Question

One end of a steel rod $(K=46J{s}^{-1}{m}^{-1\circ }{C}^{-1})$ of length 1.0 m is kept in ice at $0^{\circ }$ and the other end is kept in boiling water at ${100}^{\circ }$. The area of cross section of the rod is $0.04c{m}^2$. Assuming no heat loss to the atmosphere, find the mass of the ice melting per second. Latent heat of fusion of ice = $3.36\times {10}^{5}JK{g}^{-1}$.

Answer 1

K = 46 $J/m-s{-}^{\circ }C$ , l = 1 m ,

A = $0.04c{m}^2=4\times {10}^{-6}{m}^2$

$\frac{Q}{t}=\frac{kA(T_1-T_2)}{l}$

=$\frac{46\times 4\times {10}^6\times (100-0)}{1}$

= $184\times {10}^{-4}$

mL =$184\times {10}^{-4}$

m = $\frac{184\times {10}^{-4}}{L}$

m = $\frac{184\times {10}^{-4}}{3.36\times {10}^5}$

= 5.5 $\times {10}^{-5}g$.