A closed cubical box made of perfectly insulating materials has walls of thickness `8` cm and the only way for heat to enter or leave the box is through two solid metal plugs `A` and `B`, each of cross-sectional area `12 cm^(2)` and length `8`cm fixed in the opposite walls of the box as shown in the figure. Outer surface `A` is kept at `100^(@)C` while the outer surface `B` is kept at `4^(@)C`. The thermal conductivity of the materials of the plugs is `0.5 cals^(-1)cm^(-1) ("^(@)C^(-1))` . A source of energy generating `36 cals^(-1)` is enclosed inside the box . The equilibrium temperature of the inner surface of the box (assuming that it is same at all points on the inner surface) is :-

Explanation

The situation is show in figure. Let the temperature inside the box be (theta). The rate at which heat enters the box through the left plug is
`(DeltaQ_(1))/(Deltat)=(KA(theta_(1)-(theta)))/(x)` . The rate of heat generation in the box `=13W` . The rate at which heat fkows out of the box through the right pluyg is `(DeltaQ_(2))/(Deltat)=(KA(theta-theta_(2)))/(x)` . In the steady state
`(DeltaQ_(1))/(Deltat)+13W=(DeltaQ_(2))/(Deltat)` . or, `(KA)/(x)(theta_(1)-theta)+13W=(KA)/(x)(theta-theta_(2))` . or, `2(KA)/(x)theta=(KA)/(x)(theta_(1)+theta_(2))+13W` . or, `theta=theta_(1)+theta_(2)/(2)+((13W)xx0.08m)/(2xx(2.0Wm^(-1)`^(@)C^(-1)(12xx10^(-4)m^(2))` . `=52^(@)C+216.67^(@)C=269^(@)C` .