What is the approximate power output of a _{9 | vedclass

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(B) The number of moles in $2 \ kg$ $(2000 \ g)$ of $_{92}U^{235}$ is $n = \frac{2000 \ g}{235 \ g/mol} \approx 8.51 \ mol$.The total number of atoms

Vedclass · Accepted Answer

$59$

Vedclass · Answer

(B) The number of moles in $2 \ kg$ $(2000 \ g)$ of $_{92}U^{235}$ is $n = \frac{2000 \ g}{235 \ g/mol} \approx 8.51 \ mol$.The total number of atoms $N$ is $n 	imes N_A = \frac{2000}{235} 	imes 6.02 	imes 10^{23} \approx 5.12 	imes 10^{24} 	ext{ atoms}$.Total energy $E = N 	imes 185 \ MeV = 5.12 	imes 10^{24} 	imes 185 	imes 1.602 	imes 10^{-13} \ J \approx 1.517 	imes 10^{14} \ J$.Time $t = 30 \ days = 30 	imes 24 	imes 3600 \ s = 2.592 	imes 10^6 \ s$.Power $P = \frac{E}{t} = \frac{1.517 	imes 10^{14}}{2.592 	imes 10^6} \approx 5.85 	imes 10^7 \ W = 58.5 \ MW$.Rounding to the nearest integer, the power output is approximately $59 \ MW$.

What is the approximate power output of a $_{92}U^{235}$ reactor if it consumes $2 \ kg$ of fuel in $30 \ days$ and each fission releases $185 \ MeV$ of usable energy? (Avogadro's number $N_A = 6.02 \times 10^{26} \text{ atoms/kmol}$)

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