Question

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

Answer 1

Temperature at inner points are 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

$36=H-1+H_2$

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

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

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

$48=2T-104$

$2T=152$

$T=76°C$