Calculate the energy released in fission of 0.1 kg U235. Given that energy released per fission is 200 MeV.
SOLUTION : Here is the entire numerical calculation for the energy released in the fission of \(0.1\) kg of \(U^{235}\) with each step explained:
Given:
- Mass of \(U^{235}\) = \(0.1\) kg
- Energy released per fission = \(200\) MeV
1. Calculate Number of Moles of \(U^{235}\):
Molar mass of \(U^{235}\) = \(235\) g/mol
Number of moles \(= \frac{\text{Mass}}{\text{Molar mass}} = \frac{0.1 \, \text{kg}}{235 \, \text{g/mol}}\)
Number of moles \(= 0.1 \, \text{kg} \times \frac{1000 \, \text{g/kg}}{235 \, \text{g/mol}} = 0.4255 \, \text{mol}\)
2. Determine Number of Uranium Atoms:
Avogadro's number \(= 6.022 \times 10^{23}\) atoms/mol
Number of uranium atoms \(= 0.4255 \, \text{mol} \times 6.022 \times 10^{23} \, \text{atoms/mol} = 2.562 \times 10^{23}\) atoms
3. Calculate Total Energy Released:
Energy released \(= \text{Number of uranium atoms} \times \text{Energy released per fission}\)
Energy released \(= 2.562 \times 10^{23} \, \text{atoms} \times 200 \, \text{MeV}\)
4. Convert Energy to Joules:
\(1\) eV \(= 1.602 \times 10^{-19}\) joules
Energy released in joules \(= (2.562 \times 10^{23} \, \text{atoms} \times 200 \, \text{MeV}) \times 1.602 \times 10^{-13}\) J
Energy released in joules \(= 6.578 \times 10^{12}\) J
5. Convert Energy to Kilowatt-hours (kWh):
\(1\) kilowatt-hour (kWh) = \(3.6 \times 10^6\) joules
Energy in kilowatt-hours \(= \frac{6.578 \times 10^{12} \, \text{J}}{3.6 \times 10^6 \, \text{J/kWh}} = 1.827 \, \text{kWh}\)
Therefore, the energy released in the fission of \(0.1\) kg of \(U^{235}\) is approximately \(1.827\) kilowatt-hours.
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