Question

A slab of stone area $3500{cm}^2$ and thickness $10cm$ is exposed on the lower surface to steam at ${100}^{o}C$. A block of ice at $0^{o}C$ rests on upper surface of the slab. In one hour $4.8kg$ of ice of melted. The thermal conductivity of the stone is $J{s}^{-1}$ $m^{-1}$ $k_{-1}$ is
(Latent heat of ice $=3.36\times {10}^{5}J/kg$)

• A. $12.0$
• B. $10.5$
• C. $1.02$
• D. $1.24$

Answer 1

Answer: (D)

$1.24$

Given : $A=3500c{m}^2=0.35m^2$ $l=10cm=0.1m$ $\Delta T=100-0={100}^{o}C$
Mass of ice melted $m=4.8kg$
Time taken $t=1hr=3600s$
Latent heat of ice $L=3.36\times {10}^{5}J/kg$
Heat absorbed by ice = Heat conducted by slab
$\therefore$ $mL=\frac{KAt\Delta T}{l}$
Or $4.8\times 3.36\times {10}^5=\frac{K\left(0.35\right)\left(3600\right)\left(100\right)}{0.1}$
$⟹K=1.24J{s}^{-1}{m}^{-2}{k}^{-1}$