If a reactor produces an average power of 1000 MW for a year, how much 235 U is used up assuming 200 MeV are released per fission? O 1.75 kg O 384 kg O 0.35 kg O 1.1 x 108 kg O 3.3 × 108 kg

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### Problem Statement

**If a reactor produces an average power of 1000 MW for a year, how much \(^{235}\text{U}\) is used up assuming \(200 \text{MeV}\) are released per fission?**

### Answer Choices:

- \( o \) 1.75 kg
- \( o \) 384 kg
- \( o \) 0.35 kg
- \( o \) 1.1 \(\times\) \(10^8\) kg
- \( o \) 3.3 \(\times\) \(10^8\) kg

### Explanation:

To solve this, let's break down the problem into clear steps with relevant conversions and calculations:

1. **Convert 1000 MW to MeV per year:**
   - Power = 1000 MW = 1000 \(\times\) \(10^6\) J/s.
   - Energy consumed per year = Power \(\times\) Time = \(1000 \times 10^6 \text{ J/s} \times 365 \times 24 \times 3600 \text{ s}\).

2. **Convert joules to MeV:**
   - \(1 \text{ J} = 6.242 \times 10^{12} \text{ MeV}\).
   - Energy consumed per year in MeV = Total energy in J \(\times 6.242 \times 10^{12}\) MeV.

3. **Calculate number of fissions:**
   - Energy per fission = 200 MeV.
   - Total number of fissions = Total energy consumed in MeV per year / Energy per fission.

4. **Calculate the mass of \(^{235}\text{U}\) used up:**
   - Mass of 1 mole (Avogadro's number) = 235 g of \(^{235}\text{U}\).
   - Number of fissions (atoms) to moles conversion.

By performing these calculations, we can evaluate the correct amount of \(^{235}\text{U}\) consumed.

This problem connects fundamental principles of nuclear physics and energy conversion, and offers students a comprehensive review of physics and chemistry concepts.
Transcribed Image Text:### Problem Statement **If a reactor produces an average power of 1000 MW for a year, how much \(^{235}\text{U}\) is used up assuming \(200 \text{MeV}\) are released per fission?** ### Answer Choices: - \( o \) 1.75 kg - \( o \) 384 kg - \( o \) 0.35 kg - \( o \) 1.1 \(\times\) \(10^8\) kg - \( o \) 3.3 \(\times\) \(10^8\) kg ### Explanation: To solve this, let's break down the problem into clear steps with relevant conversions and calculations: 1. **Convert 1000 MW to MeV per year:** - Power = 1000 MW = 1000 \(\times\) \(10^6\) J/s. - Energy consumed per year = Power \(\times\) Time = \(1000 \times 10^6 \text{ J/s} \times 365 \times 24 \times 3600 \text{ s}\). 2. **Convert joules to MeV:** - \(1 \text{ J} = 6.242 \times 10^{12} \text{ MeV}\). - Energy consumed per year in MeV = Total energy in J \(\times 6.242 \times 10^{12}\) MeV. 3. **Calculate number of fissions:** - Energy per fission = 200 MeV. - Total number of fissions = Total energy consumed in MeV per year / Energy per fission. 4. **Calculate the mass of \(^{235}\text{U}\) used up:** - Mass of 1 mole (Avogadro's number) = 235 g of \(^{235}\text{U}\). - Number of fissions (atoms) to moles conversion. By performing these calculations, we can evaluate the correct amount of \(^{235}\text{U}\) consumed. This problem connects fundamental principles of nuclear physics and energy conversion, and offers students a comprehensive review of physics and chemistry concepts.
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