Answer
$Q_{cond}=30.235kW$
$m_{ice}=0.091\frac{kg}{s}$
Work Step by Step
First we calculate the area of the spherical shell:
$A=\pi D^2=\pi*(0.4m)^2=0.502655m^2$
Knowing that the heat transfer in this case is by conduction:
$k_{iron}=80.2\frac{W}{m^{\circ}C}$
$Q_{cond}=\frac{kA\Delta T}{L}=\frac{80.2\frac{W}{m^{\circ}C}*0.502655m^2*(3^{\circ}C-0^{\circ}C)}{0.004m}=30.235kW$
Taking $333.7kJ$ as the energy to melt $1k$ of ice at $0^{\circ}C$:
$m_{ice}=\frac{30.235kW}{333.7\frac{kJ}{kg}}=0.091\frac{kg}{s}$
