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
What each mark is really for — in plain English — and the wording trap that loses it.
- S1
Identify the given values: Q C = 500 J, W = 200 J
- S2
Use the formula for refrigerator COP: COP ref = Q C / W
- S3
Substitute the values: COP ref = 500 J / 200 J = 2.5
Model answer
Worked through, with each step tagged to the mark it earns.
- S1
Identify the given values: Q C = 500 J, W = 200 J
- S2
Use the formula for refrigerator COP: COP ref = Q C / W
- S3
Substitute the values: COP ref = 500 J / 200 J = 2.5
Final answer: COP ref = 2.5
Common mistakes
- Confusing the COP formula for a refrigerator with that of a heat pump. — 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. — 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. — 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. — 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). — 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. — 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
- Full worked solution (all marking points)3 marks