A nuclear power plant generates 500 MW of waste heat that must be carried away by water pumped from a lake. If the water temperature is to rise by `10^@C`, what is the required flow rate in `kg//s`?

Explanation

To solve the problem of determining the required flow rate of water in a nuclear power plant that generates 500 MW of waste heat, which raises the water temperature by 10°C, we can follow these steps:

### Step 1: Understand the given data
- Waste heat generated (Q) = 500 MW = \(500 \times 10^6\) watts (since 1 MW = \(10^6\) watts)
- Temperature rise (\(\Delta T\)) = 10°C
- Specific heat capacity of water (s) = 4200 J/(kg·°C)

### Step 2: Use the formula for heat transfer
The formula relating heat transfer to mass flow rate, specific heat, and temperature change is:
\[ Q = m \cdot s \cdot \Delta T \]
where:
- \(Q\) is the heat energy (in joules),
- \(m\) is the mass flow rate (in kg/s),
- \(s\) is the specific heat capacity (in J/(kg·°C)),
- \(\Delta T\) is the temperature change (in °C).

### Step 3: Rearrange the formula to find mass flow rate
We need to express mass flow rate \(m\) in terms of the other variables:
\[ m = \frac{Q}{s \cdot \Delta T} \]

### Step 4: Substitute the known values into the equation
Now we can substitute the known values into the equation:
\[ m = \frac{500 \times 10^6 \text{ W}}{4200 \text{ J/(kg·°C)} \cdot 10 \text{°C}} \]

### Step 5: Calculate the denominator
Calculate the denominator:
\[ 4200 \cdot 10 = 42000 \text{ J/kg} \]

### Step 6: Perform the calculation
Now substitute back into the equation:
\[ m = \frac{500 \times 10^6}{42000} \]

### Step 7: Simplify the calculation
Perform the division:
\[ m = \frac{500000000}{42000} \approx 11904.76 \text{ kg/s} \]

### Step 8: Round the answer
Rounding this to two significant figures, we get:
\[ m \approx 1.2 \times 10^4 \text{ kg/s} \]

### Final Answer
The required flow rate of water is approximately \(1.2 \times 10^4 \text{ kg/s}\).

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