Question

A closed cubicle box made of perfectly insulating material has walls of thickness $8$ cm and the only way for heat to enter and leave the box is through the metal plugs $A$ and $B$ each of cross section $12c{m}^2$ and length $8$ cm fixed in the opposite walls of the box as shown in figure. Outer surface of $A$ is kept at ${100}^{o}C$ while outer surface of $B$ is at $4^{o}C$. The thermal conductivity of the material of the plugs is $0.5cal/s/cm/{}^{o}C$. A source of energy generating $36$cal/s is enclosed inside the box. The equillibrium temperature of the inner surface of the box (assuming that it is same at all points on the inner surface) is

• A. ${38}^{o}C$
• B. ${57}^{o}C$
• C. ${76}^{o}C$
• D. ${85}^{o}C$

Answer 1

Answer: (C)

${76}^{o}C$

The principle of conservation of energy is used to solve this question.
Heat from $A$ to source + $36$ = Heat from source to $B$.
$\frac{KA}{L}(100-T)+36=\frac{KA}{L}(T-4)$
$T$=Temp of source
$K$=0.5 cal/s/cm/degC
$L=8cm$
$A=12c{m}^2$
Substituting the above values, we get $T=76{}^{o}C$