Explanation

(b) Let $L$ be the length of each rod.
Temperature of $A = 60℃$ temperature of $E = 10℃$
Let $\theta_{1},\ \theta_{2},\ \theta_{3}\ $be respective temperature of $B,\ C,\ D.$
If $Q_{1},\ Q_{2},\ Q_{3},\ Q_{4},\ Q_{5},Q_{6}$ are the amounts of heat following/sec respectively from $A\ to\ B;B\ to\ C;B\ to\ D;C\ to\ D;D\ to\ E\ and\ C\ to\ E,$ then using figure.
[image:11_image66.emf]
$Q_{1} = \ \frac{0.46A\left( 60 - \theta_{1} \right)}{L},\ Q_{2} = \frac{0.92A\left( \theta_{1} - \theta_{2} \right)}{L}$
$Q_{3} = \ \frac{0.46A\left( \theta_{1} - \theta_{3} \right)}{L},\ Q_{4} = \ \frac{0.92A\left( \theta_{2} - \theta_{3} \right)}{L}$
$Q_{5} = \ \frac{0.46A\left( \theta_{3} - 10 \right)}{L},\ Q_{6} = \ \frac{0.92A\left( \theta_{2} - 10 \right)}{L}$
Now $Q_{1} = Q_{2} + Q_{3}\ $
$\frac{0.46A\left( 60 - \theta_{1} \right)}{L} = \ \frac{0.92A\left( \theta_{1} - \theta_{2} \right)}{L} + \frac{0.46A\left( \theta_{1} - \theta_{3} \right)}{L}$
$60 - \theta_{1} = 2\left( \theta_{1} - \theta_{2} \right) + \ \theta_{1} - \theta_{3}$
Or $4\theta_{1} = 2\theta_{2} - \theta_{3} = 60{^\circ}$ ……(i)
Again , $Q_{2} = Q_{4} + Q_{6}\ $ gives
$\theta_{1} - 3\theta_{2} - \theta_{3} = 10^{{^\circ}}$ …..(ii)
Again, $Q_{5} = Q_{3} + Q_{4}\ $given
$\theta_{1} + 2\theta_{2} - 4\theta_{3} = - 10{^\circ}$ …..(iii)
Solving Eqs.(i), (ii) and (iii), we get
$\theta_{1} = 30℃,\ \theta_{2} = 20℃,\ \theta_{3} = 20℃$