In the core of a nuclear reactor, the mass of fuel decreases at a rate of 9.0 × 10⁻⁶ kg hour⁻¹ due to nuclear reactions - AQA - A-Level Physics - Question 30 - 2021 - Paper 2

To find the maximum power output of the reactor, we first need to convert the mass loss rate into a power output. The energy released from nuclear fission can be estimated using the equation:

$E = ext{mass} imes c^2$

Where $c$ is the speed of light, approximately $3.0 imes 10^8 ext{ m/s}$.

Calculating the energy released per hour:

1. The mass loss rate is given as $9.0 imes 10^{-6} ext{ kg/h}$. To find the energy released per hour, we calculate:

$E = 9.0 imes 10^{-6} ext{ kg} imes (3.0 imes 10^8 ext{ m/s})^2$

2. Simplifying that gives:

$E = 9.0 imes 10^{-6} imes 9.0 imes 10^{16} = 8.1 imes 10^{11} ext{ J/h}$

3. To convert this energy into power (watts), we note that power is energy per time. Since we calculated for one hour (3600 seconds), we convert:

$P = \frac{8.1 imes 10^{11} ext{ J}}{3600 ext{ s}} \approx 2.25 imes 10^8 ext{ W}$

Thus, the maximum power output is approximately $2.3 imes 10^8 ext{ W}$.