Three conductors of equal length with thermal conductivities k_1=60, k_2=120, k_3=135 (all in text Js ^ -1 text m ^ -1 text K ^ -1 ) are connected as shown. Conductors 1 and 2 have the same cross-sectional area A, while conductor 3 has cross-sectional area 2A. If the ends are maintained at 100°C and 0°C with the junction at theta°C (as shown), find theta.

Home › MCQs › EEJ MAIN Physics › Question #5510

EEJ MAIN Physics QUESTION #5510

Question 1 (Enter numeric answer)

Three conductors of equal length with thermal conductivities \(k_1=60\), \(k_2=120\), \(k_3=135\) (all in \(\text{Js}^{-1}\text{m}^{-1}\text{K}^{-1}\)) are connected as shown. Conductors 1 and 2 have the same cross-sectional area \(A\), while conductor 3 has cross-sectional area \(2A\). If the ends are maintained at 100°C and 0°C with the junction at \(\theta\)°C (as shown), find \(\theta\).

Correct Answer:

Correct Answer Explanation

Conductors 1 and 2 are in parallel (both connecting the 100°C end to the junction at \(\theta\)°C), and conductor 3 connects the junction to 0°C. Let length be \(L\). Heat flow rate through conductor 1: \(H_1=\frac{k_1 A(100-\theta)}{L}=\frac{60A(100-\theta)}{L}\). Heat flow through conductor 2: \(H_2=\frac{k_2 A(100-\theta)}{L}=\frac{120A(100-\theta)}{L}\). Total heat into junction: \(H_{in}=(H_1+H_2)=\frac{180A(100-\theta)}{L}\). Heat flow through conductor 3 (area \(2A\)): \(H_3=\frac{k_3\cdot2A\cdot\theta}{L}=\frac{135\times2A\theta}{L}=\frac{270A\theta}{L}\). At steady state: \(H_{in}=H_3\): \(180(100-\theta)=270\theta\Rightarrow18000-180\theta=270\theta\Rightarrow18000=450\theta\Rightarrow\theta=40\,°\text{C}\).

More Options