A partition wall has two layers A and $\mathbf{B}$ in contact, each made of a different material. They have the same thickness but the thermal conductivity of layer A is twice that of layer $\mathbf{B}$ . If the steady state temperature difference across the wall is 60K , then the corresponding difference across the layer A is
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Suppose conductivity of layer B is K, then it is 2K for layer A. Also conductivity of
combination layers A and B is K$_S$ = $\frac{2 \times 2 K \times K}{(2K + K)}$ = $\frac{4}{3}$ K Hence $\left(\frac{Q}{t}\right)_{\text{Combination}} = \left(\frac{Q}{t}\right)_A$ $\Rightarrow \frac{4}{3} \frac{K A \times 60}{2 x} = \frac{2 K . A \times (\Delta \theta)_A}{x} \Rightarrow (\Delta \theta)_A = 20 K$
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