Question

Three conductors of same length having thermal conductivity $k_1, k_2$ and $k_3$ are connected as shown in figure. Area of cross sections of $1^{st}$ and $2^{nd}$ conductor are same and for $3^{rd}$ conductor it is double of the $1^{st}$ conductor. The temperatures are given the figure. In steady state condition, the value of $\theta$ is ____ $^0C$.

(Given : $k_1 = 60Js^{-1}m^{-1}K^{-1}, k_2 = 120Js^{-1}m^{-1}K^{-1}, k_3 = 135Js^{-1}m^{-1}K^{-1}$)

Answer 1

40

Answer 2

In steady state, the rate of heat flow into a junction equals the rate of heat flow out of it. The rate of heat flow through a conductor is given by $H = \frac{kA\Delta T}{L}$. Conductors 1 and 2 are connected between $100^\circ C$ and $\theta^\circ C$, so the heat flowing into the junction from them is $H_1 + H_2$. Conductor 3 is connected between $\theta^\circ C$ and $0^\circ C$, so the heat flowing out of the junction through it is $H_3$. Thus, $H_1 + H_2 = H_3$. Substituting the given values and properties: $\frac{k_1 A (100 - \theta)}{L} + \frac{k_2 A (100 - \theta)}{L} = \frac{k_3 (2A) (\theta - 0)}{L}$ Cancelling $A/L$ from all terms and substituting $k_1=60, k_2=120, k_3=135$: $60(100 - \theta) + 120(100 - \theta) = 2 \times 135 \theta$ $180(100 - \theta) = 270 \theta$ $18000 - 180 \theta = 270 \theta$ $18000 = 450 \theta$ $\theta = \frac{18000}{450} = 40^\circ C$