Question

One end of the copper rod of length 1.0 m and cross section ${10}^{-3}{m}^2$ is immersed in boiling water at ${100}^{o}C$ and the other end at ice $0^{o}C$. If the coefficient of thermal conductivity of copper is $92cal{m}^{-1}{s}^{-1}{}^o{C}^{-1}$ and the latent heat of ice is $80cal{g}^{-1}$, then the amount of ice which will melt in one minute will be

• A. 9.2 g
• B. 8 g
• C. 6.9 g
• D. 5.4 g

Answer 1

The correct option is C 6.9 g

The rate of heat flow through conduction can be given by the equation as shown:
$\frac{dQ}{dt}=\frac{KA}{L}(T_{hot}-T_{cold})$
Substituting the given values, we get the heat flow per second as $9.2$ cal. The heat flow per minute thus becomes $552$ cal. Given the latent heat of ice is $80$ cal/g, the amount of ice that will melt = $552/80$ = $6.9$ grams.