A slab of stone of area of 0.36 $m^{2} and thickness 0.1 m is exposed on the lower surface to steam at 100^{0} C . A block of ice at 0^{0} C$ rests on the upper surface of the slab. In one hour 4.8 kg of ice is melted. The thermal conductivity of slab id
(Given latent heat of fusion of ice = $3.36 \times 10^{5} J {kg}^{- 1}$ )

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1.24 J/m/s $/^{0} C$

$2.05 J / ms /^{0} C$

$1.02 J /, / s /^{0} C$

$1.29 J / m / s^{0} C$

answer is A.

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Detailed Solution

Rate of heat given by steam $\frac{δQ}{δt} = \frac{KA}{L} (T_{1} - T_{2})$

Rate of heat taken by ice Q = $\frac{KA}{L} (T_{1} - T_{2}) t$

$Q = {mL}_{f}$

$\frac{KA}{L} (T_{1} - T_{2}) t = {mL}_{f}$

$K = \frac{{mL}_{f} L}{A (T_{1} - T_{2}) t}$

K = $\frac{4.8 \times 3.36 \times 10^{5} \times 0.1}{0.36 \times 100 \times 3600}$

= $\frac{4.8 \times 3.36}{0.36 \times 36} = 1.24 J / m / s^{0} C$

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