Specific Heat Capacity
Different substances need different amounts of energy to heat up. Specific heat capacity measures that, and one equation covers the whole topic — with the marks going on units and on which temperature to use.
1. The definition
Specific heat capacity (c) is the energy required to raise the temperature of 1 kg of a substance by 1 °C.
Unit: J/(kg °C) — joules per kilogram per degree Celsius.
Learn this definition word for word. Tutors flagged it as a guaranteed exam item, and every term matters: 1 kg, 1 °C, and energy.
Typical values:
| Substance | c (J/kg °C) |
|---|---|
| Water | 4200 |
| Aluminium | 900 |
| Copper | 385 |
| Lead | 130 |
A high specific heat capacity means the substance needs a lot of energy to warm up — and releases a lot as it cools.
Thermal capacity (heat capacity) is a related but different idea:
Thermal capacity = energy to raise the temperature of the WHOLE OBJECT by 1 °C = m × c, in J/°C.
“Specific” means per kilogram. Thermal capacity depends on the mass of the particular object; specific heat capacity is a property of the material. Confusing the two was a recorded error, as was trying to use the specific heat equation when only a thermal capacity was given.
2. The equation
E = m c Δθ energy (J) = mass (kg) × specific heat capacity (J/kg °C) × temperature CHANGE (°C)
Δθ (delta theta) means the CHANGE in temperature — the difference, not the final value.
Example: heating 2 kg of water from 20 °C to 70 °C.
- Δθ = 70 − 20 = 50 °C
- E = 2 × 4200 × 50 = 420 000 J (420 kJ)
Use the temperature DIFFERENCE, not the final temperature. Substituting 70 instead of 50 was a recorded error and is the most common mistake in the topic.
c is the specific heat capacity, not a temperature. Confusion over what “c” represents, and mixing temperature with time in the equation, were both recorded.
Rearranged forms:
- m = E / (c Δθ)
- c = E / (m Δθ)
- Δθ = E / (m c)
A temperature change in °C equals the same change in K, so you can use either for Δθ — but not for absolute temperatures.
3. Units — where the marks go
Mass must be in KILOGRAMS.
- g → kg: ÷ 1000. So 500 g = 0.5 kg
- kJ → J: × 1000. So 12 kJ = 12 000 J
- minutes → seconds: × 60
Converting grams to kilograms, and kilojoules to joules, were both directly recorded errors. Make the conversion the first line of your working — an unconverted mass makes the answer wrong by a factor of 1000.
Give answers to two or three significant figures and always include the unit.
4. Combining with electrical energy
Many questions supply the energy electrically.
E = P t (energy = power × time) P = E / t
Power is energy DIVIDED by time. A recorded error used “P = EIT”; the correct relationships are P = E/t and, for electrical circuits, P = VI.
Worked example. A 60 W heater warms 0.4 kg of water for 3 minutes. Find the temperature rise (c = 4200).
- t = 3 × 60 = 180 s
- E = P t = 60 × 180 = 10 800 J
- Δθ = E/(mc) = 10 800 ÷ (0.4 × 4200) = 10 800 ÷ 1680 = 6.4 °C
Convert minutes to seconds before using E = Pt. Recorded twice, and it changes the answer by a factor of 60.
Efficiency and percentage questions
Some questions state that only part of the energy is absorbed.
Example: a solar panel receives 2000 J and absorbs 25%.
- Energy absorbed = 0.25 × 2000 = 500 J — this is the value to use in E = mcΔθ
Read carefully whether a percentage applies to the energy supplied or absorbed. A recorded error calculated only 25% when the question needed the full incident energy, and another missed the 25% entirely. Decide which quantity the percentage modifies before substituting.
Watch for deliberately irrelevant information. Tutors warned that questions include numbers you don’t need — being given a value doesn’t mean it belongs in the equation.
5. Measuring specific heat capacity
For a solid metal block:
- Measure the mass of the block with a balance
- Insert an electric heater and a thermometer into the drilled holes
- Record the starting temperature
- Switch on and record the power and the time
- Record the final temperature and find Δθ
- Calculate: c = E / (m Δθ), with E = P t
For a liquid: heat a known mass in a well-insulated container and follow the same steps, stirring to distribute the energy evenly.
Sources of error and improvements:
Some energy is lost to the surroundings, so the measured c comes out too high — you supplied more energy than the substance actually absorbed.
- Insulate the container and add a lid to reduce losses
- Stir the liquid so the thermometer reads the average temperature
- Allow for the thermometer’s response time before reading
- Repeat and average
6. Why water’s high value matters
Water’s specific heat capacity (4200 J/kg °C) is unusually large, which is why:
- It is used as a coolant in car engines and power stations — it absorbs a lot of energy for a small temperature rise
- It is used in central heating — it carries a lot of energy round the house
- Coastal climates are milder — the sea heats and cools slowly, moderating temperatures
7. Mistakes that cost marks
Using the final temperature instead of the change.
Leaving mass in grams.
Not converting kJ to J, or minutes to seconds.
Confusing specific heat capacity with thermal capacity.
Confusing specific heat capacity with specific latent heat (which involves no temperature change).
Writing P = E × t instead of E/t.
Misapplying a percentage to the wrong energy value.
Using irrelevant given values.
Omitting units — J for energy, J/kg °C for c.
Frequently asked questions
What is specific heat capacity? The energy needed to raise the temperature of 1 kg of a substance by 1 °C.
What is the unit? J/(kg °C).
What is the equation? E = m c Δθ.
What does Δθ mean? The change in temperature — final minus initial.
What is the specific heat capacity of water? 4200 J/kg °C.
What is thermal capacity? The energy to raise the temperature of the whole object by 1 °C — equal to m × c.
How is this different from latent heat? Specific heat capacity involves a temperature change; latent heat involves a change of state at constant temperature.
How do I find the energy from a heater? E = P t, with time in seconds.
Why is my measured value too high? Energy is lost to the surroundings, so more was supplied than absorbed.
Why is water used as a coolant? Its high specific heat capacity lets it absorb a lot of energy for a small temperature rise.
Quick revision checklist
- I can state the definition precisely
- I know the unit J/(kg °C)
- I know water’s value is 4200
- I can use E = mcΔθ and rearrange it
- I use the temperature change, not the final temperature
- I convert g to kg and kJ to J
- I convert minutes to seconds for E = Pt
- I know P = E/t
- I can handle percentage-absorption questions
- I can distinguish specific heat capacity from thermal capacity
- I can distinguish it from latent heat
- I can describe the experiment and its errors
- I can explain why the measured value comes out high
- I can explain why water’s high value is useful
These notes cover specific heat capacity in the Cambridge IGCSE Physics (0625) syllabus and are written for Grade 9–11 / Year 10–11 students. They are based on teaching patterns observed across a large set of one-to-one IGCSE Physics lessons, with particular attention to the errors students make most often and the wording examiners reward. Always check the current syllabus and formula list for your own exam series.
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