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The figure shows the cross-section of the outer wall of a house built in a hill - resort to keep the house insulated from the freezing temperature of outside. The wall consists of teak wood of thickness $L_{1}$ and brick of thickness ( $L = 5 L_{1}$ ). sandwiching two layers of an unknown material with identical thermal conductivities and thickness. The thermal conductivity of teak wood is $K_{1}$ and that of brick is ( $K_{2} = 5 K$ ). Heat conduction through the wall has reached a steady-state with the temperature of three surfaces being known. $(T_{1} = 25^{\circ} C, T_{2} = 20^{\circ} C$ and $T_{5} = - 20^{\circ}$ C). Find the interface temperature $T_{4}$ and $T_{3}$ .

A

2.5

^{o}

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B

7

^{o}

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C

3.5

^{o}

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D

9

^{o}

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Solution

The correct option is A 2.5 $^{o}$
Let interface area be A. then thermal resistance of wood,
$R_{1} = \frac{L_{1}}{K_{1} A}$
and that of brick wall
$R_{2} = \frac{L_{2}}{K_{2} A} = \frac{5 L_{1}}{5 K_{1} A} = R_{1}$
Let thermal resistance of the each sand witch layer = R. Then the above wall can be visualised as a circuit
thermal current through each wall is same.
Hence $\frac{25 - 20}{R_{1}} = \frac{20 - T_{3}}{R} = \frac{T_{3} - T_{4}}{R} = \frac{T_{4} + 20}{R_{1}}$
$\Rightarrow 25 - 20 = T_{4} + 20$
$\Rightarrow$ $T_{4} = - 15^{\circ}$ C
Also, $20 - T_{3} = T_{3} - T_{4}$
$\Rightarrow$ $T_{3} = \frac{20 + T_{4}}{2} = {2.5}^{\circ}$ C

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