A clockwise engine <Carnot cycle> has four reversible steps. At hot <temperature> $T_h$, <isothermal expansion> absorbs $Q_h=T_h\Delta S>0$ and does positive <work> by the gas. <Adiabatic expansion> then cools it to $T_c$, with zero <heat> and positive <work> by the gas. <Isothermal compression> at $T_c$ emits $Q_c=-T_c\Delta S$ and requires <work> on the gas. Finally <adiabatic compression> returns it to $T_h$, with zero <heat> and <work> on the gas.

For an <ideal gas>, the $p,V$ isotherms obey $pV=\text{constant}$ and the adiabats $pV^\gamma=\text{constant}$, the latter being steeper. In the $T,S$ plane, the isotherms are horizontal and the adiabats vertical, forming a clockwise rectangle. <Heat> occurs only on the horizontal legs, while <work> occurs on all four $p,V$ legs.
$$
\boxed{W=(T_h-T_c)\Delta S,\qquad\eta_{\rm Carnot}=1-T_c/T_h.}
$$
Reversing the arrows gives a refrigerator, reversing the <heat> and <work> signs.

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Solution

Past exam of the mathematics course of the University of Cambridge / 2017 / ii / Paper 3 / 34D / a / Solution

Macroscopically, an <adiabatic process> exchanges no <heat> with its surroundings: $\delta Q=0$. A reversible adiabatic process is therefore <isentropic process>[isentropic], since $dS=\delta Q_{\rm rev}/T=0$. Microscopically, the system is thermally isolated, so changes in its energy arise from work done by changing external parameters rather than random energy exchange with a <heat bath>; in a sufficiently slow change, the populations of the corresponding microscopic states are carried along without producing entropy.

For repeated operation the working substance must execute a <thermodynamic cycle>. Reversible isothermal contact at $T_2$ absorbs heat $Q_2$, a reversible adiabat changes the temperature to $T_1$, reversible isothermal contact at $T_1$ rejects heat $Q_1>0$, and a second adiabat returns to the initial state. This is the <Carnot cycle>. Reversibility and zero net entropy change give
$$
\frac{Q_2}{T_2}-\frac{Q_1}{T_1}=0,
\qquad
\frac{Q_1}{Q_2}=\frac{T_1}{T_2}.
$$
The <thermal efficiency> is the net work output divided by the absorbed heat:
$$
\eta=\frac{W}{Q_2}=\frac{Q_2-Q_1}{Q_2}.
$$
Consequently
$$
\boxed{\eta_{\max}=1-\frac{T_1}{T_2}.}
$$
The <Carnot theorem> shows that any irreversible cycle has smaller efficiency, so the two isotherms joined by two adiabats are optimal.

Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 1 / 35D / a / Solution

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A Carnot cycle consists of two reversible <isothermal process>[isothermal changes], one at each reservoir temperature, joined by two reversible <adiabatic processes>. It attains the maximum <thermal efficiency> permitted between those reservoirs, $1-T_c/T_h$.

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