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A slab of stone of area 0.36 $m^2$ and thickness 0.1 m is exposed on the lower surface to steam at $100^\circ C$. A block of ice at $0^\circ C$ rests on the upper surface of the slab. In one hour 4.8 kg of ice is melted. The thermal conductivity of slab is (Given latent heat of fusion of ice = $3.36\times10^5$ J $kg^{-1}$)
A
1.02 J/m/s/$^\circ C$
B
1.24 J/m/s/$^\circ C$
C
1.29 J/m/s/$^\circ C$
D
2.05 J/m/s/$^\circ C$
Detailed Solution
Heat conducted through the slab in time t melts the ice: $Q = \frac{KA(\Delta T)t}{L} = mL_f$
$K = \frac{mL_f\times L}{A(\Delta T)t} = \frac{4.8\times3.36\times10^5\times0.1}{0.36\times100\times3600}$
$K = \frac{161280}{129600} = \frac{56}{45} = 1.24$ J/m/s/$^\circ C$
$K = \frac{mL_f\times L}{A(\Delta T)t} = \frac{4.8\times3.36\times10^5\times0.1}{0.36\times100\times3600}$
$K = \frac{161280}{129600} = \frac{56}{45} = 1.24$ J/m/s/$^\circ C$