If 150 J of energy is incident on area 2 m². If Q_r = 15 J, coefficient of absorption is 0.6, then amount of energy transmitted is

45 J

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 $

$⇒ \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}$

⇒   $Q_t = 45 J$