The Physics Question Bank

10 AQA GCSE Physics questions on efficiency (efficiency = useful energy out ÷ total energy in) that get harder as you go, from Foundation to Higher, each with a full worked solution.

Course: Combined Science + Separate Physics  |  Tier: Foundation + Higher  |  Equation sheet: given in the exam

The equation

efficiency = useful output energy transfer ÷ total input energy transfer

Quantity	Unit
efficiency	no unit (a decimal, or × 100 for %)
useful output energy	joules (J)
total input energy	joules (J)

Rearranged: useful output = efficiency × total input  |  total input = useful output ÷ efficiency  |  wasted = input − useful

Use FIFA for every answer

• F – Formula: write the equation as it appears on the sheet.
• I – Insert: put in the numbers (use the same units top and bottom; turn % into a decimal).
• F – Fix: rearrange to make the unknown the subject.
• A – Answer: calculate and give the unit (efficiency has none).

Foundation: aim for Q1–8.   Higher: aim for Q4–10. Most questions need rearranging.

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Questions

Q1 (F) An old filament lamp is supplied with 200 J and gives out 50 J as light. Calculate its efficiency.

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F: efficiency = useful output ÷ total input
I: efficiency = 50 ÷ 200
F: efficiency is already the subject
A: 0.25 (25%)

Q2 (F) An electric motor has an efficiency of 0.8. It is supplied with 500 J. Calculate the useful energy output.

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F: efficiency = useful output ÷ total input
I: 0.8 = useful ÷ 500
F: useful = 0.8 × 500
A: 400 J

Q3 (F) A kettle is 0.9 efficient. It transfers 1800 J usefully to the water. Calculate the total energy supplied.

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F: efficiency = useful output ÷ total input
I: 0.9 = 1800 ÷ input
F: input = 1800 ÷ 0.9
A: 2000 J

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Q4 (F) A car engine is 30% efficient. It is supplied with 60 kJ of chemical energy from fuel. Calculate the useful energy output in kJ.

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Convert first: 30% = 0.30
F: efficiency = useful output ÷ total input
I: 0.30 = useful ÷ 60
F: useful = 0.30 × 60
A: 18 kJ

Q5 (F) A TV is supplied with 500 J. 100 J is wasted as heat. Calculate its efficiency as a percentage.

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Useful output: 500 − 100 = 400 J
F: efficiency = useful output ÷ total input
I: efficiency = 400 ÷ 500 = 0.8
F: × 100 for a percentage
A: 80%

Q6 (F/H) A solar panel is 20% efficient. It produces 3.6 kJ of electrical energy. How much light energy falls on it?

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Convert first: 20% = 0.20
F: efficiency = useful output ÷ total input
I: 0.20 = 3.6 ÷ input
F: input = 3.6 ÷ 0.20
A: 18 kJ

Q7 (F/H) A lawnmower’s petrol engine is 25% efficient and is supplied with 2.0 MJ. How much energy is wasted?

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F: efficiency = useful output ÷ total input
I: 0.25 = useful ÷ 2.0
F: useful = 0.25 × 2.0 = 0.5 MJ
A: wasted = 2.0 − 0.5 = 1.5 MJ

Q8 (F/H) A gas power station is 38% efficient. It takes in 5.0 × 10¹² J of fuel energy each hour. Calculate the useful electrical energy output each hour.

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F: efficiency = useful output ÷ total input
I: 0.38 = useful ÷ 5.0 × 10¹²
F: useful = 0.38 × 5.0 × 10¹²
A: 1.9 × 10¹² J

Q9 (H) A motor that is 60% efficient lifts a 20 kg box by 3.0 m. How much electrical energy is supplied to the motor? (g = 9.8 N/kg)

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Step 1 – useful output: Eₚ = m g h = 20 × 9.8 × 3.0 = 588 J
F: efficiency = useful output ÷ total input
I: 0.60 = 588 ÷ input
F: input = 588 ÷ 0.60
A: 980 J

Q10 (H) An 85% efficient kettle heats 0.50 kg of water by 80 °C. Calculate the total energy supplied, to 2 significant figures. (c = 4200 J/kg °C)

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Step 1 – useful output: ΔE = m c Δθ = 0.50 × 4200 × 80 = 168 000 J
F: efficiency = useful output ÷ total input
I: 0.85 = 168 000 ÷ input
F: input = 168 000 ÷ 0.85 = 197 647 J
A: 2.0 × 10⁵ J (2 s.f.)

Next steps: learn the whole topic in Energy in 10 steps · more ramped questions: all 8 Energy equations · mix them up: Which equation? Energy