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Q. $150 \, J$ of energy is incident on area $2 \, m^{2}$ . If $Q_{r}= \, 15 \, J,$ coefficient of absorption is $0.6$ , then the amount of energy transmitted is

NTA AbhyasNTA Abhyas 2020Thermal Properties of Matter

Solution:

When thermal radiations $\left(\right.Q\left.\right)$ 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 $0.3 \, = \, \frac{Q_{t}}{150}$
$\Rightarrow \, Q_{t}= \, 45 \, J$