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

To calculate the mass of water in the reservoir, we can use the formula derived in the previous step:

$m = \rho \times V$

Given:

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

Now substituting the values:

$m = 998 \text{ kg/m}^3 \times 6{,}500{,}000 \text{ m}^3$

Calculating:

$m = 6{,}487{,}000{,}000 \text{ kg}$

In standard form, this is:

$m = 6.487 \times 10^{12} \text{ kg}$

The equation that links energy transferred (E), power (P), and time (t) is:

$E = P \times t$

Given:

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

Using the equation:

$E = P \times t$

Substituting the values:

$E = 1.5 \times 10^9 \text{ W} \times 18{,}000 \text{ s}$

Calculating:

$E = 2.7 \times 10^{13} \text{ J}$

1. The hydroelectric power station has a maximum power output of 1.5 × 10⁹ W, which is significantly less than the increased demand that peaks above this output during the specified time frame.

2. The demand remains high for longer than the maximum output duration of 5 hours, making it impossible for the power station to continuously meet the demand spike.