Question

One end of a copper rod of length 1.0 m and area of cross-section ${10}^{-3}$ is immersed in boiling water and the other end in ice. If the coefficient of thermal conductivity of copper is $92cal/m-s{-}^{\circ }C$ and the latent heat of ice is $8\times {10}^{4}cal/kg$, then the amount of ice which will melt in one minute is

• A. $9.2\times {10}^{-3}kg$
• B. $8\times {10}^{-3}kg$
• C. $6.9\times {10}^{-3}kg$
• D. $5.4\times {10}^{-3}kg$

Answer 1

The correct option is C

$6.9\times {10}^{-3}kg$

Heat transferred in one minute is utilised in melting the ice so, $\frac{KA({\theta }_1-{\theta }_2)t}{l}=m\times L$
$\Rightarrow m=\frac{{10}^{-3}\times 92\times (100-0)\times 60}{1\times 8\times {10}^4}=6.9\times {10}^{-3}kg$