in 2000 \mathrm{~J} of heat, and from Eq. (20.18),

\Delta S_{\mathrm{H}}=\frac{Q_{\mathrm{H}}}{T_{\mathrm{H}}}=\frac{2000 \mathrm{~J}}{500 \mathrm{~K}}=4.0 \mathrm{~J} / \mathrm{K}

During the isothermal compression c \rightarrow d at T_{\mathrm{C}}=350 \mathrm{~K}, the engine gives off 1400 \mathrm{~J} of heat, and

\begin{aligned} \Delta S_{\mathrm{C}} & =\frac{Q_{\mathrm{C}}}{T_{\mathrm{C}}}=\frac{-1400 \mathrm{~J}}{350 \mathrm{~K}} \\ & =-4.0 \mathrm{~J} / \mathrm{K} \end{aligned}

The total entropy change in the engine during one cycle is \Delta S_{\text {tot }}=\Delta S_{\mathrm{H}}+\Delta S_{\mathrm{C}}=4.0 \mathrm{~J} / \mathrm{K}+(-4.0 \mathrm{~J} / \mathrm{K})=0.

EVALUATE The result \Delta S_{\text {total }}=0 tells us that when the Carnot engine completes a cycle, it has the same entropy as it did at the beginning of the cycle. We’ll explore this result in the next subsection.

What is the total entropy change of the engine’s environment during this cycle? During the reversible isothermal expansion a \rightarrow b, the hot (500 \mathrm{~K}) reservoir gives off 2000 \mathrm{~J} of heat, so its entropy change is (-2000 \mathrm{~J}) /(500 \mathrm{~K})=-4.0 \mathrm{~J} / \mathrm{K}. During the reversible isothermal compression c \rightarrow d, the cold (350 \mathrm{~K}) reservoir absorbs 1400 \mathrm{~J} of heat, so its entropy change is (+1400 \mathrm{~J}) /(350 \mathrm{~K})=+4.0 \mathrm{~J} / \mathrm{K}. Thus the hot and cold reservoirs each have an entropy change, but the sum of these changes – that is, the total entropy change of the system’s environment-is zero.

These results apply to the special case of the Carnot cycle, for which all of the processes are reversible. In this case the total entropy change of the system and the environment together is zero. We’ll see that if the cycle includes irreversible processes (as is the case for the Otto and Diesel cycles of Section 20.3), the total entropy change of the system and the environment cannot be zero, but rather must be positive.

KEYCONCEPT For the special case of a heat engine that uses the Carnot cycle, the net entropy change of the engine and its environment (the hot and cold reservoirs) in a complete cycle is zero. This is not the case for other, less ideal cycles.