AQA GCSE Combined Science: Trilogy Physics Paper 1 (Foundation), June 2025: Question 6

11 marks · Standard Demand difficulty · Extended Answer

Calculate the specific heat capacity of an aluminium block, describe a practical method to determine it, and identify reasons why using insulation improves the accuracy of the result.

Practise this question

Question

Question 6 is divided into three parts based on an investigation into the specific heat capacity of aluminium. Table 2 provides experimental data: Mass in kilograms = 0.75, Energy transferred in joules = 13 800, and Temperature change in degrees Celsius = 20. Question 06.1 asks students to calculate the specific heat capacity of aluminium in J/kg °C using the Physics Equations Sheet for 3 marks. Question 06.2 asks students to describe a method the student could have used to obtain the results in Table 2 for 6 marks. Question 06.3 asks to tick two boxes from five options explaining why wrapping the block in insulation increases accuracy: the amount of wasted energy is less; the energy transferred to the aluminium block is greater; the insulation heats the aluminium block; the power of the electric heater increases; the temperature increase of the aluminium block is greater.
Question text

06 A student investigated the specific heat capacity of aluminium.

The student used an electric heater to increase the temperature of an

aluminium block.

The student measured the:

• mass of the block

• energy transferred to the block

• temperature change of the block.

Table 2 shows the results.

Table 2

Mass in kilograms 0.75

Energy transferred in joules 13 800

Temperature change in °C 20

06.1 Calculate the specific heat capacity of aluminium.

Use the Physics Equations Sheet.

[3 marks]

Specific heat capacity =23 J/kg °C

06.2 Describe a method the student could have used to obtain the results in Table 2.

[6 marks]

Extra space

06.3 Wrapping the aluminium block in insulation makes the value for the specific heat

capacity of aluminium more accurate.

Which are two reasons why?

[2 marks]

Tick ( ) two boxes.

The amount of wasted energy is less.

The energy transferred to the aluminium block is greater.

The insulation heats the aluminium block.

The power of the electric heater increases.

The temperature increase of the aluminium block is greater.

Mark scheme

Show the mark scheme Mark scheme for Question 6. 06.1 awards 1 mark each for: correct substitution into equation (13 800 = 0.75 × c × 20), rearrangement (c = 13 800 / (0.75 × 20)), and calculation yielding 920 J/kg °C. 06.2 provides a 3-level response mark scheme (Level 1: 1–2 marks, Level 2: 3–4 marks, Level 3: 5–6 marks) with indicative content covering measuring mass using a top pan balance, measuring energy transfer using a joulemeter or ammeter/voltmeter/stopwatch, and measuring initial and final temperature with a thermometer to calculate temperature change. 06.3 awards 1 mark each for ticking 'The amount of wasted energy is less' and 'The temperature increase of the aluminium block is greater'.

Question 6

AO /

Question Answers Extra information Mark

Spec. Ref.

06.1 13 800 = 0.75 × c × 20 1 AO2

6.1.1.3

RPA14

13 800 1

c =

0.75 × 20

c = 920 (J/kg °C) – – 8464/P1/1F –

AO /

Question Answers Mark

Spec. Ref.

06.2 Level 3: The method would lead to the production of a valid 5–6 AO1

outcome. The key steps are identified and logically sequenced. 6.1.1.3

RPA14

Level 2: The method would not necessarily lead to a valid 3–4

outcome. Most steps are identified, but the method is not fully

logically sequenced.

Level 1: The method would not lead to a valid outcome. Some 1–2

relevant steps are identified, but links are not made clear.

No relevant content 0

Indicative content

Mass

• use a top pan balance to measure mass of block

Energy transfer

• use a joulemeter to measure energy transferred to block

• connect heater to a joulemeter

• measure initial energy and switch on

• switch off and measure final energy

• calculate the difference between initial and final energy

• use an ammeter, voltmeter and stopwatch

• calculate energy transferred

Temperature change

• use a thermometer to measure temperature of block

• measure the initial temperature

• measure the final temperature

• calculate the difference between initial and final temperature

for Level 3 answers must describe all three measurements– – 8464/P/1F –

AO /

Spec. Ref.

06.3 the amount of wasted energy is 1 AO3

less 6.1.1.3

6.1.2.1

the temperature increase of the 1 RPA14

aluminium block is greater

Total Question 6 11

How to answer it

Investigating Specific Heat Capacity of Aluminium

What this question tests

  • Calculation skills (AO2): Rearranging the specific heat capacity formula ΔE = m × c × Δθ to calculate c.
  • Required practical methods (AO1): Writing a logically sequenced experimental procedure (Required Practical 14) to measure mass, electrical energy transferred, and temperature change.
  • Practical evaluation (AO3): Explaining how thermal insulation improves experimental accuracy by reducing unwanted thermal energy dissipation.
Question 06.1 · Calculation · 3 Marks

Calculating Specific Heat Capacity

Calculate the specific heat capacity of aluminium using the data provided in Table 2.

Measurement Value
Mass in kilograms (m) 0.75 kg
Energy transferred in joules (ΔE) 13 800 J
Temperature change in °C (Δθ) 20 °C

📐 Step-by-Step Calculation

1 State the formula:
Change in thermal energy = mass × specific heat capacity × temperature change
ΔE = m × c × Δθ

2 Substitute values into equation:
13 800 = 0.75 × c × 20

3 Rearrange to make c the subject:
0.75 × 20 = 15
c = 13 800 ÷ 15

4 Calculate final answer:
c = 920 J/kg °C

❌ Common Calculation Traps

  • Incorrect rearrangement: Dividing 15 by 13 800 instead of 13 800 by 15 gives 0.00109, which is physically impossible for a specific heat capacity.
  • Omitting brackets on calculator: Typing 13800 ÷ 0.75 × 20 without brackets results in 368 000 because of order of operations. Always evaluate the denominator first ( 0.75 × 20 = 15 ) or use brackets!
  • Unit confusion: Mass is already in kilograms (kg) and energy in joules (J) — no prefix conversion is needed here.

✅ Model Answer & Mark Breakdown

13 800 = 0.75 × c × 20 [1 mark]

c = 13 800 ÷ (0.75 × 20) [1 mark]

c = 920 (J/kg °C) [1 mark]

Full 3 marks awarded for final correct answer of 920 even if working is minimal, but showing substitution guarantees method marks if an arithmetic slip occurs.
Question 06.2 · Extended Response / Practical Method · 6 Marks

Designing the Method (Required Practical 14)

Describe a method the student could have used to obtain the results in Table 2.

🧠 How to Score Level 3 (5–6 Marks)

To reach Level 3, your answer must be a coherent, logically sequenced method describing how to obtain all three quantities from the table:

  • Mass (m): What piece of equipment measures it?
  • Energy transferred (ΔE): Which electrical instruments are used and how is energy determined?
  • Temperature change (Δθ): What instrument is used, and what two readings must be taken?

💡 Apparatus Needed

  • Aluminium block: Contains two drilled holes (one for heater, one for thermometer).
  • Top pan balance: Measures mass.
  • Immersion heater & power supply: Heats the block.
  • Joulemeter: Measures energy supplied directly (OR ammeter, voltmeter, stopwatch).
  • Thermometer: Measures temperature.
  • Pipette & water/oil: Put into thermometer hole to improve thermal contact.

✅ Model Answer (Level 3: 6/6 Marks)

1. Measuring mass:

  • Place the aluminium block onto a top pan balance and record its mass in kilograms.

2. Measuring temperature change:

  • Insert a thermometer into the smaller hole of the block (adding a few drops of water/oil to ensure good thermal contact).
  • Measure and record the initial temperature of the block.
  • After heating, measure and record the final (highest) temperature reached.
  • Calculate the temperature change: Δθ = final temperature − initial temperature .

3. Measuring energy transferred:

  • Place the electric immersion heater into the larger hole and connect it to a power pack via a joulemeter.
  • Record the initial joulemeter reading (or reset it to zero) and switch on the power pack.
  • Switch off the power pack after heating, record the final joulemeter reading, and calculate: energy transferred = final reading − initial reading .
  • (Alternative: connect an ammeter in series, voltmeter in parallel across heater, time with a stopwatch, then calculate E = V × I × t).

❌ Common Examiner Criticisms (Why Students Lost Marks)

  • Omitting one measurement: Many students forgot to state how to measure the mass (top pan balance), immediately capping their response below Level 3.
  • Vague energy measurement: Writing "measure energy with a heater" scores zero; a heater provides heat, but a joulemeter (or voltmeter + ammeter + stopwatch) measures the energy.
  • Only one temperature reading: Simply writing "measure temperature with a thermometer" is incomplete. You must measure initial and final temperature and calculate the difference.
Question 06.3 · Multiple Choice / Evaluation · 2 Marks

Effect of Insulation on Accuracy

Wrapping the aluminium block in insulation makes the value for the specific heat capacity of aluminium more accurate. Which are two reasons why?

✅ Correct Choices (Tick 2 Boxes)

  • ☑ The amount of wasted energy is less. [1 mark]
  • ☑ The temperature increase of the aluminium block is greater. [1 mark]

💡 Scientific Explanation

Without insulation, thermal energy transfers to the surroundings rather than staying in the block.

By adding insulation:

  • Energy losses to the surroundings are minimised (less energy is wasted).
  • More of the supplied energy goes into raising the temperature of the block, meaning for the same energy input, the temperature increase (Δθ) is greater.
  • This prevents calculated values of c from being falsely high.

❌ Incorrect Options Explained

  • ☒ The energy transferred to the aluminium block is greater — Incorrect: The energy supplied comes from the heater/power supply; insulation does not increase heater energy output.
  • ☒ The insulation heats the aluminium block — Incorrect: Insulation is a thermal insulator; it does not generate or provide heat.
  • ☒ The power of the electric heater increases — Incorrect: Heater power depends solely on current and voltage ( P = V × I ), unaffected by insulation.

Topics

Physics · P1: Energy

Question and mark scheme from the AQA GCSE Combined Science: Trilogy examination, Physics Paper 1 (Foundation), June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.