 Solution

ID: past-exam-of-the-physics-b-course-of-the-university-of-cambridge/2014/paper-2/section-c/c11/b/solution

At fixed area,

(\partial S / \partial T)_{A} = C_{A} / T

. Therefore

d S = \frac{C _{A}}{T} d T - γ^{'} d A .

For reversible adiabatic stretching,

d S = 0

and

(\frac{d T}{d A})_{S} = \frac{T γ ^{'}}{C _{A}} < 0.

The film cools, assuming positive heat capacity. Adiabaticity alone would not ensure constant entropy if the deformation were irreversible.

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