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First Law of Thermodynamics

Two walls of ...

Question

Two walls of thickness $l_{1}$ and $l_{2}$ and thermal conductivities $k_{1}$ and $k_{2}$ are in contact. In the steady state, if the temperature at the outer faces are $T_{1}$ and $T_{2}$ , the temperature at the common wall is :

A

\frac{k_{1} T_{1} l_{2} + k_{2} T_{2} l_{1}}{k_{1} l_{2} + k_{2} l_{1}}

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B

\frac{k_{1} T_{1} + k_{2} T_{2}}{l_{1} + l_{2}}

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C

(\frac{k_{1} l_{1} + k_{2} l_{2}}{T_{1} + T_{2}}) T_{1} T_{2}

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D

\frac{k_{1} l_{1} T_{1} - k_{2} l_{2} T_{2}}{k_{1} l_{1} + k_{1} l_{2}}

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Solution

The correct option is B $\frac{k_{1} T_{1} l_{2} + k_{2} T_{2} l_{1}}{k_{1} l_{2} + k_{2} l_{1}}$
Let $T$ be the intermediate temperature of the wall. So heat transfer from the wall is given as,
$Q = \frac{K A (T_{1} - T_{2})}{L}$
Here area of both the wall is equal, so for equilibrium condition equate both of them:

$\frac{K_{1} (T_{1} - T)}{l_{1}} = \frac{K_{2} (T - T_{2})}{l_{2}}$
On solving this we get, $T = \frac{K_{1} T_{1} l_{2} + K_{2} T_{2} l_{1}}{K_{1} l_{2} + K_{2} l_{1}}$

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