Two walls of thicknesses $d_1$ and $d_2$ and thermal conductivities $k_1$ and $k_2$ are in contact. In the steady state, if the temperatures at the outer surfaces are ${T_1}$ and ${T_2}$, the temperature at the common wall is
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(a) In series both walls have same rate of heat flow. Therefore
$\frac{{dQ}}{{dt}} = \frac{{{K_1}A({T_1} - \theta )}}{{{d_1}}} = \frac{{{K_2}A(\theta - {T_2})}}{{{d_2}}}$
$\frac{{dQ}}{{dt}} = \frac{{{K_1}A({T_1} - \theta )}}{{{d_1}}} = \frac{{{K_2}A(\theta - {T_2})}}{{{d_2}}}$
$ \Rightarrow {K_1}{d_2}({T_1} - \theta ) = {K_2}{d_1}(\theta - {T_2})$
$ \Rightarrow \theta = \frac{{{K_1}{d_2}{T_1} + {K_2}{d_1}{T_2}}}{{{K_1}{d_2} + {K_2}{d_1}}}$
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