Question

An aluminium rod and a copper rod of equal length 1.0 m and cross-sectional area 1 cm² are welded together as shown in figure (28−E3). One end is kept at a temperature of 20°C and the other at 60°C. Calculate the amount of heat taken out per second from the hot end. Thermal conductivity of aluminium = 200 W m⁻¹°C⁻¹ and of copper = 390 W m⁻¹°C⁻¹.

Answer 1

q₁ and q₂ are heat currents. In other words, they are the rates of flow of heat through aluminium and copper rod, respectively.

Applying KVL at the hot junction, we get
q = q₁ + q₂
Rate of heat flow, q = $\frac{KA∆T}{l}$

As q = q₁ + q₂,

$$\begin{gathered} \frac{K_{\mathrm{P}}A\left(T_1-T_2\right)}{l}=\frac{K_{1}A\left(T_1-T_2\right)}{l}+\frac{K_{2}A\left(T_1-T_2\right)}{l} \\ Kp=K_1+K_2 \\ =390+200=590\mathrm{W}/\mathrm{m}°\mathrm{C} \\ \Rightarrow q=\frac{K_{P}A\left(T_1-T_2\right)}{l} \\ q=\frac{590\times {10}^{-4}\left(60-20\right)}{1} \\ q=2.36\mathrm{W} \end{gathered}$$