An ice cube of dimensions $60\ \text{cm}\times50\ \text{cm}\times20\ \text{cm}$ is placed in an insulation box of wall thickness $1\ \text{cm}$. The box keeping the ice cube at $0^\circ\text{C}$ is brought to a room of temperature $40^\circ\text{C}$. The rate of melting of ice is approximately: (Latent heat of fusion of ice is $3.4\times10^5\ \text{J kg}^{-1}$ and thermal conductivity of the insulation wall is $0.05\ \text{W m}^{-1}\,^\circ\text{C}^{-1}$)
Answer: (B) $61\times10^{-5}\ \text{kg s}^{-1}$
Total wall area: $A = 2(0.6\times0.5 + 0.5\times0.2 + 0.6\times0.2) = 1.04\ \text{m}^2$.
$$\frac{dQ}{dt} = \frac{kA\Delta T}{d} = \frac{0.05\times1.04\times40}{0.01} = 208\ \text{W}$$
$$\frac{dm}{dt} = \frac{208}{3.4\times10^5}\approx6.1\times10^{-4} = 61\times10^{-5}\ \text{kg s}^{-1}$$
Solution by Sreeraj P, M.Sc Physics