a) Como $Q=m L_{\mathrm{F}}$ e $m=\rho A y$, temos: $Q=m L_{F}=\left[\left(1,0 \times 10^{3} \frac{\mathrm{kg}}{\mathrm{m}^{3}}\right) \times\left(50 \times 10^{6} \mathrm{~m}^{2}\right) \times 12 \mathrm{~m}\right] \times\left(80 \frac{\mathrm{cal}}{\mathrm{g}} \times \frac{4 \mathrm{~J}}{\mathrm{cal}} \times \frac{1000 \mathrm{~g}}{\mathrm{kg}}\right)=1,92 \times 10^{17} \mathrm{~J}$. b) $P=\frac{Q}{\Delta t}=\frac{k A\left(\theta_{\text {agua }}-\theta_{\text {ar }}\right)}{e}=\frac{\left(2,2 \mathrm{~W} / \mathrm{m}^{\circ} \mathrm{C}\right) \times\left(50 \times 10^{6} \mathrm{~m}^{2}\right) \times\left(15^{\circ} \mathrm{C}\right)}{5,0 \mathrm{~m}}=3,3 \times 10^{8} \mathrm{~W}$.