= Solution
Use seawater <specific heat capacity> $c_w\simeq4.0\times10^3\,\mathrm{J\,kg^{-1}\,K^{-1}}$, 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 is
$$
\boxed{Q=\rho_wc_wH\Delta T
=(1025)(4000)(50)(8.8)
=1.804\times10^9\,\mathrm{J\,m^{-2}}.}
$$
This 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>:
$$
\boxed{\frac{M_{\rm melt}}{\text{area}}
=\frac{Q}{L_f}
=\frac{1.804\times10^9}{336000}
=5.37\times10^3\,\mathrm{kg\,m^{-2}}.}
$$
For a chosen ice <mass density> $\rho_i=917\,\mathrm{kg\,m^{-3}}$, this corresponds to \b[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> $c_i=2100\,\mathrm{J\,kg^{-1}\,K^{-1}}$. 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.
Back to article page
