How much mass is converted into energy per day in a nuclear power plant operated at `10^(7)` kW?

Options

A. 9.6 g
B. 9.63 kg
C. 8.6 g
D. 7 g

Explanation

To find out how much mass is converted into energy per day in a nuclear power plant operating at \(10^7\) kW, we can follow these steps:

### Step 1: Convert Power from kW to Watts
Given that the power \(P\) is \(10^7\) kW, we need to convert this to watts.

\[ P = 10^7 \text{ kW} = 10^7 \times 10^3 \text{ W} = 10^{10} \text{ W} \]

### Step 2: Calculate Time in Seconds
We need to calculate the total time in seconds for one day.

\[ \text{Time} (t) = 1 \text{ day} = 24 \text{ hours} \times 60 \text{ minutes/hour} \times 60 \text{ seconds/minute} = 86400 \text{ seconds} \]

### Step 3: Calculate Energy Produced in One Day
Using the formula for energy, \(E = P \times t\):

\[ E = 10^{10} \text{ W} \times 86400 \text{ s} = 8.64 \times 10^{14} \text{ J} \]

### Step 4: Use Einstein's Mass-Energy Relation
According to Einstein's mass-energy equivalence principle, \(E = mc^2\), where \(c\) is the speed of light (\(c = 3 \times 10^8 \text{ m/s}\)). We can rearrange this to find mass \(m\):

\[ m = \frac{E}{c^2} \]

### Step 5: Substitute Energy and Speed of Light Values
Substituting the values we have:

\[ m = \frac{8.64 \times 10^{14} \text{ J}}{(3 \times 10^8 \text{ m/s})^2} \]

Calculating \(c^2\):

\[ c^2 = (3 \times 10^8)^2 = 9 \times 10^{16} \text{ m}^2/\text{s}^2 \]

Now substituting back into the mass equation:

\[ m = \frac{8.64 \times 10^{14}}{9 \times 10^{16}} = 9.6 \times 10^{-3} \text{ kg} \]

### Step 6: Convert Mass to Grams
To convert kilograms to grams:

\[ m = 9.6 \times 10^{-3} \text{ kg} \times 1000 \text{ g/kg} = 9.6 \text{ g} \]

### Final Answer
The mass converted into energy per day in the nuclear power plant is **9.6 grams**.

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