Question

Three rods of copper, brass and steel are welded together to form an inverted Y-shaped structure. End of copper rod is maintained at $100^\circ C$ whereas ends of brass and steel are kept at $0^\circ C.$ All the three rods have same length of $9cm$ and cross-section of $4.5cm^{2}.$ If the rate of heat flow through copper rod is $xcal/s$ . FInd $10x$ .
Given that the rods are thermally insulated from surroundings except at ends.
(Take thermal conductivities of copper, brass and steel to be $0.9,0.24$ and $0.11CGS$ units respectively)

Answer 1

$Q=Q_{1}+Q_{2}....\left(i\right)$
But $Q=KA\left(\frac{\Delta \theta }{\Delta x}\right)t$
Here, for copper rod, $\frac{Q}{t}=\frac{0 . 9 \times 4 . 5 \times \left(\right. 100 - T \left.\right)}{9}$
For brass and steel rods,
$\frac{Q_{1} + Q_{2}}{t}=\frac{\left[\right. 0 . 24 \times 4 . 5 \times \left(\right. T - 0 \left.\right) \left]\right. + \left[\right. 0 . 11 \times 4 . 5 \times \left(\right. T - 0 \left.\right) \left]\right.}{9}$
From equation (i),
$0.9\times 4.5\times \left(\right.100-T\left.\right)=\left[\right.\left(\right.0.24\times T\left.\right)+\left(0 . 11 \times T\right)\times 4.5$
$\therefore 0.9\times \left(\right.100-T\left.\right)=0.24\times T+0.11\times T$
$\therefore 90-0.9T=0.35T$
$\therefore 90=1.25T$
$\therefore T=72^\circ C$
$\therefore \frac{Q}{t}=\frac{0 . 9 \times 4 . 5 \times 28}{9}=12.6cal/s$
$10x=126$