Two identical rods AB and CD, .each of length L, cross-sectional area A and thermal conductivity k are connected as shown in Fig. Ends A, C and D are maintained at temperatures T₁ = 20°C, T₂ = 30°C and T₃ = 40°C, respectively. The temperature at B is 

 

Let $T_B$ be the temperature at B. The rate of flow of heat from C towards B is

        $\frac{Q_1}{t}=\frac{kA\left(T_2-T_B\right)}{L/2}$

The rate of flow of heat from D towards B is

        $\frac{Q_2}{t}=\frac{kA\left(T_3-T_B\right)}{L/2}$

The rate of flow of heat from B towards A is

        $\frac{Q_3}{t}=\frac{kA\left(T_B-T_1\right)}{L}$

In the steady state, the rate at which heat enters B = rate at which heat leaves B, i.e.

         $\frac{Q_1}{t}+\frac{Q_2}{t}=\frac{Q_3}{t}$

or $\frac{2kA\left(T_2-T_B\right)}{L}+\frac{2kA\left(T_3-T_B\right)}{L}=\frac{kA\left(T_B-T_1\right)}{L}$

or      $2\left(T_2-T_B\right)+2\left(T_3-T_B\right)=T_B-T_1$

which gives    $T_B=\frac{T_1+2T_2+2T_3}{5}$

                            $$\begin{gathered} =\frac{20+2\times 30+2\times 40}{5} \\ ={32}^{\circ }\mathrm{C} \end{gathered}$$