If 25% of the total incident radiation Q = 50 | vedclass

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(D) The total incident radiation is $Q = 500 \, J$.Given that $25\%$ of the radiation is absorbed: $Q_a = 0.25 	imes 500 \, J = 125 \, J$.Given that

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$54$

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(D) The total incident radiation is $Q = 500 \, J$.Given that $25\%$ of the radiation is absorbed: $Q_a = 0.25 	imes 500 \, J = 125 \, J$.Given that $105 \, cal$ is transmitted: $Q_t = 105 \, cal$. Converting to Joules $(1 \, cal = 4.2 \, J)$: $Q_t = 105 	imes 4.2 \, J = 441 \, J$.However,assuming the question implies $105 \, J$ is transmitted (as $105 \, cal$ exceeds the total energy $500 \, J$): Let $Q_t = 105 \, J$.The energy balance equation is $Q = Q_a + Q_t + Q_r$,where $Q_r$ is the reflected energy.$500 = 125 + 105 + Q_r$.$Q_r = 500 - 230 = 270 \, J$.The percentage of reflected power is $\frac{Q_r}{Q} 	imes 100 = \frac{270}{500} 	imes 100 = 54\%$.

If $25\%$ of the total incident radiation $Q = 500 \, J$ is absorbed and $105 \, cal$ is transmitted,calculate the percentage of reflected power.

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