IGCSE Physics Specific Heat Capacity Questions and Worked Solutions (2.2.2) | 0625

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IGCSE Physics 0625 · Topic 2.2.2

IGCSE Physics Specific Heat Capacity Questions and Worked Solutions

50 topic-wise practice questions on Specific Heat Capacity (Unit 2: Thermal Physics), written to the Cambridge IGCSE Physics 0625 syllabus for 2026–2028, each with a step-by-step worked solution.

50 questions2 Core38 Extended10 Challenge148 marksTextbook §2.2 · pp. 100–114
Questions tagged Core are for all candidates (Papers 1 & 3). Questions tagged Extended are Supplement content (Papers 2 & 4). Questions 41–50 are harder Challenge questions. Try each question first, then tap Show solution to check your working.

Topic guide: Specific Heat Capacity

Heating an object increases its internal energy and its temperature. Specific heat capacity tells you how much energy is needed to raise the temperature of 1 kg of a substance by 1 °C.

Key formulas

ΔE = m c Δθenergy ΔE (J), mass m (kg), specific heat capacity c (J/(kg °C)), temperature change Δθ (°C)Extended
c = ΔE ÷ (m Δθ)rearranged to find cExtended
E = P t or E = I V tenergy from a heaterExtended

Prerequisite knowledge

  • Energy and power (1.7)
  • Particle model (2.1.2)
  • Electrical energy E = IVt (4.2.5) for experiments

Exam tips & common mistakes

  • Δθ is the CHANGE in temperature (final − initial)
  • Use kg with J/(kg °C); use g with J/(g °C)
  • In mixing problems, energy lost by the hot object = energy gained by the cold

Key facts from the syllabus (2.2.2)

  • A rise in temperature increases the internal energy of an object
  • Extended a rise in temperature increases the average kinetic energy of its particles
  • Specific heat capacity is the energy needed per unit mass per unit temperature increase
  • Water has a high specific heat capacity (about 4200 J/(kg °C))
  • Extended measure c with an electric heater, thermometer, balance and joulemeter/timer; insulate to reduce losses
  • Measured values are usually too high because energy is lost to the surroundings

Textbook: §2.2 Thermal properties and temperature, pp. 100–114

1
Core[1 mark]

State what happens to the internal energy of an object when its temperature rises.

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Method: Internal energy increases.

It increases
2
Core[2 marks]

State what happens to the internal energy of a block of iron when it is heated.

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Method: Increases.

It increases
3
Extended[2 marks]

Describe what happens to particles when the temperature of an object increases.

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Method: Average KE increases.

The average kinetic energy of the particles increases
4
Extended[2 marks]

Define specific heat capacity.

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Method: Energy per unit mass per unit temperature increase.

The energy required per unit mass per unit temperature increase
5
Extended[3 marks]

Calculate the energy needed to heat 2.0 kg of water from 20 °C to 80 °C (c = 4200 J/(kg °C)).

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Method: ΔE = mcΔθ.

ΔE = 2.0 × 4200 × 60 = 504000 J
6
Extended[3 marks]

A 0.50 kg aluminium block (c = 900 J/(kg °C)) absorbs 18 000 J. Calculate its temperature rise.

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Method: Δθ = ΔE ÷ (mc).

Δθ = 18000 ÷ (0.50 × 900) = 40.0 °C
7
Extended[3 marks]

5400 J raises the temperature of a 1.2 kg block by 10 °C. Calculate its specific heat capacity.

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Method: c = ΔE ÷ (mΔθ).

c = 5400 ÷ (1.2 × 10) = 450 J/(kg °C)
8
Extended[3 marks]

Calculate the energy to heat 250 g of water by 15 °C (c = 4.2 J/(g °C)).

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Method: ΔE = mcΔθ.

ΔE = 250 × 4.2 × 15 = 15800 J
9
Extended[4 marks]

A 54 g metal object at 100 °C is put into 50 g of water at 22 °C. The final temperature is 31 °C. Calculate the specific heat capacity of the metal (water c = 4.2 J/(g °C)).

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Method: Energy lost by metal = energy gained by water.

Water gains 50 × 4.2 × 9 = 1890 J; c = 1890 ÷ (54 × 69) = 0.507 J/(g °C)
10
Extended[3 marks]

A 2.0 kW heater warms 3.0 kg of water for 5.0 minutes. Calculate the temperature rise, assuming no losses.

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Method: ΔE = Pt.

ΔE = 2000 × 300 = 600 000 J; Δθ = 600000 ÷ (3.0 × 4200) = 47.6 °C
11
Extended[4 marks]

Describe an experiment to measure the specific heat capacity of a metal block.

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Method: Electrical heater, thermometer, joulemeter or V, I, t.

Measure mass of block; insert heater and thermometer; record initial temperature; heat for a measured time; find energy (E = IVt or joulemeter); measure temperature rise; c = E ÷ (mΔθ)
12
Extended[2 marks]

State two ways to improve the accuracy of a specific heat capacity experiment.

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Method: Reduce losses.

Insulate the block; use oil in the thermometer hole; stir liquid; use the highest temperature reached
13
Extended[4 marks]

Describe an experiment to measure the specific heat capacity of water.

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Method: Kettle/immersion heater in insulated beaker.

Measure mass of water in an insulated beaker; heat with an immersion heater of known power for a measured time; stir and measure temperature rise; c = Pt ÷ (mΔθ)
14
Extended[2 marks]

Explain why the measured specific heat capacity is usually larger than the true value.

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Method: Energy lost to surroundings.

Some energy is lost to the surroundings, so less heats the object; the temperature rise is smaller, giving a larger c
15
Extended[3 marks]

An immersion heater (12 V, 4.0 A) heats 1.0 kg of metal for 300 s. The temperature rises by 32 °C. Calculate c.

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Method: E = IVt.

E = 12 × 4.0 × 300 = 14400 J; c = 14400 ÷ (1.0 × 32) = 450 J/(kg °C)
16
Extended[2 marks]

Explain why water is used in central heating radiators.

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Method: High specific heat capacity.

Its high specific heat capacity means it stores a lot of energy per kilogram per degree
17
Extended[3 marks]

Calculate the energy given out when 0.80 kg of copper (c = 390 J/(kg °C)) cools from 120 °C to 20 °C.

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Method: ΔE = mcΔθ.

ΔE = 0.80 × 390 × 100 = 31200 J
18
Extended[2 marks]

Equal masses of water and oil receive equal energy. The oil gets hotter. State which has the higher specific heat capacity.

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Method: Smaller rise → larger c.

Water
19
Extended[3 marks]

A kettle gives 336 000 J to 1.0 kg of water at 20 °C. Calculate the final temperature.

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Method: Δθ = ΔE ÷ (mc).

Δθ = 336000 ÷ 4200 = 80.0 °C; final = 100 °C
20
Extended[2 marks]

Explain why a large mass of water takes longer to heat than a small mass using the same heater.

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Method: ΔE = mcΔθ.

More mass needs more energy for the same temperature rise, and the heater supplies energy at a fixed rate
21
Extended[3 marks]

A 0.20 kg block at 85 °C cools to 25 °C, releasing 5400 J. Calculate its specific heat capacity.

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Method: c = ΔE ÷ (mΔθ).

c = 5400 ÷ (0.20 × 60) = 450 J/(kg °C)
22
Extended[2 marks]

State the unit of specific heat capacity.

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Method: J/(kg °C).

J/(kg °C)
23
Extended[3 marks]

Calculate the energy needed to heat 1.5 kg of water from 15 °C to 65 °C (c = 4200 J/(kg °C)).

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Method: mcΔθ.

ΔE = 1.5 × 4200 × 50 = 315000 J
24
Extended[3 marks]

Calculate the temperature rise when 21 000 J is supplied to 0.25 kg of water.

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Method: Δθ = E/(mc).

Δθ = 21000 ÷ (0.25 × 4200) = 20.0 °C
25
Extended[3 marks]

A 2.0 kg copper block (c = 390 J/(kg °C)) is heated from 20 °C to 70 °C. Calculate the energy supplied.

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Method: mcΔθ.

ΔE = 2.0 × 390 × 50 = 39000 J
26
Extended[3 marks]

A 0.40 kg block gains 7200 J and its temperature rises by 20 °C. Calculate its specific heat capacity.

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Method: c = E/(mΔθ).

c = 7200 ÷ 8.0 = 900 J/(kg °C)
27
Extended[3 marks]

Calculate the mass of water that can be heated by 10 °C with 84 000 J.

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Method: m = E/(cΔθ).

m = 84000 ÷ 42000 = 2.00 kg
28
Extended[3 marks]

A 500 W heater warms 1.0 kg of oil for 2.0 minutes and its temperature rises by 30 °C. Calculate c for the oil.

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Method: E = Pt.

E = 60 000 J; c = 60000 ÷ 30 = 2000 J/(kg °C)
29
Extended[3 marks]

A 3.0 kW kettle heats 1.2 kg of water from 20 °C to 100 °C. Calculate the time taken, assuming no losses.

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Method: t = mcΔθ/P.

t = 1.2 × 4200 × 80 ÷ 3000 = 134 s
30
Extended[2 marks]

Explain why coastal areas have milder climates than inland areas.

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Method: Water's high c.

Water's high specific heat capacity means the sea warms and cools slowly, moderating temperatures
31
Extended[2 marks]

Explain why a metal spoon heats up faster than the same mass of water with the same energy input.

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Method: Lower c.

The metal has a lower specific heat capacity, so the same energy gives a larger temperature rise
32
Extended[2 marks]

State the relationship between the temperature of an object and the average kinetic energy of its particles.

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Method: Higher T, higher KE.

Higher temperature means higher average kinetic energy of the particles
33
Extended[3 marks]

Two 1.0 kg blocks, aluminium (900 J/(kg °C)) and lead (130 J/(kg °C)), each receive 9000 J. Calculate each temperature rise.

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Method: Δθ = E/(mc).

Aluminium: 10.0 °C; lead: 69.2 °C
34
Extended[4 marks]

200 g of water at 80 °C is mixed with 300 g of water at 20 °C. Calculate the final temperature, assuming no energy losses.

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Method: Energy lost = energy gained.

0.2(80 − T) = 0.3(T − 20); T = 44.0 °C
35
Extended[3 marks]

An immersion heater supplies 50 W for 10 minutes to a 1.0 kg aluminium block. The temperature rises from 20 °C to 51 °C. Calculate c and explain why it is higher than the textbook value of 900 J/(kg °C).

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Method: E = Pt.

E = 30 000 J; c = 30000 ÷ 31 = 968 J/(kg °C); energy is lost to the surroundings
36
Extended[2 marks]

State one way to reduce energy losses in a specific heat capacity experiment.

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Method: Insulation.

Wrap the block in insulation (and use a lid/oil in the thermometer hole)
37
Extended[2 marks]

Explain why the thermometer hole is filled with oil in a specific heat capacity experiment on a metal block.

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Method: Better thermal contact.

To improve thermal contact between the block and the thermometer
38
Extended[3 marks]

A 0.50 kg water sample cools from 90 °C to 40 °C. Calculate the energy given out.

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Method: mcΔθ.

ΔE = 0.50 × 4200 × 50 = 105000 J
39
Extended[3 marks]

An electric heater supplies 12 V and 3.0 A for 400 s to 0.80 kg of a liquid, raising its temperature by 6.0 °C. Calculate its specific heat capacity.

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Method: E = IVt.

E = 14400 J; c = 14400 ÷ (0.80 × 6.0) = 3000 J/(kg °C)
40
Extended[2 marks]

Explain why a substance with a high specific heat capacity is useful in storage heaters.

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Method: Stores lots of energy.

It stores a large amount of energy per kilogram per degree, releasing it slowly

Challenge questions

Harder, multi-step questions at the top end of the exam. Attempt these once you are confident with the questions above.

41
ChallengeCore[3 marks]

Explain why the temperature of 1 kg of water rises less than that of 1 kg of oil when both receive the same energy.

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Method: Specific heat capacity.

Water has a higher specific heat capacity, so it needs more energy per degree
42
ChallengeExtended[4 marks]

A 2.0 kW kettle heats 1.5 kg of water from 18 °C to 100 °C in 4.5 minutes. Calculate the energy supplied, the energy gained by the water and the efficiency.

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Method: E = Pt; mcΔθ.

Supplied = 540000 J; gained = 1.5 × 4200 × 82 = 517000 J; efficiency = 95.7%
43
ChallengeExtended[5 marks]

0.50 kg of copper (c = 390 J/(kg °C)) at 200 °C is dropped into 1.0 kg of water at 20 °C. Calculate the final temperature, assuming no losses.

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Method: Energy lost = energy gained.

0.5×390×(200 − T) = 1.0×4200×(T − 20); T = 28.0 °C
44
ChallengeExtended[5 marks]

An immersion heater (12 V, 4.0 A) heats a 1.0 kg aluminium block for 10 minutes. The temperature rises from 20 °C to 48 °C. Calculate the specific heat capacity and suggest why it is larger than 900 J/(kg °C).

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Method: E = IVt.

E = 28800 J; c = 28800 ÷ 28 = 1030 J/(kg °C); energy is lost to the surroundings
45
ChallengeExtended[4 marks]

A storage heater contains 80 kg of bricks (c = 840 J/(kg °C)) heated from 20 °C to 220 °C overnight. Calculate the energy stored and how long it could supply 1.5 kW.

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Method: mcΔθ; t = E/P.

E = 80 × 840 × 200 = 1.34 × 10⁷ J; t = E ÷ 1500 = 8.96 × 10³ s ≈ 2.49 h
46
ChallengeExtended[4 marks]

Equal masses of water at 80 °C and 20 °C are mixed. Explain why the final temperature is 50 °C, and predict it if 2 kg at 80 °C is mixed with 1 kg at 20 °C.

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Method: Energy balance.

Energy lost by hot = energy gained by cold; equal masses so equal temperature changes; 2(80 − T) = (T − 20) gives T = 60.0 °C
47
ChallengeExtended[4 marks]

A 1.0 kg block at 25 °C receives 9000 J and reaches 35 °C; a 2.0 kg block of a different metal receives the same energy and reaches 30 °C. Compare their specific heat capacities.

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Method: c = E/(mΔθ).

c₁ = 900 J/(kg °C); c₂ = 900 J/(kg °C); they are the same
48
ChallengeExtended[4 marks]

A 750 W microwave heats a 0.30 kg mug of milk (c = 3900 J/(kg °C)) from 5 °C to 65 °C. Calculate the minimum time needed.

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Method: t = mcΔθ/P.

t = 0.30 × 3900 × 60 ÷ 750 = 93.6 s
49
ChallengeExtended[4 marks]

Explain, in terms of particles, what happens to the internal energy of a block as its temperature rises, and why a larger mass needs more energy for the same rise.

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Method: More particles.

Particles' average kinetic energy increases; a larger mass has more particles, each needing extra energy, so more total energy is needed
50
ChallengeExtended[4 marks]

A car's cooling system holds 6.0 kg of water. The engine transfers 30 kW to it. Calculate the temperature rise per second if no energy were removed, and explain why the radiator is essential.

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Method: Δθ = P/(mc) per second.

Δθ = 30000 ÷ (6.0 × 4200) = 1.19 °C per second; without a radiator the water would boil within about a minute
All IGCSE Physics topics← 2.2.1 Thermal Expansion of Solids, Liquids and Gases2.2.3 Melting, Boiling and Evaporation →

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