Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 72 / 4 / b / i / Solution

Use seawater specific heat capacity $c_w\simeq 4.0\times 10^3\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$, which is an additional standard material approximation: the paper supplies the ice heat capacity, not the seawater heat capacity. Relative to freezing, the assumed uniform column has $\Delta T=7-(-1.8)=8.8\mathrm{K}$. Its ocean heat content per horizontal area isThis is heat above the reference freezing state, not the absolute internal energy of seawater.

If all this heat reaches ice already at its fusion temperature, divide by the supplied latent heat:For a chosen ice mass density ${\rho }_i=917\mathrm{kg}{\mathrm{m}}^{-3}$, this corresponds to about $5.86\mathrm{m}$ of ice. Density was not specified for this conversion, so the mass per area is the result that needs no further ice-density assumption.

Cold ice must first be warmed. If its initial temperature is $T_i<T_m$, the corresponding ideal melt mass is $Q/[L_f+c_i(T_m-T_i)]$, using the supplied ice specific heat capacity $c_i=2100\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$. No initial ice temperature or salinity-dependent fusion temperature is supplied. The latent-only result is therefore an upper bound; actual melting is smaller when sensible warming, ocean or atmospheric losses and incomplete heat transfer matter.