Calculate the energy released by dissipation of one gram of U235 in kWH.

SOLUTION : The energy released by the dissipation of one gram of \(\text{U}^{235}\) can be calculated using Einstein's mass-energy equivalence formula, \(E = mc^2\), where:

\(E\) is the energy released,

\(m\) is the mass converted into energy,

\(c\) is the speed of light in a vacuum (\(c = 3 \times 10^8\) m/s).

First, let's determine the mass converted. The atomic mass of \(\text{U}^{235}\) is approximately 235 g/mol. So, for 1 gram of \(\text{U}^{235}\), the number of moles is:

\[\text{Number of moles} = \frac{\text{Mass}}{\text{Molar mass}} = \frac{1 \text{ g}}{235 \text{ g/mol}} \approx 0.00426 \text{ mol}\]

Now, we can use Avogadro's constant (\(6.022 \times 10^{23}\) mol\(^{-1}\)) to find the number of uranium atoms in one gram:

\[\text{Number of atoms} = \text{Number of moles} \times \text{Avogadro's constant} \approx 0.00426 \text{ mol} \times 6.022 \times 10^{23} \approx 2.57 \times 10^{21}\]

The mass converted into energy is the mass of one uranium atom multiplied by the number of atoms:

\[\text{Mass converted} = \text{Mass of one uranium atom} \times \text{Number of atoms}\]

The mass of one uranium atom can be calculated by dividing the molar mass by Avogadro's constant:

\[\text{Mass of one uranium atom} = \frac{\text{Molar mass}}{\text{Avogadro's constant}} \approx \frac{235 \text{ g/mol}}{6.022 \times 10^{23} \text{ mol}^{-1}} \approx 3.91 \times 10^{-22} \text{ g}\]

Finally, we can apply Einstein's formula to find the energy released:

\[E = mc^2, \quad \text{Energy} = \text{Mass converted} \times (3 \times 10^8 \, \text{m/s})^2\]

Calculations:

\[\text{Mass converted} \approx 3.91 \times 10^{-22} \text{ g/atom} \times 2.57 \times 10^{21} \text{ atoms} \approx 1.01 \times 10^{-1} \text{ g}\]

\[\text{Energy} \approx 1.01 \times 10^{-1} \text{ g} \times (3 \times 10^8 \, \text{m/s})^2 \approx 9.09 \times 10^{13} \text{ Joules}\]

To convert joules to kilowatt-hours (\(\text{kWh}\)), we use the conversion factor: \(1 \text{ kWh} = 3.6 \times 10^6 \text{ Joules}\):

\[\text{Energy in kWh} = \frac{9.09 \times 10^{13} \text{ Joules}}{3.6 \times 10^6 \text{ Joules/kWh}} \approx 25250 \text{ kWh}\]

Therefore, the energy released by the dissipation of one gram of \(\text{U}^{235}\) is approximately 25250 kilowatt-hours.

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