Question 1Warm-upFoundation
How much energy is needed to heat 2 kg of water by 10 °C? ( J/kg °C)
ΔE = mcΔθ.
Worked solution
- J
Answer: J
GCSE & IGCSE physics practice
Specific heat capacity questions and answers
Specific heat capacity, , is the energy needed to raise the temperature of 1 kg of a substance by 1 °C. The equation is , with energy in J, mass in kg, in J/kg °C and temperature change in °C. Water's is about 4200 J/kg °C, which is why it takes so long to heat.
15 questions · Foundation tier · Every answer is checked as you type it, with a full worked solution. Updated .
Where specific heat capacity comes up
| GCSE (AQA) | Physics and Combined Science, Foundation and Higher. ΔE = mcΔθ is given on the equation sheet (select and apply). Measuring c is a required practical. |
|---|---|
| Cambridge IGCSE (0625) | Supplement (Extended): define specific heat capacity and recall and use c = ΔE ÷ (mΔθ); describe experiments to measure it. |
The method
- ΔE = m × c × Δθ.
- Δθ is the change in temperature: final − initial.
- Rearranged: c = ΔE ÷ (mΔθ), Δθ = ΔE ÷ (mc), m = ΔE ÷ (cΔθ).
- With a heater: energy = power × time.
Specific heat capacity questions
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Warm-up
Question 2Warm-upFoundation
A 0.5 kg aluminium block is heated from 20 °C to 60 °C. Aluminium's specific heat capacity is 900 J/kg °C. How much energy is needed?
Δθ = 60 − 20.
Worked solution
- °C
- J
Answer: J
Question 9Warm-upFoundation
How much energy is needed to heat 5 kg of water by 20 °C? ( J/kg °C)
ΔE = mcΔθ.
Worked solution
- J
Answer: J
Exam-style
Question 3Exam-styleFoundation
It takes 7800 J to heat a 1.5 kg copper block by 13 °C. Calculate the specific heat capacity of copper.
c = ΔE ÷ (mΔθ).
Worked solution
- J/kg °C
Answer: J/kg °C
Question 4Exam-styleFoundation
63 000 J is given to 3 kg of water ( J/kg °C). By how much does its temperature rise?
Δθ = ΔE ÷ (mc).
Worked solution
- °C
Answer: °C
Question 5Exam-styleFoundation
A 250 g mug of tea cools from 80 °C to 30 °C. How much energy does it lose? (Take J/kg °C)
250 g = 0.25 kg; Δθ = 50 °C.
Worked solution
- J
Answer: J
Question 6Exam-styleFoundation
A 50 W heater warms a 1 kg block for 300 s and its temperature rises by 15 °C. Calculate the specific heat capacity of the block (assume no energy is lost).
Energy from the heater = power × time.
Worked solution
- J
- J/kg °C
Answer: J/kg °C
Question 10Exam-styleFoundation
A 2 kg copper block is heated from 15 °C to 35 °C. Copper's specific heat capacity is 385 J/kg °C. How much energy is needed?
Δθ = 35 − 15.
Worked solution
- °C
- J
Answer: J
Question 11Exam-styleFoundation
36 000 J raises the temperature of a 2 kg block by 20 °C. Calculate the specific heat capacity of the block.
c = ΔE ÷ (mΔθ).
Worked solution
- J/kg °C
Answer: J/kg °C
Question 12Exam-styleFoundation
252 000 J of energy heats some water by 30 °C. What mass of water was heated? ( J/kg °C)
m = ΔE ÷ (cΔθ).
Worked solution
- kg
Answer: kg
Question 13Exam-styleFoundation
840 J of energy is given to 0.1 kg of water at 18 °C. What is its final temperature? ( J/kg °C)
Find the temperature rise, then add it to the start.
Worked solution
- °C
- Final temperature °C
Answer: °C
Challenge
Question 7ChallengeFoundation
A 2 kW kettle heats 1.2 kg of water from 20 °C to 100 °C. Assuming no energy is wasted, how long does it take? ( J/kg °C)
Find the energy, then t = E ÷ P.
Worked solution
- J
- s
Answer: s (about 3.4 minutes)
Question 8ChallengeFoundation
The same energy is given to 1 kg of water ( J/kg °C) and 1 kg of oil ( J/kg °C). The water warms by 6 °C. How much does the oil warm by?
Same energy and mass: half the specific heat capacity gives double the temperature rise.
Worked solution
- Energy J
- Oil: °C
Answer: °C
Question 14ChallengeFoundation
A 1 kW heater is switched on for 5 minutes and warms 3 kg of oil by 40 °C. Assuming no energy is lost, calculate the specific heat capacity of the oil.
Energy = power × time, with time in seconds.
Worked solution
- J
- J/kg °C
Answer: J/kg °C
Question 15ChallengeFoundation
A 3 kW immersion heater runs for 10 minutes and heats 10 kg of water by 36 °C. Calculate the efficiency of the heater as a percentage. ( J/kg °C)
Useful energy heats the water; total energy is power × time.
Worked solution
- Total J
- Useful J
- Efficiency
Answer:
Common mistakes
- Using the final temperature instead of the temperature change.
- Leaving mass in grams.
- Forgetting to work out energy from power × time first.
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Sources
The questions are our own. The “where it comes up” notes are from each board's published specification, checked on 27 October 2026:
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