Question

A slab of stone of area $0.36m^2$ and thickness $0.1$ is exposed on the lower surface to steam at ${100}^{o}C$. A block of ice at $0^{o}C$ 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$=3.36\times {10}^{5}J{kg}^{-1}$)

• A. $1.02J/m/{s/}^{o}C$
• B. $1.24J/m/{s/}^{o}C$
• C. $1.29J/m/{s/}^{o}C$
• D. $2.05J/m/{s/}^{o}C$

Answer 1

The correct option is B $1.24J/m/{s/}^{o}C$

From Fourier's law we have,
$\frac{dQ}{dt}=\frac{KA}{L}(T_1-T_2)$
$Q=\frac{KA}{L}(T_1-T_2)t$
$Q=m{L}_f$
$\frac{KA}{L}(T_1-T_2)t=m{L}_f$
$K=\frac{m{L}_f\left(L\right)}{A(T_1-T_2)t}$

$K=\frac{4.8\times 3.36\times {10}^5\times 0.1}{0.36\times 100\times 3600}J/m/{s/}^{o}C$

$K=\frac{4.8\times 3.36}{0.36\times 36}$

$K=1.24J/m/{s/}^{o}C$