= Solution
{disambiguate=film calculations}
Reading the approximately straight trend in the supplied graph gives $\gamma(330\,\mathrm K)\simeq22\,\mathrm{mN\,m^{-1}}$ and $\gamma'\simeq-0.13\,\mathrm{mN\,m^{-1}\,K^{-1}}$. These are graph-read estimates, not more precise material constants. For $\Delta A=10^{-4}\,\mathrm{m^2}$, reversible isothermal stretching gives
$$
\boxed{W_{\rm on}=\gamma\Delta A\simeq2.2\,\mathrm{\mu J},
\qquad Q=T\Delta S=-T\gamma'\Delta A\simeq4.3\,\mathrm{\mu J}.}
$$
Both are positive: stretching does mechanical <work> on the film and requires <heat> input to prevent cooling. Its <internal energy> increases by about $6.5\,\mathrm{\mu J}$.
Write $\gamma=\alpha-\beta T$, with $\beta=-\gamma'\simeq1.3\times10^{-4}\,\mathrm{J\,m^{-2}\,K^{-1}}$. Constant $C_A$ then makes the reversible adiabatic condition
$$
\boxed{C_A\ln(T/T_*)+\beta A=\mathrm{constant},}
$$
where $T_*$ is an arbitrary fixed reference <temperature>; absorbing it into the constant gives the requested logarithm convention. Hence
$$
T_f=T_i e^{-\beta\Delta A/C_A},\qquad
\boxed{\Delta T\simeq-\frac{\beta T_i\Delta A}{C_A}
\simeq-6.1\times10^{-5}\,\mathrm K.}
$$
Here $C_A=0.070\,\mathrm{J\,K^{-1}}$. Since $\gamma-T\gamma'=\alpha$, integration gives $\Delta U=C_A(T_f-T_i)+\alpha\Delta A$. To first order this is
$$
\boxed{\Delta U\simeq\gamma(T_i)\Delta A\simeq+2.2\,\mathrm{\mu J}.}
$$
There is no <heat> exchange on this reversible adiabat. \b[Cooling and increasing <internal energy> are compatible:] positive <area> <work> more than offsets the decrease of the temperature-dependent <energy> term.
Back to article page
