= Solution
Reading the original PDF graph gives approximately
$$
\gamma(330\,\mathrm K)\simeq22\,\mathrm{mN\,m^{-1}},\qquad
\gamma'\simeq-0.13\,\mathrm{mN\,m^{-1}K^{-1}}
=-1.3\times10^{-4}\,\mathrm{J\,m^{-2}K^{-1}}.
$$
For example, the points near $330\,\mathrm K$ and $450\,\mathrm K$ are about $22$ and $6.5\,\mathrm{mN\,m^{-1}}$, respectively, giving this slope. The graph supports approximate values rather than exact fitted constants. The <area> increment is $\Delta A=10^{-4}\,\mathrm{m^2}$.
At fixed <temperature>, the <surface tension> is constant during this stretch. The <work> done \i[on] the film is positive:
$$
W_{\rm on}=\int\gamma\,dA\simeq(22\times10^{-3})(10^{-4}),
\qquad \boxed{W_{\rm on}\simeq2.2\,\mathrm{\mu J}.}
$$
For a reversible <isothermal process>, the <heat> supplied is $Q=T\Delta S=-T\gamma'\Delta A$. Thus
$$
\boxed{Q\simeq(330)(1.3\times10^{-4})(10^{-4})
=4.3\,\mathrm{\mu J}.}
$$
The film absorbs <heat> while its <temperature> stays constant. The first law gives the useful consistency check $\Delta U=Q+W_{\rm on}\simeq6.5\,\mathrm{\mu J}$, agreeing with $(\gamma-T\gamma')\Delta A$.
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