Question

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

• A. $1.29J/m/s/C$
• B. $2.05J/m/s/C$
• C. $1.02J/m/s/C$
• D. $1.24J/m/s/C$

Answer 1

The correct option is D $1.24J/m/s/C$

Given that:

Latent heat of ice, $L=3.36\times {10}^{5}J/kg$

Amount of ice melted, $m=4.8kg$

Time taken, $dt=1hour=3600sec.$

Therefore,

The amount of haet transfer through the slab is given by;

$Q=mL$

$Q=4.8\times 3.36\times {10}^{5}J$

Rate of heat transfer:

$\frac{dQ}{dt}=\frac{4.8\times 3.36\times {10}^5}{3600}$

Dimensions of the slab is given as:

Cross sectional area$A=0.36m^2$

Thickness, $L=0.1m$

Therefore, by Fourier's law:

$\frac{dQ}{dt}=-KA\frac{dT}{L}$

$\frac{4.8\times 3.36\times {10}^5}{3600}=-K\times 0.36\frac{0-100}{0.1}$

Simplifying the above equation we get;

$K=1.24W{K}^{-1}{m}^{-1}$