Question

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 $92\text{ }cal\text{/}m-s\text{ }{}^\circ C$ and the latent heat of ice is $8\times {{10}^{4}}cal\text{/}kg,$ then the amount of ice which will melt in one minute is:

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

Answer 1

Answer: (D)

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

Answer 2

From the formula $\frac{d\theta }{dt}=mL$ where t = 1 m, m = mass, L = latent heat $\frac{KA({{\theta }_{1}}-{{\theta }_{2}})t}{l}=mL$ $m=\frac{KA({{\theta }_{1}}-{{\theta }_{2}})\times 60}{1\times 8\times {{10}^{4}}}$ $m=\frac{{{10}^{-3}}\times 92\times (100-0)\times 60}{1\times 8\times {{10}^{4}}}$ $=6.9\times {{10}^{-3}}kg$