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

The Correct Option is D

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 $