Figure 4 shows a hydroelectric power station - AQA - GCSE Physics - Question 3 - 2020 - Paper 1

Using the volume of water stored in the reservoir, we can calculate the mass as follows:

1. Given:

   • Volume, $V = 6,500,000 \text{ m}^3$
   • Density, $\rho = 998 \text{ kg/m}^3$

2. From the formula, we have: $m = \rho \times V$ $m = 998 \text{ kg/m}^3 \times 6,500,000 \text{ m}^3$ $m = 6,487,000,000 \text{ kg}$

3. Converting the mass into standard form: $m = 6.487 \times 10^9 \text{ kg}$.

The equation linking energy transferred, power, and time is:

$E = P \times t$.

To find the maximum energy, we can use the equation from the previous step:

1. Given:

   • Power, $P = 1.5 \times 10^9 \text{ W}$
   • Time, $t = 5 \text{ hours} = 5 \times 60 \times 60 = 18,000 \text{ seconds}$

2. Now substituting the values into the equation: $E = P \times t$ $E = 1.5 \times 10^9 \text{ W} \times 18000 \text{ s}$ $E = 2.7 \times 10^{13} \text{ J}$.

1. The demand for electricity during this period is significantly higher than the maximum output of the hydroelectric power station, which is limited to 1.5 × 10⁹ W.
2. The power station is also constrained by the maximum operation time of 5 hours, which further limits its ability to meet sustained high demand.