Two bars of same length and same cross -sectional area but of different thermal conductivities $K_1$ and $K_2$ are joined end to end as shown in the figure .One end of the compound bar is at temperature $T_1$ and the opposite end at temperature $T_2$ $(where T_1 > T_2)$.

The temperature of the junction is

Let $L$ and $A$ be length and area of cross-section of each bar respectively.

$\therefore$ Heat current through the bar 1 is

$H_1=\dfrac{K_1A(T_1-T_0)}{L}$

At steady state,$H_1=H_2$

$\therefore \dfrac{K_1A(T_1-T_0)}{L}=\dfrac{K_2A(T_0-T_2)}{L}$

$K_1(T_1-T_0)=K_2(T_0-T_2)$

$K_1T_1-K_1T_0=K_2T_0-K_2T_2$

$K_1T_0+K_2T_0=K_1T_1+K_2T_2$

$T_0(K_1+K_2)=K_1T_1+K_2T_2$

$T_0=\dfrac{K_1T_1+K_2T_2}{(K_1+K_2)}$