Specific Heat Capacity

In this lesson you will learn what specific heat capacity means, how to use the equation linking energy, mass, specific heat capacity, and temperature change, and why different materials heat up at different rates. This is part of AQA GCSE Combined Science Trilogy (8464), Section 6.1.

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What Is Specific Heat Capacity?

Specific heat capacity is the amount of energy required to raise the temperature of 1 kg of a substance by 1 °C (or 1 K).

Different materials have different specific heat capacities. A material with a high specific heat capacity requires more energy to increase its temperature and also releases more energy when it cools.

Material	Specific Heat Capacity (J/kg °C)
Water	4200
Aluminium	900
Copper	390
Iron	450
Lead	130
Oil	2000

Exam Tip: Water has a very high specific heat capacity. This is why it is used in central heating systems — it can store and release large amounts of energy. AQA frequently asks you to explain practical applications using this fact.

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The Equation

$\Delta E = m \, c \, \Delta \theta$ΔE=mcΔθ

Where:

• $\Delta E$ΔE = change in energy (or energy transferred) in joules (J)
• $m$m = mass in kilograms (kg)
• $c$c = specific heat capacity in J/(kg °C)
• $\Delta \theta$Δθ = change in temperature in °C

Exam Tip: This equation is on the AQA equation sheet. Make sure you can rearrange it to find any of the four variables.

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Worked Examples

Worked Example 1

How much energy is needed to heat 3 kg of water from 20 °C to 80 °C? ($c$c for water = 4200 J/kg °C)

$\Delta E = 3 \times 4200 \times (80 - 20) = 3 \times 4200 \times 60 = 756{,}000 \text{ J} = 756 \text{ kJ}$ΔE=3×4200×(80−20)=3×4200×60=756,000 J=756 kJ

Worked Example 2

A 2 kg copper block absorbs 15,600 J of energy. What is its temperature rise? ($c$c for copper = 390 J/kg °C)

$\Delta \theta = \frac{\Delta E}{m \, c} = \frac{15{,}600}{2 \times 390} = \frac{15{,}600}{780} = 20 \text{ °C}$Δθ=mcΔE​=2×39015,600​=78015,600​=20 °C

Worked Example 3

156,000 J of energy raises the temperature of a substance by 40 °C. Its mass is 5 kg. What is the specific heat capacity?

$c = \frac{\Delta E}{m \, \Delta \theta} = \frac{156{,}000}{5 \times 40} = \frac{156{,}000}{200} = 780 \text{ J/(kg °C)}$c=mΔθΔE​=5×40156,000​=200156,000​=780 J/(kg °C)

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Why Specific Heat Capacity Matters

Heating and Cooling Rates

A material with a low specific heat capacity heats up quickly but also cools down quickly. A material with a high specific heat capacity heats up slowly but retains energy for longer.

graph LR
    A["Low c material\n(e.g. copper)"] -->|"Heats up quickly"| B["High temperature\nreached faster"]
    C["High c material\n(e.g. water)"] -->|"Heats up slowly"| D["Stores more energy\nper degree rise"]

Practical Applications

Application	Material	Why
Central heating radiators	Water	High $c$c — stores lots of energy
Cooking pans	Copper / aluminium	Low $c$c — heats up quickly
Car engine coolant	Water	High $c$c — absorbs lots of heat without boiling
Storage heaters	Concrete / brick	Moderate $c$c — stores energy at night, releases during the day

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Linking to Internal Energy

When you heat a substance, you increase its internal (thermal) energy. Internal energy is the total kinetic energy and potential energy of all the particles within the substance.

• Increasing temperature → particles move faster → kinetic energy of particles increases → internal energy increases.
• The energy transferred to a substance using $\Delta E = mc\Delta\theta$ΔE=mcΔθ goes into the internal energy store.

Exam Tip: AQA may ask you to link specific heat capacity to changes in internal energy. Always state that heating increases the kinetic energy of particles, which increases the internal (thermal) energy of the substance.

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Common Mistakes

Mistake	Correction
Using temperature instead of temperature change	$\Delta\theta$Δθ is the change, not the final temperature
Confusing heat and temperature	Heat is energy transferred; temperature measures how hot something is
Using grams instead of kilograms	Always convert mass to kg before substituting
Forgetting units on the answer	Energy in J (or kJ), temperature in °C

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Summary

• Specific heat capacity ($c$c) is the energy needed to raise 1 kg of a substance by 1 °C.
• Equation: $\Delta E = mc\Delta\theta$ΔE=mcΔθ.
• Water has a very high specific heat capacity (4200 J/kg °C).
• Materials with low $c$c heat up and cool down quickly; materials with high $c$c store more energy.
• Heating a substance increases the internal (thermal) energy of its particles.
• This is tested in AQA GCSE Combined Science Trilogy (8464), Section 6.1.

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Extended Worked Examples

Worked Example 4 — Mixing Problem

500 g of water at 20 °C is mixed with 200 g of water at 80 °C. Assuming no heat loss, what is the final temperature? ($c$c for water = 4200 J/(kg °C).)

Let the final temperature be $T$T.

Energy lost by hot water = Energy gained by cold water:

$0.2 \times 4200 \times (80 - T) = 0.5 \times 4200 \times (T - 20)$0.2×4200×(80−T)=0.5×4200×(T−20)

$0.2 \times (80 - T) = 0.5 \times (T - 20)$0.2×(80−T)=0.5×(T−20)

$16 - 0.2T = 0.5T - 10$16−0.2T=0.5T−10

$26 = 0.7T \implies T \approx 37.1 \text{ °C}$26=0.7T⟹T≈37.1 °C

Worked Example 5 — Rearrangement for Mass

An immersion heater delivers 252,000 J to water at 20 °C, raising its temperature to 80 °C. What mass of water was heated?

$m = \frac{\Delta E}{c \Delta\theta} = \frac{252{,}000}{4200 \times 60} = \frac{252{,}000}{252{,}000} = 1 \text{ kg}$m=cΔθΔE​=4200×60252,000​=252,000252,000​=1 kg

Worked Example 6 — Comparing Two Materials

You supply 10,000 J of energy to 1 kg of aluminium ($c = 900$c=900) and 1 kg of copper ($c = 390$c=390). Starting at 20 °C, what temperature does each reach?

Aluminium: $\Delta\theta = 10{,}000 / (1 \times 900) = 11.1 \text{ °C}$Δθ=10,000/(1×900)=11.1 °C, final 31.1 °C.

Copper: $\Delta\theta = 10{,}000 / (1 \times 390) = 25.6 \text{ °C}$Δθ=10,000/(1×390)=25.6 °C, final 45.6 °C.

Copper heats up more than twice as much for the same energy — it has a lower specific heat capacity.

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Comparison Table — High vs Low Specific Heat Capacity