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(B) The total incident heat energy rate is $P_i = 1000 \ J \ min^{-1}$.According to the law of conservation of energy for heat radiation,the sum of th

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$500$

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(B) The total incident heat energy rate is $P_i = 1000 \ J \ min^{-1}$.According to the law of conservation of energy for heat radiation,the sum of the coefficients of reflection $(r)$,absorption $(a)$,and transmission $(t)$ is equal to $1$,i.e.,$r + a + t = 1$.Given $r = 0.1$ and $a = 0.8$,we can find the coefficient of transmission $(t)$:$t = 1 - (r + a) = 1 - (0.1 + 0.8) = 1 - 0.9 = 0.1$.The rate of transmitted heat energy is $P_t = t 	imes P_i = 0.1 	imes 1000 \ J \ min^{-1} = 100 \ J \ min^{-1}$.For a time duration of $5 \ minutes$,the total transmitted heat energy $(Q_t)$ is:$Q_t = P_t 	imes 	ext{time} = 100 \ J \ min^{-1} 	imes 5 \ min = 500 \ J$.

Heat energy is incident on the surface of a material at the rate of $1000 \ J \ min^{-1}$. If the coefficient of absorption is $0.8$ and the coefficient of reflection is $0.1$,then the amount of heat energy transmitted through the material in $5$ minutes is: (in $J$)

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