Specific heat capacity: the equation, units and worked examples

Specific heat capacity is how much energy it takes to warm 1 kg of a substance by 1 °C. Its equation is ΔE = mcΔθ: energy = mass × specific heat capacity × temperature change. Water has a specific heat capacity of about 4200 J/kg °C, so it takes 4200 J to warm 1 kg of water by 1 °C. In the US it is usually called specific heat, and the same equation is written Q = mcΔT.

What is specific heat capacity?

Different materials warm up at different rates. Put the same amount of energy into 1 kg of water and 1 kg of aluminium, and the aluminium gets hotter, nearly five times as much. That is because aluminium has a lower specific heat capacity: each kilogram needs less energy for each degree.

The specific heat capacity definition used in UK exams is: the amount of energy needed to raise the temperature of one kilogram of a substance by one degree Celsius. A high value means a material is slow to heat up and slow to cool down. A low value means it heats and cools quickly.

The specific heat capacity equation

The Greek letter θ (theta) stands for temperature and Δ (delta) means "change in", so Δθ is the final temperature minus the starting temperature. Always use the change, never just the final temperature.

Rearranging the formula

The specific heat capacity formula rearranges to find any of the four quantities:

The Specific Heat Capacity Calculator does any of these rearrangements for you and shows the working, so you can check your answers.

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Specific Heat Capacity Calculator

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Specific heat capacity units

The unit of specific heat capacity is the joule per kilogram per degree Celsius, J/kg °C. A-level papers write it as J kg⁻¹ K⁻¹. These are the same size, because a change of 1 °C is the same as a change of 1 K: only the starting point of the scales differs. So water's value is 4200 J/kg °C or 4200 J kg⁻¹ K⁻¹.

US chemistry courses often use grams instead, so water's specific heat becomes 4.18 J/(g °C). Divide a value in J/kg °C by 1000 to get J/(g °C).

Specific heat capacity of water and other materials

These are approximate values at around room temperature, from the HyperPhysics table at Georgia State University. Exam questions give you the value they want you to use, which for water is usually 4200 J/kg °C.

Water's specific heat capacity is unusually high, which is why it is so useful. A hot water bottle or a radiator full of water holds a lot of energy and releases it slowly. The sea warms up and cools down far more slowly than the land, which is why coastal places have milder winters. Metals have low values, which is why a metal spoon in hot tea warms up so quickly.

Here is a quick check before the examples.

How to calculate specific heat capacity: worked examples

Example 1: finding the energy

How much energy is needed to heat 2 kg of water from 20 °C to 70 °C? Use c = 4200 J/kg °C.

The temperature change is 70 − 20 = 50 °C.

The energy needed is 420 000 J, or 420 kJ.

Example 2: finding the specific heat capacity

A 0.5 kg metal block is heated with 9000 J of energy and its temperature rises by 20 °C. Find its specific heat capacity, and suggest what the metal might be.

The specific heat capacity is 900 J/kg °C, which matches aluminium in the table.

Example 3: finding the temperature rise

A 0.8 kg copper block absorbs 6160 J. Copper has c = 385 J/kg °C. By how much does its temperature rise?

The temperature rises by 20 °C. If it started at 18 °C, it ends at 38 °C.

Example 4: a kettle, using power and time

A 2.0 kW kettle heats 1.5 kg of water from 15 °C to 100 °C. Assuming all the energy goes into the water, how long does it take?

First the energy, with Δθ = 100 − 15 = 85 °C:

Then use energy = power × time, so time = energy ÷ power. Remember 2.0 kW = 2000 W.

It takes about 268 s, which is about 4½ minutes. A real kettle takes a little longer, because some energy heats the kettle itself and the air around it.

Example 5: mixing hot and cold water

0.2 kg of water at 80 °C is mixed with 0.3 kg of water at 20 °C. Ignoring energy lost to the surroundings, what is the final temperature T?

The energy lost by the hot water equals the energy gained by the cold water. Both are water, so c cancels:

The final temperature is 44 °C. It is closer to 20 °C than to 80 °C because there was more cold water.

Example 6: the method of mixtures (A-level)

A 0.25 kg metal block is heated to 100 °C and dropped into 0.40 kg of water at 18 °C. The water and block end at 22 °C. Find the specific heat capacity of the metal, ignoring the container and any energy lost.

Energy gained by the water, which rises by 4 °C:

The block falls from 100 °C to 22 °C, a change of 78 °C, and gives out the same 6720 J:

The metal's specific heat capacity is about 345 J/kg °C. In a real experiment, energy lost to the air would make this value come out too low, because the water would warm less than it should.

Temperature against energy: what the graph shows

The graph shows the temperature of 0.5 kg of water, starting at 20 °C, as energy is supplied. Rearranging ΔE = mcΔθ gives Δθ = ΔE ÷ (mc), so the graph is a straight line.

Here x is the energy supplied in kJ and y is the temperature in °C. The gradient is 1 ÷ (mc), so a material with a lower specific heat capacity, or a smaller mass, would give a steeper line. This is how the specific heat capacity practical works: you plot temperature against energy, measure the gradient, and work out c from it.

Heat capacity and specific heat capacity

Heat capacity is the energy it takes to warm a whole object by 1 °C, whatever its mass. Its unit is J/°C. Specific heat capacity is measured for each kilogram, which makes it a property of the material rather than of one object.

A 2 kg aluminium pan has a heat capacity of 2 × 900 = 1800 J/°C. A 1 kg aluminium pan has a heat capacity of 900 J/°C. Both have the same specific heat capacity, 900 J/kg °C, because both are aluminium.

Specific heat capacity and latent heat

The equation ΔE = mcΔθ only works while the substance stays in the same state. When ice melts or water boils, the temperature stops rising even though energy is still going in. That energy breaks bonds between particles instead, and is worked out with E = mL, the specific latent heat equation. A heating curve shows both: sloping parts where ΔE = mcΔθ applies, and flat parts where E = mL applies. Our States of Matter Simulator lets you heat particles and watch the change of state happen.

Common mistakes

• Using the final temperature instead of the change. Heating water from 20 °C to 70 °C is Δθ = 50 °C, not 70 °C.
• Mass in grams. With c in J/kg °C, the mass must be in kg. 250 g is 0.25 kg.
• Energy in kJ. 4.5 kJ is 4500 J. Convert before you substitute.
• Power in kW. 2.0 kW is 2000 W, and time must be in seconds: 5 minutes is 300 s.
• Using ΔE = mcΔθ across a change of state. Split the problem: one part for heating, one part for melting or boiling with E = mL.
• Rounding too early in a two-step problem such as the kettle. Keep the full energy value until the end.

Is the specific heat capacity equation on the equation sheet?

Yes, on AQA's GCSE sheet and in AQA's A-level booklet. This was checked against AQA's documents in November 2026; other boards publish their own sheets, so check yours.

The sheet gives you the equation, not the method. You still need to choose it, rearrange it, convert units and know the definition, which is often asked for in words. In England, Ofqual has confirmed equation sheets for GCSE Physics and Combined Science for the life of the current specifications, and AQA has published its sheets for 2027.

Practise thermal physics in a short timed paper

Specific heat capacity comes up in GCSE Physics and again in A-level thermal physics. GCSE Physics Higher and A-level Physics: Year 2 each give you a short timed paper every day, with 8 questions worth 25 to 35 marks across the course, so thermal physics turns up regularly but not in every paper.

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