Question

A slab of stone area 3600 cm${}^2$ and thickness 10 cm is exposed on the lower surface to steam at 100${}^{0}C$. A block of ice at $0^{0}C$ rest on upper surface of the slab. In one minute, 4.8 kg of ice is melted. The thermal conductivity of the stone in $J{s}^{-1}{m}^{-1}{k}^{-1}$ is (latent heat of ice =$3.36\times {10}^5$J/kg)

Answer 1

In Time $t=60s,4.8kg$ of ice in melted.

Heat consumed in process $=mL=4.8\times 3.36\times {10}^5$

$△Q=1.6128\times {10}^{6}J$

From Thermal transfer also,

$\frac{△Q}{△t}=\frac{KA}{L}△T$ Here $△T=$ temp diff $=100K$.

$\Rightarrow K=\frac{△Q}{△t}.\left(\frac{L}{A}\right).\frac{1}{△T}$ given,

$=\frac{1.6128\times {10}^6}{60}\times \frac{0.1}{0.36}\times \frac{1}{100}=74.67J/smK$