These eleven questions cover specific heat capacity, graph gradients, heating curves, latent energy and energy losses from heat calculations. All data is invented for practice and is not taken from any exam paper. They go from easier to harder.
Write your answer with its unit on paper first, then open the answer. Use c (water) = 4200 J/kg °C, specific latent heat of fusion of ice = 3.3 × 10⁵ J/kg and specific latent heat of vaporisation of water = 2.3 × 10⁶ J/kg. Whether latent heat is required depends on your syllabus year, so check the Cambridge page for your exam year.
Questions
1. How much energy is needed to warm 1.5 kg of water from 20 °C to 30 °C?
Show answer
Δθ = 10 °C. Q = 1.5 × 4200 × 10 = 63 000 J.
2. How much energy is needed to warm 250 g of water from 20 °C to 80 °C?
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m = 0.25 kg and Δθ = 60 °C. Q = 0.25 × 4200 × 60 = 1050 × 60 = 63 000 J.
3. A 3.0 kg mass of water absorbs 252 kJ. Find its temperature rise.
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252 kJ = 252 000 J. m × c = 3.0 × 4200 = 12 600 J/°C. Δθ = 252 000 ÷ 12 600 = 20 °C.
4. A 0.80 kg metal block warms from 40 °C to 65 °C when it absorbs 16 kJ. Calculate its specific heat capacity.
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Δθ = 25 °C and Q = 16 000 J. c = 16 000 ÷ (0.80 × 25) = 16 000 ÷ 20 = 800 J/kg °C.
5. A graph of energy (vertical) against temperature (horizontal) for a 0.50 kg sample is a straight line through (20 °C, 0 J) and (60 °C, 30 000 J). Find the specific heat capacity.
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Gradient = 30 000 ÷ 40 = 750 J/°C = m × c. c = 750 ÷ 0.50 = 1500 J/kg °C.
6. A 600 W heater warms 0.60 kg of a solid to its melting point and keeps heating steadily. The temperature stays flat for 6.0 minutes while it melts. Find the specific latent heat of fusion.
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t = 6.0 × 60 = 360 s. E = 600 × 360 = 216 000 J. l = 216 000 ÷ 0.60 = 360 000 J/kg (3.6 × 10⁵ J/kg).
7. How much energy melts 0.25 kg of ice at 0 °C?
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Q = m × l = 0.25 × 330 000 = 82 500 J.
8. 0.25 kg of ice at 0 °C is melted and the water is warmed to 20 °C. Find the total energy.
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Melting: 82 500 J. Warming: 0.25 × 4200 × 20 = 21 000 J. Total = 82 500 + 21 000 = 103 500 J.
9. A 1500 W heater warms 2.0 kg of water from 20 °C to 70 °C. If no energy is lost, how long does it take?
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Q = 2.0 × 4200 × 50 = 420 000 J. t = 420 000 ÷ 1500 = 280 s, which is 4 minutes 40 seconds.
10. In question 9 the heater actually takes 350 s. Find the energy supplied, the energy lost and the efficiency of the heating.
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Supplied = 1500 × 350 = 525 000 J. Lost = 525 000 − 420 000 = 105 000 J. Efficiency = 420 000 ÷ 525 000 = 0.80, so 80%.
11. (a) How much energy turns 0.10 kg of water at 100 °C into steam at 100 °C? (b) How long does a 2000 W heater take if all its energy is used? (c) Explain why the temperature stays at 100 °C during this time.
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(a) Q = 0.10 × 2 300 000 = 230 000 J.
(b) t = 230 000 ÷ 2000 = 115 s.
(c) The energy is used to separate the particles against the forces between them, which increases their potential energy. The average kinetic energy does not change, so the temperature stays constant while the water boils.
If you got these wrong
Match the kind of error to the lesson that fixes it.
- Answers a thousand times too large or small, or wrong units: return to using specific heat capacity with compatible units. Convert grams and kilojoules first.
- Gradient questions (5), or an answer that is the upside-down of the right one: return to interpreting an energy-temperature slope. Check which axis holds energy.
- Explanations of flat sections (6, 11c): return to distinguishing a temperature change from a phase change.
- Latent heat and combined stages (6, 7, 8, 11): return to calculating latent energy where in scope.
- Losses, time and efficiency (9, 10): return to energy losses as a limitation of a model.
Log any repeated slip in the mistake log and retest queue and retry a fresh version a few days later. If the same error keeps returning, online one-to-one Physics tuition lets a teacher watch the step where it happens.