Question

A closed cubical box is made of a perfectly insulating material walls of thickness 8 cm and the only way for heat to enter or leave the box is through two solid metallic cylindrical plugs, each of cross - sectional area $12c{m}^2$ and length 8 cm, fixed in the opposite walls of the box. The outer surface A on one plug is maintained at ${100}^{\circ }$C while the outer surface B of the other plug is maintained at $4^{\circ }$ C. The thermal conductivity of the material of each plug is $0.5cal{/}^{\circ }C/cm$. A source of energy generating 36 cal/s is enclosed inside the box. Assuming the temperature to be the same at all points on the inner surface, the equilibrium temperature of the inner surface of the box is

• A. ${62}^{\circ }$C
• B. ${46}^{\circ }$C
• C. ${76}^{\circ }$C
• D. ${52}^{\circ }$ C

Answer 1

Answer: (C)

${76}^{\circ }$C

Temperature at inner points is same and be T

Total heat current =$H_1+H_2$ (1)

We know that heat current =$\frac{KA\Delta T}{l}$

Where, K= Thermal conductivity

A=Area

$\Delta$T=Temperature difference

l=Length along heat traveled

We are given total heat current = 36cal/second.

$36=H_1+H_2$

$36=\frac{KA(T-100)}{8}+\frac{KA(T-4)}{8}$

$36=\frac{0.5\times 12(T-100)}{8}+\frac{0.5\times 12(T-4)}{8}$

$36=\frac{6}{8}(T-100)+\frac{6}{8}(T-4)$

$48=2T-104$

$2T=152$

$T=76°C$