Three conductors of same length having thermal conductivity $k_1$, $k_2$ and $k_3$ are connected as shown in the figure. Area of cross sections of $1^{\text{st}}$ and $2^{\text{nd}}$ conductor are same and for $3^{\text{rd}}$ conductor it is double of the $1^{\text{st}}$ conductor. The temperatures are given in the figure. In steady state condition, the value of $\theta$ is ______ $^\circ\text{C}$. (Given: $k_1 = 60\ \text{J s}^{-1}\text{m}^{-1}\text{K}^{-1}$, $k_2 = 120\ \text{J s}^{-1}\text{m}^{-1}\text{K}^{-1}$, $k_3 = 135\ \text{J s}^{-1}\text{m}^{-1}\text{K}^{-1}$)
Numerical value type. Enter your answer.
Answer: 40
Conductors 1 and 2 are in parallel between $100^\circ\text{C}$ and $\theta$; conductor 3 (area $2A$) carries the same total heat current from $\theta$ to $0^\circ\text{C}$. With equal lengths $l$:
$$\frac{(k_1+k_2)A(100-\theta)}{l} = \frac{k_3(2A)(\theta-0)}{l}$$
$$180(100-\theta) = 270\,\theta \Rightarrow 18000 = 450\,\theta \Rightarrow \theta = 40^\circ\text{C}$$
Solution by Sreeraj P, M.Sc Physics