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

To calculate the mass (m) of water, we can rearrange the density equation:

$m = \rho \times V$

Substituting the given values:

• Density (ρ) = 998 kg/m³
• Volume (V) = 6,500,000 m³

$m = 998 \times 6,500,000 = 6,487,000,000 \, \text{kg}$

In standard form, this is: $m = 6.487 \times 10^9 \, \text{kg}$

The equation that links energy transferred, power, and time is:

$E = P \times t$

where E is energy transferred, P is power, and t is time.

Given:

• Power (P) = 1.5 x 10⁹ W
• Time (t) = 5 hours = 5 x 60 x 60 seconds = 18,000 seconds

Using the equation: $E = P \times t = (1.5 \times 10^9) \times (18,000)$

Calculating this gives: $E = 2.7 \times 10^{13} \, J$

1. The variation in demand is (much) greater than 1.5 x 10⁹ W, exceeding the maximum output of the hydroelectric power station.
2. Demand remains high for longer than 5 hours; specifically, from 04:00 to 16:00, which is 12 hours.