Two metal slabs of same cross-sectional area have thicknesses $d_1$ and ; and thermal conductivities $K_1$…
Two metal slabs of same cross-sectional area have thicknesses $d_1$ and ; 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 $T_2\left(T_1>T_2\right)$. The temperature $T$ of their common junction is
$\frac{K_1 T_1 d_2+K_2 T_2 d_1}{K_1 d_2+K_2 d_1}$
$\frac{K_1 T_1+K_2 T_2}{K_1+K_2}$
$\frac{K_1 T_1+K_2 T_2}{T_1+T_2}$
$\frac{K_1 T_1 d_1+K_2 T_2 d_2}{K_1 d_2+K_2 d_1}$
Solution
Heat current, $\dot{Q}_1=\frac{K_1\left(T_1-T\right) A}{d_1}$
For second slab,
Heat current, $\dot{Q}_2=\frac{K_2\left(T-T_2\right) A}{d_2}$
As slabs are in series same heat current flows through them
$\dot{Q}_1=\dot{Q}_2$
$\therefore \frac{K_1\left(T_1-T\right) A}{d_1}=\frac{K_2\left(T-T_2\right) A}{d_2}$
Therefore, temperature $T$ of their common junction is
$T=\frac{K_1 T_1 d_2+K_2 T_2 d_1}{K_2 d_1+K_1 d_2}$