Question

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

• 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

Heat transferred to ice in time ‘t’ is
$Q=\left(\frac{KA\Delta \theta }{l}\right)\times t$
If ‘m’ is the mass of ice melted, then θ = mL. Therefore,
$mL=\left(\frac{KA\Delta \theta }{l}\right)t$
$\Rightarrow m=\frac{92\times {10}^{-3}\times 100}{1.0}\times \frac{60}{80}=6.9gm$