Question

One end of a $0.25 \,m$ long metal bar is in steam and the other is in contact with ice. If $12\, g$ 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 $80 \,cal \,g^{-1}$).

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

Answer 1

Answer: (D)

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