10 AQA GCSE Physics questions on specific heat capacity (E = mc x change in temperature) 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
📌 KEY DEFINITION – learn it word for word
Specific heat capacity is the amount of energy needed to raise the temperature of 1 kg of a substance by 1 °C. Unit: J/kg °C.
The equation
change in thermal energy = mass × specific heat capacity × temperature change ΔE = m c Δθ
| Quantity | Symbol | Unit |
|---|---|---|
| change in thermal energy | ΔE | joules (J) |
| mass | m | kilograms (kg) |
| specific heat capacity | c | joules per kilogram per degree Celsius (J/kg °C) |
| temperature change | Δθ | degrees Celsius (°C) |
Rearranged: Δθ = ΔE ÷ (m c) | m = ΔE ÷ (c Δθ) | c = ΔE ÷ (m Δθ) | Water: c = 4200 J/kg °C
Use FIFA for every answer
- F – Formula: write the equation as it appears on the sheet.
- I – Insert: put in the numbers (convert to standard units first).
- F – Fix: rearrange to make the unknown the subject.
- A – Answer: calculate and always give the unit.
Foundation: aim for Q1–8. Higher: aim for Q4–10. Most questions need rearranging.
Questions
Q1 (F) 2.0 kg of water in a pan is heated by 10 °C. Calculate the energy transferred. (c = 4200 J/kg °C)
Show solution
F: ΔE = m c Δθ
I: ΔE = 2.0 × 4200 × 10
F: ΔE is already the subject
A: 84 000 J
Q2 (F) 42 000 J is transferred to 0.50 kg of water in a baby’s bottle warmer. Calculate the temperature rise. (c = 4200 J/kg °C)
Show solution
F: ΔE = m c Δθ
I: 42 000 = 0.50 × 4200 × Δθ, so 42 000 = 2100 × Δθ
F: Δθ = 42 000 ÷ 2100
A: 20 °C
Q3 (F) A 1.0 kg aluminium block needs 9000 J to warm up by 10 °C. Calculate the specific heat capacity of aluminium.
Show solution
F: ΔE = m c Δθ
I: 9000 = 1.0 × c × 10
F: c = 9000 ÷ 10
A: 900 J/kg °C
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Q4 (F) 250 g of water in a mug starts at 20 °C and receives 63 kJ of energy. Calculate its final temperature. (c = 4200 J/kg °C)
Show solution
Convert first: 250 g = 0.25 kg; 63 kJ = 63 000 J
F: ΔE = m c Δθ
I: 63 000 = 0.25 × 4200 × Δθ, so 63 000 = 1050 × Δθ
F: Δθ = 63 000 ÷ 1050 = 60 °C
A: final temperature = 20 + 60 = 80 °C
Q5 (F) A kettle transfers 336 kJ to heat water from 20 °C to 100 °C. Calculate the mass of water. (c = 4200 J/kg °C)
Show solution
Convert first: 336 kJ = 336 000 J; Δθ = 100 − 20 = 80 °C
F: ΔE = m c Δθ
I: 336 000 = m × 4200 × 80, so 336 000 = 336 000 × m
F: m = 336 000 ÷ 336 000
A: 1.0 kg
Q6 (F/H) A 1.2 kg copper saucepan absorbs 46.2 kJ as it heats from 20 °C to 120 °C. Calculate the specific heat capacity of copper.
Show solution
Convert first: 46.2 kJ = 46 200 J; Δθ = 120 − 20 = 100 °C
F: ΔE = m c Δθ
I: 46 200 = 1.2 × c × 100, so 46 200 = 120 × c
F: c = 46 200 ÷ 120
A: 385 J/kg °C
Q7 (F/H) A storage heater brick (c = 880 J/kg °C) releases 92.4 kJ as it cools from 60 °C to 25 °C. Calculate its mass.
Show solution
Convert first: 92.4 kJ = 92 400 J; Δθ = 60 − 25 = 35 °C
F: ΔE = m c Δθ
I: 92 400 = m × 880 × 35, so 92 400 = 30 800 × m
F: m = 92 400 ÷ 30 800
A: 3.0 kg
Q8 (F/H) A swimming pool holds 5.0 × 105 kg of water. The heating system transfers 4.2 × 109 J. Calculate the temperature rise. (c = 4200 J/kg °C)
Show solution
F: ΔE = m c Δθ
I: 4.2 × 109 = 5.0 × 105 × 4200 × Δθ, so 4.2 × 109 = 2.1 × 109 × Δθ
F: Δθ = 4.2 × 109 ÷ 2.1 × 109
A: 2.0 °C
Q9 (H) A 2.0 kW kettle heats 1.5 kg of water from 20 °C to 100 °C. Assuming no energy is wasted, how long does it take? (c = 4200 J/kg °C)
Show solution
Step 1 – energy needed: ΔE = m c Δθ = 1.5 × 4200 × 80 = 504 000 J
Step 2 – time
F: P = E / t
I: 2000 = 504 000 ÷ t
F: t = 504 000 ÷ 2000
A: 252 s (4 minutes 12 seconds)
Q10 (H) A 3.0 kW kettle heats 1.2 kg of water from 15 °C to 100 °C in 160 s. Calculate the efficiency of the kettle, to 2 significant figures. (c = 4200 J/kg °C)
Show solution
Useful energy: ΔE = m c Δθ = 1.2 × 4200 × 85 = 428 400 J
Total input: E = P t = 3000 × 160 = 480 000 J
F: efficiency = useful output ÷ total input
I: efficiency = 428 400 ÷ 480 000
A: 0.8925 = 0.89 (89%) (2 s.f.)
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