A closed cubical box is made of perfectly insulating material and the only way for heat to enter or leave the box is through two solid cylindrical metal plugs, each of cross–sectional area $12 c m^{2}$ and length 8 cm fixed in the opposite walls of the box. The outer surface of one plug is kept at a temperature of $100 ° C,$ while the outer surface of other plug is maintained at a temp of $4 ° C$ .A source of energy 13W is enclosed inside the box. Find the equilibrium temperature of the inner surface of the box assuming that it is same at all points on the inner surface .
(Thermal conductivity of the material of the plug is $2.0 W / m - ° C$ )

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$268.67 ° C$

$364 ° C$

$568 ° C$

$292 ° C$

answer is A.

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Detailed Solution

$H_{1} = \frac{K A (θ_{1} - θ)}{x}, H_{2} = \frac{K A (θ - θ_{2})}{x}$

$H_{1} + P = H_{2}$

$\frac{K A (θ_{1} - θ)}{x} + P = \frac{K A (θ - θ_{2})}{x}$

$2 \times \frac{(x \times 10^{- 4}) (100 - θ)}{0.08} + 13 = \frac{2 \times (12 \times 10^{- 4}) \times (θ - 4)}{0.08}$

$3 \times 10^{- 2} (100 - θ) = 3 \times 10^{- 2} (θ - 4) - 13$

$\Rightarrow θ = 268.67 ° C$

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