A rod CD of thermal resistance 5.0 K/W is joined at the middle of an identical rod AB as shown in fig. The ends A, B and D are maintained at 100°C, 0°C and 25°C respectively. The temperature of junction C will be
 

We know that thermal resistance $\mathrm{R}=\mathrm{l}/\mathrm{KA}$
Let heat currents between AC, CB and CD be i, i₁ and. i₂ respectively as shown in the fig. (3).Then $\mathrm{i}={\dot{\mathrm{I}}}_1+{\mathrm{i}}_2$
So,  $\frac{100-\mathrm{T}}{(\mathrm{R}/2)}=\frac{\mathrm{T}-0}{(\mathrm{R}/2)}+\frac{\mathrm{T}-25}{\mathrm{R}}$
or     $200-2\mathrm{T}=2\mathrm{T}+(\mathrm{T}-25)$
or     $225=5\mathrm{T}\text{ or }\mathrm{T}={45}^{\circ }\mathrm{C}$