= Solution
The horizontal area is $A=10^6\,\mathrm{m^2}$ and the total mass from part (a) is $m=2.55\times10^{11}\,\mathrm{kg}$. The <iceberg melt heat budget> includes warming to $0^\circ\mathrm C$ followed by fusion. With temperature $T(z)=-0.6z$ in the upper $50$ m and $-30^\circ\mathrm C$ below,
$$
Q_{\rm sensible}=Ac_i\left[\int_0^{50}(300+12z)(0.6z)\,dz+\int_{50}^{300}900(30)\,dz\right].
$$
The two integrals are $525000$ and $6750000\,\mathrm{kg\,K\,m^{-2}}$, so using the stated ice <specific heat capacity>,
$$
Q_{\rm sensible}=10^6(2100)(7275000)=1.52775\times10^{16}\,\mathrm J.
$$
The <latent heat of fusion> contribution is
$$
Q_{\rm latent}=mL_f=(2.55\times10^{11})(336000)=8.568\times10^{16}\,\mathrm J.
$$
Therefore
$$
\boxed{Q=Q_{\rm sensible}+Q_{\rm latent}=1.009575\times10^{17}\,\mathrm J\simeq1.01\times10^{17}\,\mathrm J.}
$$
This produces liquid water at $0^\circ\mathrm C$; warming the meltwater further would require additional heat.
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