Question

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

• A. $75\text{W}$
• B. $200\text{W}$
• C. $400\text{W}$
• D. $4\text{W}$

Answer 1

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

We need $i_b$

$i=\frac{\Delta T}{R_{eq}}$, $R=\frac{l}{KA}$

$\frac{1}{R_{eq}}=\frac{1}{R_{A1}}+\frac{1}{R_{cu}}+\frac{1}{R_{A1}}$

$\frac{1}{R_q}=\frac{k_{Al}A}{l}+\frac{k_{cu}A}{l}+\frac{k_{Ak}A}{I}$

$\frac{1}{R_{eq}}=\frac{A}{l}\times 800$

$i_b=100\times \left(\frac{A}{l}\right)\times 800$

But given $i_a=40=\frac{\Delta T}{R_{eq}}$

$\begin{array}{cc}{R}_{eq}& =R_{A1}+R_{cu}+R_{Al}\\ & =2R_{Al}+R_{Cu}\\ & =\frac{2l}{K_{Al}A}+\frac{l}{K_{cu}A}=\frac{l}{80A}\end{array}$

$⟹\frac{A}{l}=0.005$

$\therefore i_b=100\times 0.005\times 800=400W$