Energy changes in systems specific heat capacity & RP1

SPEC 4.1.1.3 • ENERGY — SHC & RP1

Specific Heat Capacity

The equation, the values, and Required Practical 1 — the most-examined practical of the qualification's first paper. One equation, one experiment, two ways examiners trap students.

Examined Every Series

Specific heat capacity (SHC) is one of the most reliable Paper 1 calculations. RP1 is the practical assessed alongside it — and it appears in roughly half of recent paper-1 series. Both together are about 6–8 marks per paper.

1

Equation

RP1

First Required Practical

⭐

High-Priority Topic

By the end of this lesson

You'll define SHC, apply ΔE = m c Δθ in both directions (energy from data, or c from data), describe RP1's apparatus and method, identify why your measured c usually comes out too high, and avoid the unit-conversion trap that cost marks in Jun 2024.

Specific Heat Capacity

Different materials need different amounts of energy to warm up. Specific heat capacity quantifies that — per kilogram, per degree.

Specific Heat Capacity (c)

The amount of energy required to raise the temperature of 1 kilogram of a substance by 1 degree Celsius.

Units: J/kg°C (joules per kilogram per degree Celsius).

"Specific" matters

"Specific heat capacity" (per kg, per °C) is not the same as "heat capacity" (total energy per °C, no per-kg). Examiner reports flag students who use them interchangeably — they're different physical quantities.

Common SHC Values

Materials with high SHC need lots of energy to warm up — and store lots of thermal energy. That's why water is used in central heating systems and why coastal towns have milder climates than inland ones.

Water

4200

J/kg°C

Aluminium

900 J/kg°C

Iron

~450

J/kg°C

Copper

~400

J/kg°C

Why water is special

Water's SHC (4200) is roughly 10× higher than most metals. To raise 1 kg of water by 1°C takes 4200 J; to raise 1 kg of aluminium by 1°C takes only 900 J. This is the physics behind hot-water bottles, central heating, and even why your body tolerates rapid temperature changes (you're mostly water).

Values can vary

These are textbook approximations. The exact SHC depends on the sample and alloy — AQA past papers have used 450 or 500 J/kg°C for iron, for example. The question will always provide the value you need to substitute. The table above is for sanity-checking your final answer, not for substitution.

The Thermal Energy Equation

Take the SHC definition and scale it: any mass, any temperature change. The result is the central equation of this lesson.

Equation (on equation sheet)

ΔE = m c Δθ

ΔE = change in thermal energy in joules (J)

m = mass in kilograms (kg)

c = specific heat capacity in J/kg°C

Δθ = temperature change in degrees Celsius (°C)

Rearranging for c

The same equation, rearranged: c = ΔE / (m × Δθ). This is the form you'll use in RP1 to measure a material's SHC from experimental data.

When the equation applies

Only when the substance is not changing state. During melting or boiling, energy goes into breaking bonds rather than raising temperature, so latent heat (Lesson 3.04) takes over. ΔE = m c Δθ describes the sloped sections of a heating curve, not the flat ones.

The Δθ Trap — It's the CHANGE

The single most common error on this equation: substituting the final temperature instead of the temperature change. They're not the same.

✗ Wrong substitution

"Heat 0.5 kg of water from 20°C to 100°C."

ΔE = 0.5 × 4200 × 100

= 210 000 J

Used the final temperature. Answer is too big by a factor of 100/80.

✓ Correct

"Heat 0.5 kg of water from 20°C to 100°C."

Δθ = 100 − 20 = 80°C

ΔE = 0.5 × 4200 × 80

= 168 000 J

Δθ is the change. Calculate it on its own line, then substitute.

Habit to build

Whenever the question gives two temperatures, write "Δθ = (final − initial) = (X − Y) = Z °C" as a separate line before substituting into the main equation. The Δ symbol means "change in" — never the final value.

Worked Example — Heating Water

A kettle contains 0.50 kg of water at 20 °C. The water is heated to 100 °C. The specific heat capacity of water is 4200 J/kg°C. Calculate the energy needed. 3 marks

EEquation:ΔE = m c Δθ

SSubstitute:Δθ = 100 − 20 = 80 °C. ΔE = 0.50 × 4200 × 80

UUnit: joules (J)

AAnswer:ΔE = 168 000 J = 168 kJ

Sanity check

168 kJ is the energy for half a kettle. A 2 kW kettle (= 2000 J/s) would deliver this in 168 000 / 2000 = ~84 seconds — about a minute and a half. That matches reality.

Worked Example — Find c

An aluminium block of mass 1.0 kg is heated using 9000 J of energy. The temperature rises from 20 °C to 30 °C. Calculate the specific heat capacity. 4 marks

EEquation:ΔE = m c Δθ → c = ΔE / (m × Δθ)

SSubstitute:Δθ = 30 − 20 = 10 °C. c = 9000 / (1.0 × 10)

UUnit: J/kg°C

AAnswer:c = 900 J/kg°C

Recognise the value

900 J/kg°C is the standard SHC of aluminium. If the answer is close to a known value (4200 water, 900 Al, ~450 Fe, ~400 Cu), it's a good sign you've done the calc correctly. If you got 9000 or 90, recheck the arithmetic.

Required Practical 1

The first required practical of the course. The aim, in spec wording:

"An investigation to determine the specific heat capacity of one or more materials. The investigation will involve linking the decrease of one energy store (or work done) to the increase in temperature and subsequent increase in thermal energy stored." — Spec 8.2.1, Required Practical 1

In plain English

Use a known amount of electrical energy to heat a metal block. Measure the temperature change. Use ΔE = m c Δθ in reverse (c = ΔE / m Δθ) to find c.

RP1 Apparatus

A standard set-up: power supply, joulemeter (measures the electrical energy delivered), insulated metal block, immersion heater, thermometer, and a stopwatch.

Why this set-up

The block has two holes drilled into it — one for the heater, one for the thermometer. The joulemeter measures electrical energy delivered directly, avoiding errors from V × I × t calculations. The insulating jacket reduces heat loss to the surroundings.

No joulemeter? Use V × I × t

If the apparatus has a voltmeter and ammeter instead of a joulemeter, calculate energy with ΔE = V × I × t — voltage from the voltmeter, current from the ammeter, time from the stopwatch. This is the same energy the joulemeter would measure, just calculated rather than read directly. Past papers use both versions.

Variables & Controls

TypeWhat & how

Independent (IV)The material of the block (e.g. aluminium block vs copper block) — or, when finding c of a single material, the energy supplied at fixed time intervals.

Dependent (DV)The temperature change Δθ, measured with the thermometer.

Control 1Mass of each block — weighed on a balance — kept the same so the result per kg is comparable.

Control 2Energy input — same heater, same time — so ΔE is identical for fair comparison.

Control 3Insulation — block wrapped in an insulating jacket — to reduce heat loss to the surroundings.

Mark-scheme tip

Every control needs a reason. "Same mass" earns 1 mark. "Same mass so c is per kg comparable across materials" earns 2 marks. Reason-giving is the most-cited improvement in RP examiner reports.

Method

1

Measure the mass of the metal block on a digital balance. Record m .

2

Wrap the block in an insulating jacket to reduce heat loss to the surroundings.

3

Insert the heater and thermometer into the two holes. Connect the heater to a joulemeter and power supply (or measure V and I if no joulemeter is available).

4

Record the initial temperature only after it has stabilised at room temperature (don't rush this step).

5

Switch on the heater for a fixed time (e.g. 5 minutes). Read the energy from the joulemeter (or calculate ΔE = V × I × t).

6

Record the final temperature when it stops rising. Calculate Δθ = final − initial .

7

Calculate c = ΔE / (m × Δθ) . Repeat for different materials to compare.

Reading the Graph

Two ways to extract c from experimental data — and exam questions use both.

Method 1 — Single calculation

Read off Δθ from a temperature-vs-time graph (initial to final). Use the given energy and mass: c = ΔE / (m × Δθ).

Most common in AQA past papers.

Method 2 — Gradient (Level 3)

Take multiple readings; plot ΔE on the y-axis vs Δθ on the x-axis. Best-fit line passes through origin. Gradient = m × c, so c = gradient ÷ m.

More rigorous — averages out random error.

Why ΔE on the y-axis

For Method 2, convention says the dependent variable goes on the y-axis. To make the gradient = m × c (which lets you find c by dividing by m), plot ΔE on y, Δθ on x. Wrong axes lose graph marks.

Which method does AQA use?

Recent past papers usually give a temperature-vs-time graph and ask you to read off Δθ, then calculate c (Method 1). The gradient approach (Method 2) is the more rigorous version and earns Level-3 marks on extended-response questions about analysis.

Why c Comes Out Too High

In every real experiment, your measured c will usually be higher than the textbook value. Understanding why is worth a mark in itself.

The reason

Some of the energy delivered by the heater is dissipated to the thermal store of the surroundings — not all of it goes into heating the block. So the actual temperature rise (Δθ) is smaller than expected for the energy supplied.

In c = ΔE / (m × Δθ), a smaller Δθ on the bottom gives a bigger c. The lost energy makes c look higher than its true value.

✗ Weak answer

"There was an error in the experiment."

Vague — what error? In which direction?

✓ Mark-scheme answer

"Some energy is dissipated to the thermal store of the surroundings, so Δθ is smaller than expected. This makes the calculated c larger than the true value."

Names the destination, the direction of the error, and the reason.

How to reduce the error

Use a thicker insulating jacket; minimise gaps in the insulation around the heater and thermometer holes; minimise the heating time so less energy escapes. Mention at least one specific improvement when asked — generic "improve the experiment" loses marks.

Examiner Alert

Jun 2024 P1H — kJ → J conversion trap

Q8.3: the question gave "specific heat capacity of air = 1.01 kJ/kg°C". Students who substituted 1.01 directly (instead of converting to 1010 J/kg°C) lost marks. The mark scheme explicitly notes ECF for an "incorrectly / not converted value of c".

Q10.4 same paper: the same trap repeated for latent heat — "specific latent heat of fuse wire = 60 kJ/kg". The unit-conversion habit applies across the whole topic, not just SHC.

Convert ALL units to SI before substituting. If c is given in kJ/kg°C, multiply by 1000. If L is given in kJ/kg, multiply by 1000. If mass is in grams, divide by 1000. Do it on a separate line so the examiner sees you've done it. — AQA P1H Jun 2024 mark scheme, Q8.3 & Q10.4

Two-trap defence

The same habit defends both KE/EPE (squaring) and SHC (kJ → J): convert and show all working on separate lines before the main substitution. The examiner can't credit work they can't see, and they can't penalise a clean conversion line you've shown explicitly.

Common Errors That Cost Marks

Students often writeCorrect version

ΔE = m c × (final temperature)ΔE = m c × Δθ, where Δθ = final − initial

Substituting c in kJ/kg°C without convertingConvert to J/kg°C (× 1000) on a separate line first

"Heat capacity"Specific heat capacity (per kg, per °C) — different physical quantity

"Energy was lost""Energy was dissipated to the thermal store of the surroundings"

Plotting Δθ on the y-axisΔE on the y-axis, Δθ on the x-axis (gradient = m × c)

"Heat loss" as the only source of errorName the direction of the error too — c comes out too high because Δθ is smaller than expected

Mass in grams substituted directlyConvert to kg first (e.g. 250 g = 0.25 kg)

Practice Question

A 50 W immersion heater is placed in 0.20 kg of cooking oil. The heater is switched on for 4.0 minutes. The temperature of the oil rises from 18 °C to 48 °C.

Calculate the specific heat capacity of the oil. Assume no heat is lost to the surroundings. 4 marks

M1

Energy supplied: ΔE = P × t = 50 × (4.0 × 60) = 50 × 240 = 12 000 J

M2

Temperature change: Δθ = 48 − 18 = 30 °C

M3

Equation rearranged: c = ΔE / (m × Δθ) = 12 000 / (0.20 × 30) = 12 000 / 6

A

Answer: c = 2000 J/kg°C

Checkpoint

✅ You can now

Define specific heat capacity as the energy needed to raise 1 kg of substance by 1 °C.

Apply ΔE = m c Δθ in both directions (energy → temp, or temp → c).

Always calculate Δθ = final − initial on a separate line.

Describe the RP1 apparatus — power supply, joulemeter, insulated metal block, heater, thermometer, stopwatch.

State the IV, DV and controls for RP1, with reasons.

Plot ΔE vs Δθ with gradient = m × c, then divide by m to find c.

Explain why measured c is higher than the true value (heat dissipated to surroundings → Δθ smaller → c larger).

Avoid the kJ/kg°C trap — convert to J/kg°C before substituting.

Coming Next

Lesson 1.04 — Power. P = E/t and P = W/t. Two equations, several unit conversions (W → kW, ms → s), and a small set of question patterns that come up nearly every series.

Summary

The One Equation

ΔE = m c Δθ

Δθ = final − initial. Always calculate this on its own line first.

Rearranged for c: c = ΔE / (m × Δθ). This is the form you use in RP1.

Equation is on the equation sheet (since June 2022).

RP1 essentials

Insulated metal block, heater, thermometer

Measure ΔE (joulemeter or VIt)

Measure Δθ = final − initial

c = ΔE / (m × Δθ)

Graph: ΔE vs Δθ → gradient = mc

Errors to avoid

Final temp instead of Δθ

kJ/kg°C without converting

Grams instead of kg

"Heat capacity" instead of SHC

"Energy lost" instead of dissipated

Exam Tip

For SHC calculations, write three lines before the main substitution: (1) the equation in symbols, (2) Δθ = final − initial, (3) any unit conversions. Then substitute, calculate, and write the answer with its unit. Each line is its own potential mark.