Question

One end of a $0.25m$ long metal bar is in steam and the other is in contact with ice .If $12g$ of ice melts per minute, then the thermal conductivity of the metal is (Given cross section of the bar$=5\times {10}^{-4}{m}^2$ and latent heat of ice is $80cal{g}^{-1}$)

• A. $20cal{s}^{-1}{m}^{-1}{}^o{C}^{-1}$
• B. $10cal{s}^{-1}{m}^{-1}{}^o{C}^{-1}$
• C. $40cal{s}^{-1}{m}^{-1}{}^o{C}^{-1}$
• D. $30cal{s}^{-1}{m}^{-1}{}^o{C}^{-1}$

Answer 1

The correct option is D $30cal{s}^{-1}{m}^{-1}{}^o{C}^{-1}$

Given: The length of the metal bar is $x=0.25m$

The change temperature across the ends of the rod is $T_1-T_2=100-0={100}^{o}C$
The time taken to melt the ice is $t=1min=60s$

$A=5\times {10}^{-4}{m}^2$ $,L=80cal{g}^{-1}$

The amount of heat required by the ice is:

$Q=mL=12\times 80=960cal$

But,$Q=\frac{KA(T_1-T_2)t}{x}$ or

$960=\frac{K\times 5\times {10}^{-4}\times 100\times 60}{0.25}$

$\therefore K=\frac{960\times 0.25}{5\times {10}^{-2}\times 60}$

$\Rightarrow 30cal{s}^{-1}{m}^{-1}{}^o{C}^{-1}$