Specific heat capacity, explained with examples
Specific heat capacity tells you how much energy it takes to warm up a material. It explains why a metal spoon heats quickly in hot tea while the sea stays cool on a hot day.
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The definition
The specific heat capacity of a substance is the amount of energy needed to raise the temperature of 1 kg of the substance by 1 °C. Its unit is joules per kilogram per degree Celsius, J/kg °C.
Water has a specific heat capacity of about 4200 J/kg °C, which is very high. Copper is about 385 J/kg °C and aluminium about 900 J/kg °C. So it takes more than ten times as much energy to warm a kilogram of water by one degree as to warm a kilogram of copper.
The equation
ΔE = m c Δθ
- ΔE is the change in thermal energy, in joules (J)
- m is the mass, in kilograms (kg)
- c is the specific heat capacity, in J/kg °C
- Δθ is the temperature change, in °C
The Δ symbol means “change in”. Δθ is the final temperature minus the starting temperature. Rearranged forms: c = ΔE/(mΔθ), m = ΔE/(cΔθ) and Δθ = ΔE/(mc). It is provided on the GCSE equations sheet, but you need to be able to choose and use it quickly.
Worked examples
Example 1: energy needed
How much energy is needed to heat 2 kg of water from 20 °C to 70 °C? (c = 4200 J/kg °C)
- Δθ = 70 − 20 = 50 °C.
- ΔE = 2 × 4200 × 50 = 420,000 J, or 420 kJ.
Example 2: temperature rise
A 0.5 kg aluminium block (c = 900 J/kg °C) absorbs 9000 J. By how much does its temperature rise?
- Δθ = ΔE/(mc) = 9000/(0.5 × 900) = 20 °C.
Example 3: finding c
A 1.5 kg metal block is heated with 27,000 J and its temperature rises from 18 °C to 58 °C. Find c.
- Δθ = 40 °C. c = 27,000/(1.5 × 40) = 450 J/kg °C, close to iron.
Example 4: grams to kilograms
How much energy raises 250 g of water by 80 °C?
- m = 0.25 kg. ΔE = 0.25 × 4200 × 80 = 84,000 J.
Example 5: using power and time
A 2 kW kettle heats 1.2 kg of water from 15 °C to 100 °C. Assuming no energy is wasted, how long does it take?
- ΔE = 1.2 × 4200 × 85 = 428,400 J.
- Energy = power × time, so t = 428,400/2000 ≈ 214 s, about 3.6 minutes. In reality it takes longer, because some energy heats the kettle and the air.
Why water's high value matters
Water's high specific heat capacity means it can store a lot of energy and changes temperature slowly. That is why central heating systems use water to carry energy around a house, why hot water bottles stay warm for hours, and why coastal areas have milder climates than inland areas: the sea warms up and cools down much more slowly than land. It also helps living things, which are mostly water, keep a steady body temperature.
The required practical
The aim is to find the specific heat capacity of a material, usually a metal block, using an electric heater.
- Measure the mass of the block with a balance.
- Place a heater in one hole in the block and a thermometer in the other. Add a drop of water or oil to the thermometer hole for good thermal contact.
- Wrap the block in insulation to reduce energy loss.
- Connect the heater to a power supply with a joulemeter, or with an ammeter and voltmeter so you can calculate power with P = IV.
- Record the starting temperature, switch on, and record the temperature and energy supplied every minute for about ten minutes.
- Plot temperature against energy supplied. The gradient equals 1/(mc), so c = 1/(m × gradient).
Errors and improvements
- Energy is lost to the surroundings, so the measured temperature rise is smaller than it should be and the calculated c is too high. Use more insulation, and a lid if heating a liquid.
- The thermometer may lag behind the true temperature. Keep recording for a short time after switching off and use the highest reading.
- Not all the heater's energy goes into the block; some heats the heater itself.
Example 6: analysing results
In the practical, a 1 kg block received 15,000 J and its temperature rose by 16 °C. Find c and explain why it is higher than the accepted value of 900 J/kg °C.
- c = 15,000/(1 × 16) ≈ 938 J/kg °C.
- Some of the 15,000 J was lost to the surroundings, so less energy heated the block than was assumed. The temperature rise was smaller, which made the calculated c too large.
Specific heat capacity versus specific latent heat
Specific heat capacity is about changing temperature. Specific latent heat is about changing state without changing temperature, for example ice melting at 0 °C. Its equation is E = mL. When a substance melts or boils, energy breaks the bonds between particles rather than speeding them up, so the temperature stays constant until the change of state is complete. Questions sometimes combine the two, such as heating ice to 0 °C, melting it, and then heating the water.
Common mistakes
- Using grams instead of kilograms.
- Using the final temperature instead of the temperature change.
- Mixing up specific heat capacity (temperature change) and specific latent heat (change of state).
- Saying energy losses make the calculated value too low; they make it too high.
Practice questions
Try each one before opening the answer.
Q1. How much energy is needed to heat 3 kg of water by 10 °C?
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Q2. A 2 kg copper block (c = 385 J/kg °C) cools from 80 °C to 30 °C. How much energy does it release?
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Q3. Which heats up faster with the same energy input: 1 kg of water or 1 kg of aluminium?
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Q4. A 50 W heater runs for 5 minutes and heats 0.4 kg of oil by 18.75 °C. Find c.
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Frequently asked questions
Is specific heat capacity in Combined Science?
Yes. The equation and the required practical are in the energy topic for both Combined Science and separate Physics.
Why do the units sometimes say J/kg K?
A temperature change of 1 °C is the same size as a change of 1 kelvin, so J/kg °C and J/kg K mean the same thing. A level uses kelvin.
Key takeaways
- Specific heat capacity is the energy to raise 1 kg by 1 °C.
- ΔE = mcΔθ, with mass in kg.
- Water's high value makes it a good store of thermal energy.
- In the practical, energy losses make c come out too high.
- Specific latent heat is for changes of state, not temperature.