 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ii/paper-4/35a/b/iii/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

Direct contact permits no work extraction, so conservation of internal energy gives

C_{V} (T_{f} - T_{1}) + C_{V} (T_{f} - T_{2}) = 0, T_{f} = \frac{T _{1} + T _{2}}{2} .

The total entropy change is

Δ S = C_{V} lo g \frac{T _{f}}{T _{1}} + C_{V} lo g \frac{T _{f}}{T _{2}} = C_{V} lo g \frac{( T _{1} + T _{2} ) ^{2}}{4 T _{1} T _{2}} \geq 0.

The arithmetic-geometric mean inequality makes the last expression positive when

T_{1} \neq = T_{2}

, with equality only when the bodies were already at the same temperature. This is the entropy production of irreversible heat flow.

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