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Conduction Law

If two conduc...

Question

If two conducting slabs of thickness $d_{1}$ and $d_{2}$ , and thermal conductivity $K_{1}$ and $K_{2}$ are placed together face to face as shown in figure in the steady state temperatures of outer surfaces are $O_{1}$ , and $O_{2}$ . The temperature of common surface is

A

\frac{K_{1} O_{1} d_{1} + K_{2} O_{2} d_{2}}{K_{1} d_{1} + K_{2} d_{2}}

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B

\frac{K_{1} O_{1} + K_{2} O_{2}}{K_{1} + K_{2}}

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C

\frac{K_{1} O_{1} + K_{2} O_{2}}{O_{1} + O_{2}}

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D

\frac{K_{1} O_{1} d_{2} + K_{2} O_{2} d_{1}}{K_{1} d_{1} + K_{2} d_{2}}

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Solution

The correct option is D $\frac{K_{1} O_{1} d_{2} + K_{2} O_{2} d_{1}}{K_{1} d_{1} + K_{2} d_{2}}$
In steady state of temperature , the rate of flow of heat in both the slabs will be same .
Let $θ$ be the temperature of common surface ,
then ,
$Q / t = K_{1} A (O_{1} - θ) / d_{1} = K_{2} A (θ - O_{2}) / d_{2}$ ,
or $\frac{K_{1}}{d_{1}} . (O_{1} - θ) = \frac{K_{2}}{d_{2}} . (θ - O_{2})$ ,
or $\frac{K_{1} O_{1}}{d_{1}} + \frac{K_{2} O_{2}}{d_{2}} = θ (\frac{K_{1}}{d_{1}} + \frac{K_{2}}{d_{2}})$ ,
or $θ = \frac{K_{1} O_{1} d_{2} + K_{2} O_{2} d_{1}}{K_{1} d_{1} + K_{2} d_{2}}$

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