To find the power output of a uranium-235 reactor, we can follow these steps:
### Step 1: Understand the given data
- The reactor uses 2 kg of uranium-235 in 30 days.
- Each fission of uranium-235 releases 185 MeV of usable energy.
### Step 2: Convert energy per fission from MeV to Joules
1 MeV = \(1.6 \times 10^{-13}\) Joules. Therefore, the energy released per fission in Joules is:
\[
E_{\text{fission}} = 185 \, \text{MeV} \times 1.6 \times 10^{-13} \, \text{J/MeV} = 2.96 \times 10^{-11} \, \text{J}
\]
### Step 3: Calculate the number of fissions in 2 kg of uranium-235
First, we need to find the number of moles of uranium-235 in 2 kg:
- Molar mass of uranium-235 = 235 g/mol
- Number of moles in 2 kg (2000 g):
\[
\text{Number of moles} = \frac{2000 \, \text{g}}{235 \, \text{g/mol}} \approx 8.51 \, \text{mol}
\]
Now, calculate the number of atoms in 2 kg:
\[
\text{Number of atoms} = \text{Number of moles} \times N_A = 8.51 \, \text{mol} \times 6.022 \times 10^{23} \, \text{atoms/mol} \approx 5.12 \times 10^{24} \, \text{atoms}
\]
### Step 4: Calculate the total energy released from all fissions
Total energy released can be calculated as:
\[
E_{\text{total}} = \text{Number of fissions} \times E_{\text{fission}} = 5.12 \times 10^{24} \times 2.96 \times 10^{-11} \, \text{J} \approx 1.52 \times 10^{14} \, \text{J}
\]
### Step 5: Convert the time from days to seconds
30 days can be converted to seconds:
\[
\text{Time} = 30 \, \text{days} \times 24 \, \text{hours/day} \times 60 \, \text{minutes/hour} \times 60 \, \text{seconds/minute} = 2,592,000 \, \text{seconds}
\]
### Step 6: Calculate the power output
Power is defined as energy per unit time:
\[
P = \frac{E_{\text{total}}}{\text{Time}} = \frac{1.52 \times 10^{14} \, \text{J}}{2,592,000 \, \text{s}} \approx 58.6 \, \text{W}
\]
### Step 7: Convert power to megawatts
1 MW = \(10^6\) W, so:
\[
P \approx 58.6 \, \text{W} \approx 58.6 \times 10^{-6} \, \text{MW} \approx 58.6 \, \text{MW}
\]
### Final Answer
The power output of the reactor is approximately **58.6 MW**.
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