12.1.7Particle Model

Specific Heat Capacity — Edexcel GCSE Combined Science 1SC0 - Physics Revision Notes & Topical Questions

5 min read 9 parts Lesson 8 of 13

📚 Edexcel GCSE Combined Science 1SC0 - Physics Hub / Topical Revision

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Part 1 of 7Why different materials heat differently

If you supply the same energy to 1 kg of water and 1 kg of copper, the copper gets much hotter.

Specific heat capacity

Different substances need different amounts of energy to raise their temperature, and this property is the specific heat capacity.

It is defined as the energy required to raise the temperature of 1 kg of the substance by 1 °C (or 1 K).

The symbol is c and the unit is joules per kilogram per degree Celsius, J kg⁻¹ °C⁻¹.

Part 2 of 7The equation

change in thermal energy = mass × specific heat capacity × change in temperature

ΔQ = m c Δθ

where ΔQ 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⁻¹ °C⁻¹ and Δθ is the temperature change in °C (final minus initial).

The equation

Rearranged: c = ΔQ ÷ (m Δθ) and Δθ = ΔQ ÷ (mc).

The equation applies only while the substance stays in one state: during melting or boiling the temperature does not change,

so the equation cannot be used there.

Part 3 of 7Typical values

Water has a very high specific heat capacity of about 4200 J kg⁻¹ °C⁻¹.

Ice is about 2100, aluminium 900, iron 450, copper 390 and lead 130 J kg⁻¹ °C⁻¹.

A high value means the substance stores a lot of energy for each degree and is slow to heat and slow to cool.

Water is therefore good for central heating and cooling systems: it carries a lot of thermal energy per kilogram.

Coastal regions have milder climates than inland ones because the large mass of sea water warms and cools slowly.

Part 4 of 7Using electrical energy

If a heater of power P (watts) is switched on for time t, the electrical energy transferred is E = Pt.

Using electrical energy

If all of it goes to the substance, Pt = mcΔθ.

In reality some energy is lost to the surroundings,

so the measured temperature rise is smaller than the calculated value and c obtained from an experiment is too large unless the apparatus is insulated.

On a graph of temperature against energy supplied, the line is straight with a gradient of 1 ÷ (mc). Doubling the mass halves the temperature rise for the same energy; a steeper line therefore means a smaller mc.

Part 5 of 7Common calculation steps

1. Identify m in kg (convert from grams). 2. Find Δθ by subtracting. 3. Write c in the right unit. 4. Multiply: ΔQ = mcΔθ. For problems where hot and cold objects exchange energy, assume energy lost by the hot object equals energy gained by the cold one if no energy escapes: m₁c₁Δθ₁ = m₂c₂Δθ₂.

Part 6 of 7Using specific heat capacity in context

Specific heat capacity explains why a metal spoon in hot soup heats up much faster than the soup itself: the spoon has small mass and low c, so a small energy transfer raises its temperature a lot.

Using specific heat capacity in context

It explains why sand on a beach becomes hot in daylight but cools quickly at night, whereas the sea changes temperature slowly.

Storage heaters use blocks of dense material with high specific heat capacity, heated at night using cheap electricity and releasing the energy during the day.

Mixing problemsa 0.20 kg block of copper at 100 °C (c = 390) is dropped into 0.50 kg of water at 20 °C (c = 4200).

Using specific heat capacity in context

Equating energy lost by copper and gained by water with final temperature T: 0.20 × 390 × (100 − T) = 0.50 × 4200 × (T − 20).

That gives 7800 − 78T = 2100T − 42 000, so 49 800 = 2178T and T ≈ 22.9 °C.

The water temperature rises only about 3 °C, which shows its large heat capacity.

Part 7 of 7Worked examples

How much energy is needed to heat 1.5 kg of water from 20 °C to 80 °C? (c = 4200 J kg⁻¹ °C⁻¹)
A 0.50 kg aluminium block (c = 900 J kg⁻¹ °C⁻¹) is heated by a 60 W heater for 3.0 minutes. Ignoring energy losses, find its temperature rise.
A 2.0 kg metal object absorbs 36 kJ and rises from 15 °C to 55 °C. Find its specific heat capacity.

Key terms

Term	Meaning
Specific heat capacity	Energy needed to raise the temperature of 1 kg of a substance by 1 °C.
Thermal energy change	Energy gained or lost by heating or cooling, ΔQ.
Temperature change	Δθ, final temperature minus initial temperature.
Joule	Unit of energy.
Insulation	Material used to reduce heat loss in an experiment.

Avoid these mistakes

Using grams instead of kilograms.

Using the final temperature rather than the temperature change.

Applying ΔQ = mcΔθ during a change of state when the temperature is constant.

Saying the high specific heat capacity of water means water is heated quickly; it means it heats slowly.

Forgetting that energy lost to the surroundings makes an experimental c value too big.

Last updated: 20 October 2026

Part 1 of 7Why different materials heat differently

Part 2 of 7The equation

Part 3 of 7Typical values

Part 4 of 7Using electrical energy

Part 5 of 7Common calculation steps

Part 6 of 7Using specific heat capacity in context

Part 7 of 7Worked examples

How much energy is needed to heat 1.5 kg of water from 20 °C to 80 °C? (c = 4200 J kg−1 °C−1)

A 0.50 kg aluminium block (c = 900 J kg−1 °C−1) is heated by a 60 W heater for 3.0 minutes. Ignoring energy losses, find its temperature rise.

A 2.0 kg metal object absorbs 36 kJ and rises from 15 °C to 55 °C. Find its specific heat capacity.

Key terms

Avoid these mistakes

How much energy is needed to heat 1.5 kg of water from 20 °C to 80 °C? (c = 4200 J kg⁻¹ °C⁻¹)

A 0.50 kg aluminium block (c = 900 J kg⁻¹ °C⁻¹) is heated by a 60 W heater for 3.0 minutes. Ignoring energy losses, find its temperature rise.