AQA GCSE Physics Physics Paper 1 (Foundation), June 2023: Question 4

10 marks · Low Demand difficulty · Short Answer

Calculate percentages of renewable electricity, determine gravitational potential energy, and relate electrical power, energy, and seasonal output variations for a hydroelectric generator.

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Question

Question text

04 The UK uses renewable energy resources to generate some of its electricity.

Figure 4 shows the proportion of electricity generated by different renewable energy

resources in the UK in 2020.

Figure 4

04.1 Calculate the percentage of electricity generated using hydroelectric power.

[2 marks]

Percentage = %

A remote village in the UK uses a hydroelectric generator to provide electricity.

04.2 The mass of water that passes through the hydroelectric generator each day

is 2 500 000 kg.

The change in vertical height of the water is 15.0 m.

gravitational field strength = 9.8 N/kg

Calculate the decrease in gravitational potential energy of the water.

Use the equation:

gravitational potential energy = mass × gravitational field strength × height

[2 marks]

Decrease in gravitational potential energy = J

Use the Physics Equations Sheet to answer questions 04.3 and 04.4.

04.3 Write down the equation which links energy (E), power (P) and time (t).

[1 mark]

04.4 The hydroelectric generator transfers electrical power of 3000 W to the village.

Calculate the energy transferred to the village in 60 minutes.

[3 marks]

*17* 19

Energy transferred = J

04.5 The hydroelectric generator is turned by falling river water.

Figure 5 shows how the power output of the hydroelectric generator varied during

one year.

Figure 5

Explain one reason why the power output varied.

[2 marks]

Mark scheme

Show the mark scheme

Question 4

AO /

Question Answers Extra information Mark

Spec. Ref.

04.1 other energy resources 1 AO2

= 95 (%) 4.1.3

hydroelectric = 5 (%) 1

AO /

Spec. Ref.

04.2 Ep = 2 500 000 × 9.8 × 15 1 AO2

4.1.1.2

Ep = 367 500 000 (J) allow 370 000 000 (J) 1

or or

E = 3.675 × 108 (J) E = 3.7 × 108 (J)

p p

AO /

Spec. Ref.

04.3 energy = power × time 1 AO1

or 4.2.4.2

E = P × t 4.1.1.4

AO /

Spec. Ref.

04.4 t = 3600 (s) 1 AO2

4.2.4.2

E = 3000 × 3600 allow a correct substitution using 1 4.1.1.4

an incorrectly/not converted

value for t

E = 10 800 000 (J) allow an answer consistent with 1

or their incorrectly/not converted

E = 1.08 × 107 (J) value for t

allow a correct answer given to–HYSICS– –

2 s.f.

AO /

Spec. Ref.

04.5 the level of the water in the river 1 AO3

varies 4.1.3

the amount of rainfall varies 13

and is lower in the summer allow specified months or range 1

months of months eg April to September

MP2 dependent on scoring MP1

Total Question 4 10

How to answer it

Renewable Energy Resources & Hydroelectric Calculations

📋 What this question tests

This question assesses your understanding of renewable energy resources, reading data from pie charts and graphs, and performing fundamental energy calculations:

Data extraction: Reading percentages from a pie chart and identifying trends from annual seasonal graphs.
Gravitational Potential Energy (GPE): Applying the formula E_p = m × g × h with large values.
Energy and Power: Recalling and rearranging E = P × t and converting time units (minutes to seconds).
Real-world evaluation: Explaining environmental and weather-related factors affecting hydroelectric power output.

Part 04.1

Calculating Percentage from a Pie Chart

Question: Calculate the percentage of electricity generated using hydroelectric power. [2 marks]

📐 Step-by-Step Calculation

Sum the given renewable percentages:
Total of other sources = Wind (56%) + Biofuel (29%) + Solar (10%) = 95%
Subtract from the total percentage (100%):
Hydroelectric = 100% − 95% = 5%

✅ Mark Scheme Breakdown

Mark 1: Summing the other energy resources = 95 (%)
Mark 2: Hydroelectric = 5 (%)

💡 A correct final answer of 5% scores both marks directly, even without working.

❌ Common Errors

Misreading the chart numbers (e.g. confusing 29% for 20%).
Simple arithmetic errors when adding 56 + 29 + 10 under exam pressure.

Part 04.2

Gravitational Potential Energy Calculation

Question: Calculate the decrease in gravitational potential energy of the water. [2 marks]

📐 Step-by-Step Calculation

Identify given values:
Mass ( m ) = 2 500 000 kg
Height ( h ) = 15.0 m
Gravitational field strength ( g ) = 9.8 N/kg
Substitute into formula:
E_p = m × g × h
E_p = 2 500 000 × 9.8 × 15.0
Calculate final value:
E_p = 367 500 000 J (or 3.675 × 10⁸ J )

✅ Mark Scheme Breakdown

Mark 1: Correct substitution: 2 500 000 × 9.8 × 15
Mark 2: Correct value: 367 500 000 (J) or 3.675 × 10⁸ (J)

💡 Also allowed: Rounding to 2 s.f. gives 370 000 000 (J) or 3.7 × 10⁸ (J) .

🧠 Exam Technique: Large Numbers

When keying large numbers into your calculator, count the zeros carefully (5 zeros in 2 500 000). Writing your final answer in standard form ( 3.675 × 10⁸ ) prevents losing marks from accidentally dropping a zero on the answer line.

Part 04.3

Equation Recall: Energy, Power, and Time

Question: Write down the equation which links energy (E), power (P) and time (t). [1 mark]

✅ Correct Answer

Either of the following forms is awarded full credit:

energy = power × time
E = P × t

Rearranged forms such as P = E / t or t = E / P are also accepted.

❌ Common Errors

Writing incorrect rearrangements like E = P / t or E = t / P .
Confusing energy ( E ) with efficiency or electrical potential.

Part 04.4

Calculating Energy Transferred

Question: Calculate the energy transferred to the village in 60 minutes when power = 3000 W. [3 marks]

📐 Step-by-Step Calculation

Convert time to standard SI units (seconds):
t = 60 minutes × 60 seconds = 3600 s
Substitute into formula:
E = P × t = 3000 W × 3600 s
Calculate the answer:
E = 10 800 000 J (or 1.08 × 10⁷ J )

✅ Mark Scheme Breakdown

Mark 1: Converting time to seconds: t = 3600 (s)
Mark 2: Substitution: E = 3000 × 3600
Mark 3: Correct answer: 10 800 000 (J) or 1.08 × 10⁷ (J)

💡 Rounding to 2 significant figures ( 1.1 × 10⁷ J ) is also accepted.

❌ The #1 Unit Trap

Many students calculate 3000 × 60 = 180 000 J . While error-carried-forward (ECF) allows 2 marks for this, you lose the first mark because time must always be converted to seconds when power is in watts (W = J/s)!

Part 04.5

Interpreting Seasonal Trends in Hydroelectric Power

Question: Explain one reason why the power output varied during the year. [2 marks]

✅ Model Answer

Point 1 (Cause): The amount of rainfall varies / the water level in the river varies across the year. [1 mark]

Point 2 (Linked detail): There is less rainfall / lower river levels during the summer months (e.g. June/July). [1 mark]

💡 Key Knowledge

Hydroelectric power relies on the volume and flow rate of water falling through the turbine.
From Figure 5, the power output drops significantly between April and September, reaching a minimum in July (~300 W).
In the UK, summer typically has lower precipitation and higher evaporation, reducing river volume.

🧠 Examiner Tip: Dependent Marks

Notice that Mark Point 2 is dependent on Mark Point 1. Simply saying "it was summer" will not score marks unless you clearly connect it to water level / rainfall.

❌ Common Misconceptions

"People use less electricity in summer" — This explains electrical demand, not generator power output capacity driven by river flow.
Vague statements like "the weather changed" without naming rain, water volume, or river level.

Topics

Physics · P1: Energy · P2: Electricity

Question and mark scheme from the AQA GCSE Physics examination, Physics Paper 1 (Foundation), June 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.