Past exam of the physics B course of the University of Cambridge / 2014 / Paper 2 / Section C / C10

A heat engine operates cyclically, absorbing heat $Q_h$ from a hotter source, rejecting heat $Q_c$ to a colder sink and delivering net work $W=Q_h-Q_c$. Its efficiency is $\eta =W/Q_h$. The working substance returns to its original state; the reservoirs and surroundings need not.

Suppose an engine were more efficient than a reversible Carnot engine between the same reservoirs. Couple it to a reversed Carnot engine, scaling the cycles to have equal work. With no net external work, their combined effect would be to transfer heat from the colder reservoir to the hotter one, contradicting the Second law of thermodynamics. No such engine can exceed ${\eta }_C=1-T_c/T_h$. Equivalently, the cyclic entropy inequality gives $-Q_h/T_h+Q_c/T_c\ge 0$, hence the same bound.

The reversible working substance changes entropy by $Q_h/T_h$ on its hot isotherm and $-Q_c/T_c$ on its cold isotherm, with zero change on the two reversible adiabats. Closure requires $Q_h/T_h=Q_c/T_c$. The substance has zero entropy change over a cycle, while the reservoirs have changes $-Q_h/T_h$ and $Q_c/T_c$. Thus $\Delta S_{\mathrm{universe}}=0$ for the complete Carnot cycle.

Put $P_{\ast }=p_1{V}_1=nR{T}_1$. For the specified ideal gas, $C_V=3nR/2$, $C_p=5nR/2$ and $\gamma =5/3$. Isobaric expansion doubles temperature, so $T_2=2T_1$. Reversible adiabatic closure gives $T{V}^{\gamma -1}=\mathrm{constant}$, henceThere is no heat input on that compression. The other stages giveTherefore $W_{\mathrm{net}}^{\mathrm{by}}=P_{\ast }-W_{31}^{\mathrm{on}}=0.44494P_{\ast }$ andThe hot reservoir changes entropy by $-Q_{12}/(2.5T_1)=-P_{\ast }/T_1$. The cold reservoir changes by $-Q_{23}/(0.5T_1)=4.11012P_{\ast }/T_1$. Since the gas completes a cycle,Although the mechanical paths can be quasistatic, heat flows across finite temperature differences in both contacts. This entropy generation explains why the efficiency is well below the Carnot value $1-0.5/2.5=80\%$ between these reservoirs.

For an isothermal final compression at $T_1$, the preceding constant-volume cooling must instead end at $T_3=T_1$; keeping the old state 3 would not close the cycle. Compression then requires work $P_{\ast }\ln 2$, while the heat input remains $5P_{\ast }/2$. Thuswhich is lower: this isothermal compression occurs at a higher temperature throughout than the original adiabat and consumes more work. It also rejects heat during compression.