An ice cube of dimensions 60   cm × 50   cm × 20   cm is placed in an insulation…

Using the equation of conduction,

$\frac{\mathrm{d}Q}{\mathrm{d}t}=\frac{KA\mathrm{\Delta }T}{l}$

Now the total area of the box will be,

$A=2\left(0.6\times 0.5+0.5\times 0.2+0.2\times 0.6\right)$

$A=2\left(0.3+0.1+0.12\right)$

$A=2\left(0.52\right)=1.04 {\mathrm{m}}^2$

Therefore,

$\frac{\mathrm{d}Q}{\mathrm{d}t}=\frac{KA\mathrm{\Delta }T}{l}=\frac{0.05\times 1.04\times 40}{0.01}=208 \mathrm{J} {\mathrm{s}}^{-1}$

This heat will melt the ice. Therefore,

$$\begin{gathered} \frac{\mathrm{d}Q}{\mathrm{d}t}=\frac{dm}{dt}L \\ \Rightarrow 208=\frac{dm}{dt}\times 3.4\times {10}^5 \\ \Rightarrow \frac{dm}{dt}=61\times {10}^{-5} \mathrm{kg} {\mathrm{s}}^{-1} \end{gathered}$$