Q. One end of a copper rod of length $1.0\ m$ and area of cross-section $10^{- 3}m^{2}$ is immersed in boiling water and the other end in ice. If the coefficient of thermal conductivity of copper is $92\ cal/m$-$s$-$℃$ 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) $5.4 \times 10^{- 3}kg$
Explanation
(c) Heat transferred in one minute is utilised in melting the ice so, $\frac{KA\left( \theta_{1} - \theta_{2} \right)t}{l} = m \times L$
$\Rightarrow m = \frac{10^{- 3} \times 92 \times (100 - 0) \times 60}{1 \times 8 \times 10^{4}} = 6.9 \times 10^{- 3}kg$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.