= Solution
For the specified increment,
$$
\frac{\beta\Delta A}{C_A}
=\frac{(1.3\times10^{-4})(10^{-4})}{0.070}
\simeq1.86\times10^{-7}.
$$
Using the adiabat, or its first-order <Taylor expansion>,
$$
\Delta T=330\left(e^{-1.86\times10^{-7}}-1\right)\,\mathrm K
\simeq-330(1.86\times10^{-7})\,\mathrm K,
\qquad
\boxed{\Delta T\simeq-6.1\times10^{-5}\,\mathrm K.}
$$
This is a very small cooling, about $61\,\mathrm{\mu K}$. There is no <heat> exchange, and the film is stretched against positive <surface tension>, so the <internal energy> \b[increases], with
$$
\boxed{\Delta U=W_{\rm on}=\int_{A_i}^{A_f}\gamma(T(A))\,dA
\simeq2.2\,\mathrm{\mu J}>0.}
$$
The cooling does not contradict this increase. In the linear model, $\gamma-T\gamma'=\gamma_0\simeq0.065\,\mathrm{J\,m^{-2}}$, so
$$
\Delta U=C_A\Delta T+\gamma_0\Delta A
\simeq-4.3\,\mathrm{\mu J}+6.5\,\mathrm{\mu J}.
$$
Creation of surface raises the <internal energy> more than the cooling lowers its temperature-dependent part. The change in <surface tension> during this tiny <temperature> fall changes the <work> only at a negligible higher order.
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