# A slab of stone of area 0.36 m^2 and thickness 0.1 is exposed on the lower surface to steam at 100 C. A block of ice at 0 C rests on the upper surface of the slab. In one hour 4.8kg of ice is melted. The thermal conductivity of slab is :(Given latent heat of fusion of ice=3.36×10^5 Jkg−1).

Rate of heat given by steam, dA/dt = KA/L (T1 – T2)

Rate of heat taken by ice, Q = KA/L (T1 – T2)t

Since,

Q=mLf

KA/L (T1 – T2)t = mLf

$$\begin{array}{l} K=\frac{m L_{f} L}{A\left(T_{1}-T_{2}\right) t} \\ K=\frac{4.8 \times 3.36 \times 10^{5} \times 0.1}{0.36 \times 100 \times 3600} \\ =\frac{4.8 \times 3.36}{0.36 \times 36}=1.24 \mathrm{~J} / \mathrm{m} / \mathrm{s}^{\circ} \mathrm{C} . \end{array}$$