Calculate the power output of a {}_{92}^{235} | vedclass

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(C) Number of $^{235}U$ atoms in $2 \, 	ext{kg}$ of fuel $= \frac{6 	imes 10^{23}}{235} 	imes 2000 \approx 5.106 	imes 10^{24} \, 	ext{atoms}$.En

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$58$

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(C) Number of $^{235}U$ atoms in $2 \, 	ext{kg}$ of fuel $= \frac{6 	imes 10^{23}}{235} 	imes 2000 \approx 5.106 	imes 10^{24} \, 	ext{atoms}$.Energy released per fission $= 185 \, 	ext{MeV} = 185 	imes 1.6 	imes 10^{-13} \, 	ext{J} = 2.96 	imes 10^{-11} \, 	ext{J}$.Total energy released for $2 \, 	ext{kg}$ of fuel $= (5.106 	imes 10^{24}) 	imes (2.96 	imes 10^{-11} \, 	ext{J}) \approx 1.511 	imes 10^{14} \, 	ext{J}$.Time taken $= 30 \, 	ext{days} = 30 	imes 24 	imes 60 	imes 60 \, 	ext{s} = 2.592 	imes 10^{6} \, 	ext{s}$.Power output $= \frac{	ext{Total Energy}}{	ext{Time}} = \frac{1.511 	imes 10^{14} \, 	ext{J}}{2.592 	imes 10^{6} \, 	ext{s}} \approx 5.83 	imes 10^{7} \, 	ext{W} = 58.3 \, 	ext{MW}$.

Calculate the power output of a ${}_{92}^{235}U$ reactor, if it takes $30 \, \text{days}$ to consume $2 \, \text{kg}$ of fuel, and if each fission releases $185 \, \text{MeV}$ of usable energy. (Given: Avogadro's number $= 6 \times 10^{23} \, \text{mol}^{-1}$) .......... $\text{MW}$ (in $.3$)

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