Question

In a nuclear reactor, fuel is consumed at the rate of \(2 \times 10^{-3}\) g/s. If 100 % of the mass is converted to energy, find the power output of the reactor in kilowatts (kW).

Answer 1

Step 1: Use Einstein's mass-energy equivalence: \[ E = mc^2, \] where \(m\) is the mass converted per second (mass flow rate), \(c = 3 \times 10^8 \, m/s\) is the speed of light. 
Step 2: Convert the mass flow rate to kilograms per second: \[ m = 2 \times 10^{-3} \, g/s = 2 \times 10^{-6} \, kg/s. \] 
Step 3: Calculate power output \(P\) in watts (J/s): \[ P = m c^2 = 2 \times 10^{-6} \times (3 \times 10^8)^2 = 2 \times 10^{-6} \times 9 \times 10^{16} = 1.8 \times 10^{11} \, W. \] 
Step 4: Convert watts to kilowatts: \[ P = \frac{1.8 \times 10^{11}}{1000} = 1.8 \times 10^{8} \, kW. \] 
Answer: \[ \boxed{1.8 \times 10^{8} \, \text{kW}}. \]