Explanation

(b) When thermal radiations $(Q)$ fall on a body, they are partly reflected, partly absorbed and partly transmitted.
$\ \ \ \ \ \ \ \ \ \ \ Q\ = \ Q_{a} + Q_{r} + Q_{t}$
And $\ \ \ \ \ \ \ \ \ \ \frac{Q_{a}}{Q}\ + \frac{Q_{r}}{Q} + \frac{Q_{t}}{Q} = \ a\ + \ r + \ t\ = 1$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \frac{\ 15}{150}\ + \ 0.6\ + \ x\ = \ 1$
or $\ \ \ \ \ \ \ \ \ \ \ 0.1\ + \ 0.6\ + \ x\ = 1$
or $\ x = 0.3$
Transmitting power, $t\ = \ \frac{Q_{t}}{Q}$
Or $0.3\ = \ \frac{Q_{t}}{150}$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ Q_{t} = \ 45\ J$