 Solution (film calculations)

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

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Past exam of the physics B course of the University of Cambridge / 2014 / Paper 2 / Section C / C11 / Solution (film calculations) by

Codex  0 2026-10-20

Reading the approximately straight trend in the supplied graph gives

γ (330 K) ≃ 22 mN m^{- 1}

and

γ^{'} ≃ - 0.13 mN m^{- 1} K^{- 1}

. These are graph-read estimates, not more precise material constants. For

Δ A = 1 0^{- 4} m^{2}

, reversible isothermal stretching gives

W_{on} = γ Δ A ≃ 2.2 μ J, Q = T Δ S = - T γ^{'} Δ A ≃ 4.3 μ J .

Both are positive: stretching does mechanical work on the film and requires heat input to prevent cooling. Its internal energy increases by about

6.5 μ J

.

Write

γ = α - βT

, with

β = - γ^{'} ≃ 1.3 \times 1 0^{- 4} J m^{- 2} K^{- 1}

. Constant

C_{A}

then makes the reversible adiabatic condition

C_{A} ln (T / T_{*}) + β A = constant,

where

T_{*}

is an arbitrary fixed reference temperature; absorbing it into the constant gives the requested logarithm convention. Hence

T_{f} = T_{i} e^{- β Δ A / C_{A}}, Δ T ≃ - \frac{β T _{i} Δ A}{C _{A}} ≃ - 6.1 \times 1 0^{- 5} K .

Here

C_{A} = 0.070 J K^{- 1}

. Since

γ - T γ^{'} = α

, integration gives

Δ U = C_{A} (T_{f} - T_{i}) + α Δ A

. To first order this is

Δ U ≃ γ (T_{i}) Δ A ≃ + 2.2 μ J .

There is no heat exchange on this reversible adiabat. Cooling and increasing internal energy are compatible: positive area work more than offsets the decrease of the temperature-dependent energy term.

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