= Solution
First estimate the excess <pressure> as weight divided by contact area:
$$
\Delta p=\frac{mg}{S}
=\frac{50(9.81)}{10^{-4}}\ \mathrm{Pa}
\simeq4.91\times10^6\ \mathrm{Pa}.
$$
For the equilibrium estimate, use the <Clapeyron equation> for <melting>. The <specific volume> change on melting is $\Delta v=1/\rho_w-1/\rho_i$, and the <latent heat of fusion> $L$ gives $\Delta s=L/T_m$. Thus the <common-pressure melting-point slope> is
$$
\frac{dT_m}{dp}=\frac{T_m}{L}
\left(\frac1{\rho_w}-\frac1{\rho_i}\right).
$$
Take $T_m\simeq273.15\ \mathrm K$ and the usual rounded <water> density $\rho_w\simeq1000\ \mathrm{kg\,m^{-3}}$, in addition to the supplied $\rho_i=920\ \mathrm{kg\,m^{-3}}$ and $L=3.3\times10^5\ \mathrm{J\,kg^{-1}}$. Then
$$
\frac{dT_m}{dp}\simeq-7.20\times10^{-8}\ \mathrm{K\,Pa^{-1}},
\qquad \Delta T_m\simeq-0.353\ \mathrm K,
$$
so
$$
\boxed{T_m\simeq272.80\ \mathrm K\simeq-0.35\,{}^\circ\mathrm C.}
$$
The sign is negative because <ice> contracts on melting. This is a linear equilibrium estimate treating the load as a common local <pressure> on coexisting ice and water. The actual contact <stress> is nonuniform, and frictional heating, deformation and drainage also affect skating; the estimate is not a complete account of the lubricating water film.
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