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 Incoming links: Entropy generated by cooling a finite body in a reservoir

Past exam of the physics B course of the University of Cambridge / 2019 / Paper 2 / Section A / A1 / a / Solution 2026-10-18

The units supplied for 130 indicate specific heat capacity, so the block's total heat capacity is

C = m c = 0.5 (130) = 65 J K^{- 1}

. Treat it as constant, and the lake as a thermal reservoir at

T_{0} = 283.15 K

. The initial block temperature is

T_{i} = 373.15 K

.

Compute the block's entropy change using a reversible comparison path, even though the actual cooling is irreversible:

Δ S_{block} = \int_{T_{i}}^{T_{0}} \frac{C d T}{T} = C ln \frac{T _{0}}{T _{i}} = - 17.940 J K^{- 1} .

The lake receives

Q = C (T_{i} - T_{0}) = 5850 J

at constant temperature, so

Δ S_{lake} = \frac{Q}{T _{0}} = 20.660 J K^{- 1} .

Therefore

Δ S_{Universe} = 2.72 J K^{- 1} .

This is entropy generated by cooling a finite body in a reservoir. The positive total is consistent with the Second law of thermodynamics, although the cooling block's own entropy decreases.

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