Question

The three rods shown in the figure have identical dimensions. Heat flows from the hot end at a rate of $40\text{W}$ in the arrangement (a). Find the rate of heat flow when the rods are joined as in the arrangement shown in (b).
[Assume $K_{AI}=200{\text{W/m}}^{\circ }\text{C}$ and $K_{cu}=400{\text{W/m}}^{\circ }\text{C}$]

• A. $75\text{W}$
• B. $400\text{W}$
• C. $180\text{W}$
• D. $4\text{W}$

Answer 1

The correct option is B $400\text{W}$

Let the area of cross-section of each rod be $A$ and length be $L$.

Case (a) When the rods are connected in series.
So each rod would have a heat resistance of $R=\frac{L}{KA}$
Since the rods are connected in series,
$R_{eff}=R_1+R_2+R_3=\frac{L}{K_{Al}A}+\frac{L}{K_{Cu}A}+\frac{L}{K_{Al}A}$
But, $R_{eff}=\frac{3L}{K_{eq}A}$
$\therefore$ We can write the above equation as
$\frac{3}{K_{eq}}=\frac{1}{K_{Al}}+\frac{1}{K_{Cu}}+\frac{1}{K_{Al}}$
$\Rightarrow \frac{3}{K_{eq}}=\frac{1}{200}+\frac{1}{400}+\frac{1}{200}=0.0125{\text{W/m}}^{\circ }\text{C}$
We know that, rate of heat flow $H=\frac{K_{eq}A(T_2-T_1)}{3L}$
$\Rightarrow H=\frac{A\times 100}{L\times 0.0125}$
$\Rightarrow H=\frac{8000\times A}{L}$
Given, $H=40\text{W}$
Thus, $\frac{A}{L}=0.005\dots \left(1\right)$

(b) When the rods are connected in parallel,
$\frac{1}{R_{eff}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}$
But, $R_{eff}=\frac{L}{K_{eq}\times 3A}$
$\Rightarrow 3K_{eq}=K_{Cu}+2K_{Al}$
$\Rightarrow K_{eq}=(400+400)/3=\frac{800}{3}{\text{W/m}}^{\circ }\text{C}$
We know that, rate of heat flow
$H=\frac{K_{eq}\times 3A(T_2-T_1)}{L}$
​​​​​​​From the figure, we can say that
$H=\frac{800\times 3A\times 100}{3L}$
$\Rightarrow H=800\times 0.005\times 100$ (using (1))
$\Rightarrow H=400\text{W}$
Thus, option (b) is the correct answer.