The big idea: A metal pan heats up in seconds on the hob, but a pan of water on the same ring takes ages to get hot — some things need far more energy to warm up than others.
Specific heat capacity
- The specific heat capacity (c) of a substance tells you how much energy it takes to warm 1 kg of it by 1 degree.
- Unit: J kg⁻¹ K⁻¹ (joules per kilogram per kelvin: the energy to raise 1 kg by 1 K, i.e. 1 °C).
Animated graph
Watch the graph build step by step in study mode.
Big c vs small c: A big c means the substance is hard to heat up — it soaks up lots of energy per degree.
Water has a very big c (about 4200 J kg⁻¹ K⁻¹), so it warms and cools slowly. A metal has a small c, so it heats up fast.
Free preview
This is the free notes preview
You're reading the free notes. Aimnova Pro unlocks the full study experience — and you can try it with your first topic free to keep:
- FlashcardsLock in vocabulary and key terms with spaced repetition.
- Practice questionsAnswer exam-style questions and get instant AI marking.
- Mock exams & past-paper vaultSit full mocks and see exactly how examiners award marks.
- Personalised study planA daily plan built around your exam date and weak areas.
The energy you need depends on three things: how much stuff there is (mass m), how hard it is to heat (c), and how big a temperature rise you want (ΔT).
Multiply them together:
- thermal energy added or removed (J)
- mass of the substance (kg)
- specific heat capacity (J kg⁻¹ K⁻¹)
- temperature change (K, or °C — same size)
ΔT is a temperature CHANGE: ΔT means final temperature − start temperature (the Greek letter Δ, 'delta', means 'change in').
A change of 1 K is the same size as a change of 1 degree C, so you can use either — never convert to kelvin for ΔT.
| Symbol | Quantity | Unit |
|---|---|---|
| Q | thermal energy added/removed | J (joules) |
| m | mass | kg |
| c | specific heat capacity | J kg⁻¹ K⁻¹ |
| ΔT | temperature change (final − start) | K or degrees C |
How much energy is needed to heat 0.50 kg of water from 20 degrees C to 80 degrees C? Water has c = 4200 J kg⁻¹ K⁻¹.
Model answer plan
See the answer broken into clear points, with guidance on how each point earns marks.
Energy from an electric heater: A heater of power P running for a time t supplies Q = Pt (or Q = IVt from the current and the p.d.). Time in seconds.
A 50 W heater for 3.0 minutes: Q = 50 × 180 = 9.0 × 10³ J. If it warms 0.20 kg of oil by 25 K, c = 9.0 × 10³ ÷ (0.20 × 25) = 1.8 × 10³ J kg⁻¹ K⁻¹.
Energy lost = energy gained: When a hot object and a cold object reach the same temperature with no energy lost to the surroundings, the energy the hot one loses equals the energy the cold one gains:
m₁c₁ΔT₁ = m₂c₂ΔT₂, each with its own temperature change.
0.20 kg of a metal at 100 °C is put into 0.50 kg of water at 20 °C. The final temperature is 24 °C. Water has c = 4200 J kg⁻¹ K⁻¹. Determine the specific heat capacity of the metal, assuming no energy is lost.
Model answer plan
See the answer broken into clear points, with guidance on how each point earns marks.
Practice with exam-style questions
Answer exam-style questions and get AI feedback that shows you exactly what examiners want to see in a full-marks response.
How this is tested — Q = mcΔT is one of the most-used equations in Theme B:
Paper 1A
- Quick MCQs — compare the energy two substances need.
- Or spot that ΔT in K equals ΔT in degrees C.
Paper 2
- Rearrange the formula to find an unknown — most often the specific heat capacity c from a heating experiment, sometimes the mass or the energy.
The classic trap: Putting the actual temperature into ΔT instead of the change, or forgetting that cooling means Q comes out (negative).
Rearranging for c: To find a substance's specific heat capacity from an experiment, rearrange the given formula:
c = Q ÷ (m × ΔT) — energy supplied, divided by mass and by the temperature rise.
The words to know
| Word | What it means |
|---|---|
| Specific heat capacity c | Energy to raise the temperature of 1 kg by 1 K (J kg⁻¹ K⁻¹) |
| Thermal equilibrium | Same temperature: no net energy transfer between the bodies |
| Energy supplied by a heater | Q = Pt = IVt (t in seconds) |
In an experiment, 9.0 × 10³ J of energy is supplied to a 0.40 kg block of aluminium. Its temperature rises by 25 K. Calculate the specific heat capacity of aluminium.
Model answer plan
See the answer broken into clear points, with guidance on how each point earns marks.
The experiment
Why the measured value comes out wrong: Some energy always escapes to the surroundings, so the block gains less than IVt. Using all of IVt as Q makes the calculated c too big. In a mixing experiment the same loss makes the final temperature lower than predicted.
Improvements: insulate the block or use a lid, stir so the temperature is even, and use a bigger temperature rise so reading errors matter less.
Reading a cooling curve: A hot object loses energy faster when it is much hotter than the room. As it cools, the temperature difference shrinks, so it cools more slowly: equal time intervals give smaller and smaller temperature drops, and the curve flattens out.