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. $20cal{s}^{-1}{m}^{-1}{}^{\circ }{C}^{-1}$
• B. $10cal{s}^{-1}{m}^{-1}{}^{\circ }{C}^{-1}$
• C. $40cal{s}^{-1}{m}^{-1}{}^{\circ }{C}^{-1}$
• D. $80cal{s}^{-1}{m}^{-1}{}^{\circ }{C}^{-1}$

Answer 1

Answer: (D)

Here, $x=0.25m$,
$T_1-T_2=100-0=100{}^{\circ }C$
$t=1min=60s$,
$A=5\times 10^{-4}{m}^2$,
$L=80cal{g}^{-1}$
$Q=mL=12\times 80$
$=960cal$
But, $Q=\frac{KA\left(T_1-T_2\right)t}{x}$
or $960=\frac{K\times 5\times 10^{-4}\times 100\times 60}{0.25}$
$\therefore K=\frac{960\times 0.25}{5\times 10^{-2}\times 60}$
$=80cal{s}^{-1}{m}^{-1}{}^{\circ }{C}^{-1}$