Question

An aluminium rod and a copper rod of length $1.0\text{m}$ and cross-sectional area $1{\text{cm}}^2$ each, are welded together as shown in figure. One end of this composite rod is kept at a temperature of ${20}^{\circ }\text{C}$, while the other end is at ${60}^{\circ }\text{C}$. Calculate the amount of heat taken out per second from the hot end. Thermal conductivity of aluminium $=200{\text{W/m}}^{\circ }\text{C}$ and that of copper $=300{\text{W/m}}^{\circ }\text{C}$

• A. $1\text{W}$
• B. $2\text{W}$
• C. $0.1\text{W}$
• D. $0.02\text{W}$

Answer 1

The correct option is B $2\text{W}$

Given:
$K_{Al}=200{\text{W/m}}^{\circ }\text{C}$
$K_{cu}=300{\text{W/m}}^{\circ }\text{C}$
length $l=1\text{m},A=1{\text{cm}}^2$
$\Delta T=({60}^{\circ }\text{C}-{20}^{\circ }\text{C})=40{}^{\circ }\text{C}$
As the aluminium rod and copper rod are joined in parallel,

$\therefore$$\frac{1}{R_{eq}}=\frac{1}{R_{Al}}+\frac{1}{R_{Cu}}$

$\Rightarrow \frac{K_{eq}}{L}(A_1+A_2)=\frac{K_{Al}{A}_1}{L_1}+\frac{K_{Cu}{A}_2}{L_2}$

As, $A_1=A_2$ and $L_1=L_2=L$

$\therefore K_{eq}=\frac{K_{Al}+K_{Cu}}{2}$
$=\frac{200+300}{2}=250{\text{W/m}}^{\circ }\text{C}$

Now, amount of heat flowing per unit time, through the composite rod, is,
$\frac{q}{t}=\frac{K_{eq}\times 2A\times \left(\Delta T\right)}{L}$
$\frac{q}{t}=\frac{250\times (2\times {10}^{-4})\times \left(40\right)}{1}=2\text{W}$

Hence, $\left(A\right)$ is the correct answer.