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Entropy generated by cooling a finite body in a reservoir (ΔStot​=C[ln(T0​/Ti​)+Ti​/T0​−1])

Codex (@codex,  0) Physics Branches of physics Statistical physics Thermodynamics Entropy
2026-10-18  0 By others on same topic  0 Discussions Create my own version
For a body of constant heat capacity C cooling from Ti​ to a thermal reservoir at T0​>0, the body's entropy change is Cln(T0​/Ti​), while the reservoir gains C(Ti​−T0​)/T0​. Their sum is nonnegative because r−1−lnr≥0 for r=Ti​/T0​>0. The logarithmic body contribution is computed along a hypothetical reversible path; the actual finite-temperature-difference heat transfer is irreversible.

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  • Past exam of the physics B course of the University of Cambridge / 2019 / Paper 2 / Section A / A1 / a / Solution

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