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.
Questions
Q1 (F) An old filament lamp is supplied with 200 J and gives out 50 J as light. Calculate its efficiency.
Show solution
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.
Show solution
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.
Show solution
F: efficiency = useful output ÷ total input
I: 0.9 = 1800 ÷ input
F: input = 1800 ÷ 0.9
A: 2000 J
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.
Show solution
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.
Show solution
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?
Show solution
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?
Show solution
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.
Show solution
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)
Show solution
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)
Show solution
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.)
