A-Level · Physics · AQA · Mark scheme decoded
AQA A-Level Physics: Heat Pumps and Refrigerators: Coefficients of Performance — mark scheme explained
The short answer
In the realm of engineering physics, understanding the principles and uses of heat pumps and refrigerators is crucial. These devices are essential in various applications, from household appliances to industrial processes. The key concept that ties these systems together is the coefficient of performance (COP), which measures their efficiency.
The question
A refrigerator extracts 500 J of heat from the cold reservoir and requires 200 J of work input. Calculate its COP.
[Paraphrased for study — not reproduced from any exam paper.]
Mark scheme, decoded
How the examiner actually awards the marks on this topic.
Gradora's own decode of the marking approach — not the exam board's published mark scheme.
How marks are awarded
For calculation questions, marks are typically awarded for identifying correct values, using the right formula, showing working steps, and providing the final answer. For conceptual questions, marks are given for clear explanations that demonstrate understanding of key principles.
What the command words demand
- Calculate
- Perform a numerical calculation using given data and appropriate formulas.
- Explain
- Provide a detailed account of how or why something happens, including relevant principles and concepts.
- Compare
- Identify and describe the similarities and differences between two or more items or concepts.
- Evaluate
- Assess the advantages and disadvantages of a particular method or system.
Model answer
A full-mark response to the question above, worked through step by step.
Timing: Allocate about 2-3 minutes per mark to ensure you have enough time to carefully read the question, perform calculations, and check your work.
- Identify the given values: Q C = 500 J, W = 200 J0 marks
- Use the formula for refrigerator COP: COP ref = Q C / W1 mark
- Substitute the values: COP ref = 500 J / 200 J = 2.52 marks
Final answer: COP ref = 2.5
Work through every step correctly and you earn all 3 marks.
Another worked example
A heat pump delivers 1200 J of heat to the hot reservoir and requires 400 J of work input. Calculate its COP.
- Identify the given values: Q H = 1200 J, W = 400 J0 marks
- Use the formula for heat pump COP: COP hp = Q H / W1 mark
- Substitute the values: COP hp = 1200 J / 400 J = 32 marks
Final answer: COP hp = 3
Work through every step correctly and you earn all 3 marks.
Common mistakes
Confusing the COP formula for a refrigerator with that of a heat pump.
Why it happens: Students often mix up the formulas, leading to incorrect calculations. It's crucial to remember that Q C is used for refrigerators and Q H for heat pumps.
Fix: Memorize and understand the specific COP formula for each device: COP ref = Q C / W and COP hp = Q H / W.
Using the wrong temperatures in the ideal (Carnot) COP formula.
Why it happens: Students sometimes use the temperature of the hot reservoir for T C or vice versa, leading to incorrect results.
Fix: Always double-check that you are using the correct temperatures: T C for the cold reservoir and T H for the hot reservoir.
Forgetting to convert temperatures to Kelvin before calculating COP.
Why it happens: Students often use Celsius or Fahrenheit instead of Kelvin, which can lead to significant errors in calculations.
Fix: Always ensure that temperatures are converted to Kelvin (K) before using them in the COP formula.
Misinterpreting the meaning of a COP value greater than 1 for a heat pump.
Why it happens: Students may not understand that a COP greater than 1 means the heat pump is delivering more heat to the hot reservoir than the work input required.
Fix: Understand that a COP > 1 indicates efficiency, as the device is outputting more energy (heat) than it consumes (work).
Believing a refrigerator's COP must be less than 1 (confusing COP with thermal efficiency).
Why it happens: Students assume COP behaves like the efficiency of a heat engine (always C / W has no such upper bound and is commonly greater than 1.
Fix: Remember that refrigerator COP = Q C / W is generally greater than 1 (typically 2–4 for a domestic fridge) and can be any positive value. Unlike thermal efficiency, the COP is not capped below 1.
Failing to consider the practical implications of temperature differences on COP.
Why it happens: Students might not realize how a smaller temperature difference can significantly affect the efficiency of heat pumps and refrigerators.
Fix: Understand that a smaller temperature difference between the hot and cold reservoirs generally results in a higher COP, making the device more efficient.
Where the marks go
The question types you’ll meet on this topic and the marks each one carries — so you know what to expect and where to focus.
| Question type | What you’re asked to do | Marks |
|---|---|---|
| Calculate COP | Calculate a refrigerator's coefficient of performance from heat extracted and work input. | 3 |
| Calculate COP | Find a heat pump's coefficient of performance from heat delivered and work input. | 3 |
| Calculate Carnot COP | Use the reservoir temperatures to calculate a refrigerator's ideal coefficient of performance. | 4 |
| Calculate Carnot COP | Use reservoir temperatures to calculate the ideal coefficient of performance for a heat pump. | 4 |
| Total across these question types | 14 | |
Question types and mark tariffs are Gradora’s guidance based on how this topic is typically examined — not the board’s official paper structure.