Explanation

(a) The situation is given in the figure. Let $\theta$ be the temperature at point C.
[image:11_image63.emf]
We know that the rate flow of heat
$\frac{Q}{t} = \frac{KA(\theta_{1} - \theta_{2})}{d}$
Here, K= coefficient of thermal conductivity
A=area of cross-section
$\Rightarrow$ $\frac{KA(100 - \theta)}{40} = \frac{KA(\theta - 10)}{60}$
$\frac{100 - \theta}{2} = \frac{\theta - 10}{3}$
$300 - 3\theta = 2\theta - 20$
$5\theta = 320$
$\theta = \frac{320}{5}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \theta = 64℃$