 Solution

First estimate the excess pressure as weight divided by contact area: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 isTake $T_m\simeq 273.15\mathrm{K}$ and the usual rounded water density ${\rho }_w\simeq 1000\mathrm{kg}{\mathrm{m}}^{-3}$, in addition to the supplied ${\rho }_i=920\mathrm{kg}{\mathrm{m}}^{-3}$ and $L=3.3\times 10^5\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}$. ThensoThe 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.