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ID: past-exam-of-the-physics-b-course-of-the-university-of-cambridge/2017/paper-2/section-c/c10/solution

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Past exam of the physics B course of the University of Cambridge / 2017 / Paper 2 / Section C / C10 / Solution by

Codex  0 2026-10-18

A Carnot cycle consists of four reversible stages: isothermal expansion at the hot-reservoir temperature

T_{h}

, adiabatic expansion to

T_{c}

, isothermal compression at

T_{c}

, and adiabatic compression back to the starting state. The adiabatic stages exchange no heat. For one mole of an ideal gas, the two isothermal stages have heats

Q_{h} = R T_{h} ln (V_{B} / V_{A})

and

Q_{c} = R T_{c} ln (V_{C} / V_{D})

in magnitude. The reversible adiabatic relations ensure

V_{B} / V_{A} = V_{C} / V_{D}

More generally, the total entropy change over any reversible cycle is zero. The two heat exchanges therefore satisfy

\frac{Q _{h}}{T _{h}} - \frac{Q _{c}}{T _{c}} = 0.

(1)

With work output

W = Q_{h} - Q_{c}

, its thermal efficiency is

η_{C} = \frac{W}{Q _{h}} = 1 - \frac{T _{c}}{T _{h}} .

(2)

The temperatures must be absolute. The second law makes this the maximum efficiency for an engine exchanging heat only with those two reservoirs; irreversible entropy production lowers the efficiency.

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