Two metal slabs of same cross-sectional area | vedclass

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Question

(A) The heat current through the first slab is given by $\dot{Q}_1 = \frac{K_1 A (T_1 - T)}{d_1}$.For the second slab,the heat current is $\dot{Q}_2 =

Vedclass · Accepted Answer

$\frac{K_1 T_1 d_2 + K_2 T_2 d_1}{K_1 d_2 + K_2 d_1}$

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(A) The heat current through the first slab is given by $\dot{Q}_1 = \frac{K_1 A (T_1 - T)}{d_1}$.For the second slab,the heat current is $\dot{Q}_2 = \frac{K_2 A (T - T_2)}{d_2}$.Since the slabs are connected in series,the same heat current flows through both,so $\dot{Q}_1 = \dot{Q}_2$.Equating the two expressions: $\frac{K_1 A (T_1 - T)}{d_1} = \frac{K_2 A (T - T_2)}{d_2}$.Canceling $A$ from both sides: $\frac{K_1 (T_1 - T)}{d_1} = \frac{K_2 (T - T_2)}{d_2}$.Cross-multiplying gives: $K_1 d_2 (T_1 - T) = K_2 d_1 (T - T_2)$.Expanding: $K_1 d_2 T_1 - K_1 d_2 T = K_2 d_1 T - K_2 d_1 T_2$.Rearranging to solve for $T$: $K_1 d_2 T_1 + K_2 d_1 T_2 = T (K_1 d_2 + K_2 d_1)$.Thus,$T = \frac{K_1 T_1 d_2 + K_2 T_2 d_1}{K_1 d_2 + K_2 d_1}$.

Two metal slabs of same cross-sectional area have thicknesses $d_1$ and $d_2$,and thermal conductivities $K_1$ and $K_2$ respectively,are connected in series. The free ends of the two slabs are kept at temperatures $T_1$ and $T_2$ $(T_1 > T_2)$. The temperature $T$ of their common junction is

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