AQA A-Level Physics Paper 3 (3BC), June 2025: Question 4

5 marks · Medium difficulty · Short Answer

Define the coefficient of performance of a heat pump, evaluate claims regarding COP variation with outside temperature, and calculate the electrical energy required for heating.

Practise this question

Question

Question 04 consists of three parts. Part 04.1 asks to state what is meant by the coefficient of performance of a heat pump for 1 mark. A scenario then describes replacing a gas boiler with an electric heat pump providing underfloor heating at 40 degrees Celsius using outside air between 12 and -8 degrees Celsius. Two salesperson claims are presented: first, that the COP will vary greatly depending on outside temperature; second, that colder temperatures lead to lower energy running costs. Part 04.2 asks to comment on these claims for 3 marks. Part 04.3 gives an average COP of 7.5 delivering 2880 kilowatt-hours of heating and asks to select the electrical energy used from four options: 333 kWh, 384 kWh, 2500 kWh, and 21 600 kWh for 1 mark.
Question text

04.1 State what is meant by the coefficient of performance of a heat pump.

[1 mark]

A jeweller hopes to reduce workshop heating costs by replacing a gas-fired boiler with

an electric heat pump. The heat pump will provide underfloor heating using water

at a temperature of 40 °C.

The air outside the building provides the low-temperature reservoir for the heat pump.

Over the winter period the outside temperature will vary between 12 °C and −8 °C.

The salesperson for a heat pump company claims that:

• the coefficient of performance will vary greatly, depending on the outside

temperature

• the colder the outside temperature, the lower the energy costs will be for running

the heat pump.

04.2 Comment on the salesperson’s claims.

[3 marks]

04.3 A heat pump is installed. The average COPhp over the winter period is 7.5

In this period the heat pump delivers 2880 kW h of underfloor heating.

What is the electrical energy used by the heat pump in this period?

Tick ( ) one box.

[1 mark]

333 kW h

384 kW h

2500 kW h

21 600 kW h

Mark scheme

Show the mark scheme Mark scheme for Question 04. For 04.1: energy delivered to hot space divided by work input (1 mark). For 04.2: refers to COP formula or dependence on temperature difference (1 mark); assesses claim 1 as correct with calculated COP values from 11.2 to 6.5 or temperature variation argument (1 mark); assesses claim 2 as incorrect because lower outside temperature decreases COP and requires more electrical work (1 mark). For 04.3: tick in second box only (384 kWh) for 1 mark. Total 5 marks.

Question Answers Additional comments/Guidance Mark AO

Direction of energy transfer is needed.

04.1 energy delivered to hot space 1 AO1

Idea of the ratio ✓ Q

work input Do not accept H on without the terms defined,

W

or answers in terms of temperatures only.

Treat mention of efficiency as neutral.

04.2 TH If no other mark given: award MAX 1 for idea that 3 1× AO1

Refers to / uses COPhp =

(TH −TC ) first statement correct AND second statement

2× AO3

incorrect.

OR states that COPhp depends on/inversely proportional to

temperature difference ✓1

Allow a.e. in one calculation of COPhp but no ecf

1st bullet: EITHER shows COPhp varies from 11.2 to 6.5 for conclusion. An a.e. does not include failure to

AND idea that statement is correct convert the TH on the top of the fraction to kelvin.

OR states that COPhp will vary greatly because

temperature varies greatly AND idea that statement is

correct ✓2

For ✓3 :

2nd bullet: (TH is constant and) idea that the lower TC is, allow reverse argument

the lower the COPhp accept idea that for lower TC more work needed

So more (electrical) energy needed AND idea that (per second) to raise water temp to 40 ºC, so

statement is incorrect ✓3 greater energy cost.

– A-LEVEL PHYSICS – –

04.3 Tick in second box only ✓ 1 AO2

384 kW h

Total 5

How to answer it

Thermodynamics: Heat Pumps & Coefficient of Performance

📋 What This Question Tests

This question assesses your understanding of reversed heat engines (heat pumps), specifically:

  • Definition: Formulating an exact definition of the coefficient of performance (COP) for a heat pump.
  • Evaluation & Application: Using the theoretical maximum COP relationship COPhp = TH / (TH − TC) to evaluate commercial claims critically.
  • Temperature Conversions: Remembering to convert temperatures from Celsius to Kelvin in theoretical thermodynamic formulas.
  • Energy Calculations: Rearranging the defining ratio to determine electrical energy consumption from thermal energy delivered.

Question 04.1

Definition of Coefficient of Performance (1 Mark)

✅ Model Answer

The ratio of the thermal energy delivered to the hot space (useful heat output) to the work input (energy supplied to drive the pump).

COPhp = QH / W

🧠 Exam Technique: Be Specific with Direction

The direction of energy transfer is essential. You must state "heat delivered to the hot space/reservoir". Stating simply "energy transferred" or "heat moved" is ambiguous and loses the mark because a refrigerator transfers heat from a cold space, whereas a heat pump delivers heat to a warm space.

❌ Common Pitfalls

  • Writing unqualified symbols: Writing just QH / W without explicitly defining what QH and W stand for scores 0 marks.
  • Defining in terms of temperature: Defining COP purely as TH / (TH − TC) is incorrect. That is the formula for the theoretical maximum COP of a Carnot heat pump, not the fundamental definition of COP.
Mark Scheme Breakdown [1 Mark]:
• [1 Mark, AO1]: Idea of ratio: (energy delivered to hot space) / (work input).

Question 04.2

Evaluating Salesperson's Claims (3 Marks)

💡 Key Knowledge

For an ideal heat pump, the maximum theoretical coefficient of performance is:

COPhp = TH / (TH − TC)

where temperatures must be expressed in Kelvin (K).

  • Underfloor heating: TH = 40 °C = 40 + 273.15 = 313 K
  • Mild winter day: TC = 12 °C = 12 + 273.15 = 285 K
  • Cold winter day: TC = −8 °C = −8 + 273.15 = 265 K

📐 Step-by-Step Evaluation

Claim 1: "The coefficient of performance will vary greatly, depending on the outside temperature."

  1. Calculate COP at 12 °C:
    COPhp, max = 313 / (313 − 285) = 313 / 28 ≈ 11.2
  2. Calculate COP at −8 °C:
    COPhp, max = 313 / (313 − 265) = 313 / 48 ≈ 6.5
  3. Judgement on Claim 1: COP drops significantly from 11.2 to 6.5 (almost halving). Therefore, the claim is correct.

Claim 2: "The colder the outside temperature, the lower the energy costs will be for running the heat pump."

  1. As outside temperature (TC) falls, the temperature lift (TH − TC) increases.
  2. Because COP = QH / W, a lower COP means more electrical work ( W = QH / COP ) is required to deliver each unit of heat.
  3. More electricity consumed per unit of heating means running costs will increase, not decrease.
  4. Judgement on Claim 2: The claim is incorrect.

✅ Model Answer Structure

  • Refers to theoretical formula COPhp = TH / (TH − TC) (or states COP is inversely proportional to temperature difference).
  • Shows COP varies between 11.2 and 6.5 (or argues large ΔT range causes large COP change), confirming Claim 1 is correct.
  • Explains that as TC falls, COP decreases, meaning more electrical work is needed per unit of heat delivered, so running costs increase; therefore, Claim 2 is incorrect.

❌ Common Mistakes

  • Celsius in numerator: Using 40 / 28 = 1.43 instead of Kelvin ( 313 / 28 = 11.2 ). Temperatures in thermodynamic ratio formulas must always be absolute (Kelvin).
  • Missing conclusion: Stating equations or values without clearly saying "Claim 1 is correct" and "Claim 2 is incorrect".
Mark Scheme Breakdown [3 Marks]:
• Mark 1 [AO1]: States or uses COPhp = TH / (TH − TC) OR states COP depends on / is inversely proportional to temperature difference.
• Mark 2 [AO3]: Shows COP varies from 11.2 to 6.5 AND states that claim 1 is correct (or argues large temperature difference causes large COP variation).
• Mark 3 [AO3]: Explains that lower TC gives lower COP, requiring more work/energy input, so claim 2 is incorrect.
Note: If no marks scored above, award MAX 1 mark if candidate correctly identified claim 1 is correct AND claim 2 is incorrect.

Question 04.3

Electrical Energy Calculation (1 Mark)

📐 Step-by-Step Calculation

1. Identify given values:

  • Average COPhp = 7.5
  • Heat delivered (QH) = 2880 kW h

2. Rearrange the formula:

COPhp = QH / W ⟹ W = QH / COPhp

3. Calculate:

W = 2880 / 7.5 = 384 kW h

4. Select the matching option: Tick the second box ( 384 kW h ).

❌ Distractor Analysis

  • 333 kW h : Calculation trap from incorrect rounding or formula misapplication.
  • 2500 kW h : Subtracting or misusing ratios.
  • 21 600 kW h : Multiplication trap: 2880 × 7.5 = 21 600 . Always sanity-check: electrical work input must be less than heat output for a heat pump!
Mark Scheme Breakdown [1 Mark]:
• [1 Mark, AO2]: Tick in second box only (384 kW h).

Topics

Optional topics · 3.11 Engineering physics (A-level only)

Question and mark scheme from the AQA A-Level Physics examination, Paper 3 (3BC), June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.