 Solution

ID: past-exam-of-the-physics-b-course-of-the-university-of-cambridge/2022/paper-2/section-c/c8/b/solution

Because the working substance returns to its original thermodynamic state, its net entropy change is zero. There is no entropy change on either reversible ideal-gas adiabat; on each isothermal process the entropy transfer is the exchanged heat divided by the bath's absolute temperature. Hence

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

The first law of thermodynamics over a complete cycle gives

W = Q_{h} - Q_{c}

. The thermal efficiency is therefore

η_{Carnot} = \frac{W}{Q _{h}} = 1 - \frac{Q _{c}}{Q _{h}} = 1 - \frac{T _{c}}{T _{h}} = 1 - \frac{T _{1}}{T _{2}} .

The use of kelvins is essential. This result does not depend on the working substance. For the ideal gas illustration, the two reversible ideal-gas adiabats imply

V_{C} / V_{B} = V_{D} / V_{A} = (T_{h} / T_{c})^{1/ (γ - 1)}

, so

ln (V_{B} / V_{A}) = ln (V_{C} / V_{D})

; the explicit heat expressions in (a) give the same ratio.

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