Solution:
Heat transferred $frac{Q}{t} = frac{K A Delta T}{x}$
$frac{m L}{t} = frac{K A (100 - 0)}{x}$
$K = frac{m L x}{t A Delta T} = frac{4.8 times 3.36 times 10^5 times 0.1}{3600 times 0.36 times 100} = 1.24 text{ J/m/s/}^circ text{C}$
A slab of stone of area $0.36 text{ m}^2$ and thickness $0.1 text{ m}$ is exposed on the lower surface to steam at $100^circ text{C}$. A block of ice at $0^circ text{C}$ rests on the upper surface of the slab. In one hour $4.8 text{ kg}$ of ice is melted. The thermal conductivity of slab is: (Given latent heat of fusion of ice $= 3.36 times 10^5 text{ J/kg}$)
(2012 Mains)
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