4.1.1.3 Specific Heat Capacity — Physics with Kate

4.1.1.3REQUIRED PRACTICAL

AQA GCSE Physics · Topic 4.1 Energy

Energy Changes in Systems: Specific Heat Capacity

In this lesson you'll learn to: say what specific heat capacity means, use ΔE = m c Δθ, and carry out the required practical — heating a liquid, plotting temperature against time, and getting c from the gradient of the graph.

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Start here — run the experiment

Do the practical before you read about it. Pick a liquid, set the mass and the heater power, and press Start heating. Watch the thermometer and the graph together — and notice that the graph is flat for the first minute or so before it becomes a straight line. Then swap water for cooking oil and see how much faster the oil heats up.

Liquid:

Mass 0.5 kg Heater power 60 W

The working

Required practical · the run is sped up 60× · readings taken at 3 min and 8 min

▲ The whole 10-minute experiment, sped up 60×, with the working done for you.

Now find out why the oil heats faster

It's not because oil is "thinner" — it's because oil has a lower specific heat capacity. The notes below explain what that means, give you the equation, and walk through the practical step by step.

Read the notes ↓

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Revision notes

What specific heat capacity means

Some substances are easy to heat up and some are stubborn. Specific heat capacity is the number that tells you which is which.

Learn this definition word for word

The specific heat capacity of a substance is the amount of energy needed to raise the temperature of one kilogram of the substance by one degree Celsius.

Water has a specific heat capacity of 4200 J/kg°C. That means you must put in 4200 joules to warm 1 kg of water by just 1 °C. Cooking oil only needs about 2000 J for the same job — which is why the oil in the experiment heats up more than twice as fast as the water.

⭐ Key fact — a high specific heat capacity cuts both ways

A substance with a high specific heat capacity is slow to heat up and equally slow to cool down. That's why water is used in central heating and car radiators, and why the sea is still cold in May and still warm in October.

The equation

ΔE = m c Δθ

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

Symbol	Quantity	Unit
ΔE	change in thermal energy	J (joules)
m	mass	kg
c	specific heat capacity	J/kg°C
Δθ	temperature change	°C

💡 Exam tip — Δθ is a change, not a temperature

The Greek letter Δ ("delta") means change in. So if water goes from 18 °C to 100 °C, then Δθ = 100 − 18 = 82 °C, not 100. Subtracting is the step people forget.

Some values to get a feel for

Substance	Specific heat capacity (J/kg°C)	Heats up…
Water	4200	very slowly
Ethanol	2440	fairly slowly
Ice	2100	fairly slowly
Cooking oil	2000	fairly quickly
Aluminium	900	quickly
Glass	670	quickly
Copper	385	very quickly
Lead	130	almost instantly

You are never asked to memorise these — they're given in the question. But it's worth knowing that water's is unusually high, and that metals are low.

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Required practical: finding c

The aim is to measure the specific heat capacity of a liquid (or a metal block) by heating it with a known power and timing how fast the temperature rises.

Method

Measure the mass of the liquid in kg using a balance. (Weigh the empty beaker first and subtract.)
Insulate the beaker with lagging and add a lid, so as little energy as possible escapes to the room.
Put an immersion heater and a thermometer into the liquid. Use a joulemeter, or an ammeter and voltmeter, to find the power of the heater.
Record the starting temperature, switch the heater on and start a stopwatch.
Record the temperature every minute for about 10 minutes.
Plot a graph of temperature (y) against time (x).
Draw a line of best fit through the straight part and find its gradient.

Getting c from the gradient

Two equations describe the same energy. The heater supplies:

E = P t

energy supplied = power × time

and that energy warms the liquid:

E = m c Δθ

Put them equal and rearrange:

c = P ÷ (m × gradient)

where the gradient is Δθ ÷ Δt, in °C per second

💡 Exam tip — minutes into seconds, every time

Your graph has minutes on the x-axis, but the power is in watts — joules per second. Convert before you divide: 5.0 min × 60 = 300 s. Forget this and your answer is 60 times too small.

💡 Exam tip — why the graph starts flat

Nothing happens to the temperature for the first minute or so, because the heater itself has to warm up before it starts transferring energy to the liquid. Always take your gradient from the straight section, never from the whole line.

⭐ Key fact — why the measured value is always too high

You assume every joule from the heater goes into the liquid, but some warms the beaker, the thermometer and the room. So the real Δθ is smaller than it should be, and your calculated c comes out higher than the true value. Improve it by lagging the container and using a lid.

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Worked examples

EXAMPLE 1A kettle heats 250 g of water from 18 °C to 100 °C. The specific heat capacity of water is 4200 J/kg°C. Calculate the energy transferred.

Convert the mass, and work out the temperature change 250 g ÷ 1000 = 0.25 kg Δθ = 100 − 18 = 82 °C

Substitute into ΔE = m c Δθ ΔE = 0.25 kg × 4200 J/kg°C × 82 °C = 86 100 J

ΔE = 86 000 J (86 kJ) To 2 s.f. Writing it in kJ is fine — just don't drop the unit.

EXAMPLE 2A 2.0 kg aluminium block (c = 900 J/kg°C) is given 45 000 J of energy. Calculate its temperature rise.

Rearrange ΔE = m c Δθ for Δθ Δθ = ΔE ÷ (m c) = 45 000 ÷ (2.0 × 900) = 45 000 ÷ 1800 = 25 °C

Δθ = 25 °C A rise of 25 °C. If it started at 20 °C it is now at 45 °C.

EXAMPLE 3The practical. A 60 W heater warms 0.40 kg of oil. The temperature rises by 30 °C in 400 s. Calculate the specific heat capacity of the oil.

Energy supplied by the heater E = P t = 60 W × 400 s = 24 000 J

Rearrange ΔE = m c Δθ for c c = ΔE ÷ (m Δθ) = 24 000 ÷ (0.40 × 30) = 24 000 ÷ 12 = 2000 J/kg°C

c = 2000 J/kg°C Close to the book value for cooking oil — and notice the unit: J/kg°C.

📘 Now do it in your workbook

Specific heat capacity calculations and the full required-practical write-up, with worked answers so you can mark your own work.

AQA GCSE Physics Workbook · ENERGY · pages 9 & 10

⬇ Free: pages 9 & 10 (PDF) Get the full workbook on Amazon

The full workbook covers the whole of Topic 1 Energy with exam-style questions and worked answers.

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Prefer to watch? Here's the whole thing

The practical demonstrated end to end, including how to draw the gradient triangle and get c out of it.

▲ In GHL you can also use the lesson's built-in video field instead of this embed.

Now put it into practice

Pages 9 & 10 of the workbook are free. Required practicals are worth a lot of marks and come up every year — the workbook has exam-style questions on the method, the graph and the calculation, with full worked answers.

⬇ Free: pages 9 & 10 (PDF) Full workbook on Amazon