Reversed heat engines (A-level only) — AQA A-Level Physics

This topic explores reversed heat engines, which are devices that transfer thermal energy from a cold region to a hotter region using an external work input. Unlike heat engines that generate work from a temperature difference, reversed heat engines consume work to create a temperature difference, forming the basis of refrigerators, heat pumps, and air conditioners. Understanding their operation is crucial for analysing real-world applications and for linking to the second law of thermodynamics.

In the AQA A-Level Physics specification, reversed heat engines are studied in the context of thermodynamics and energy transfer. You will learn to define and calculate the coefficient of performance (COP) for both refrigerators and heat pumps, and to apply the Carnot cycle in reverse to determine maximum theoretical efficiencies. This topic also reinforces the concept of entropy and the limits imposed by the second law, which are essential for understanding energy resources and environmental impact.

• →A reversed heat engine uses work input to transfer heat from a cold reservoir to a hot reservoir, opposite to the natural direction of heat flow.
• →The coefficient of performance (COP) measures the effectiveness of a reversed heat engine: COP_ref = Q_c / W for a refrigerator, and COP_hp = Q_h / W for a heat pump.
• →The maximum theoretical COP is achieved by a Carnot reversed cycle and depends only on the absolute temperatures of the reservoirs: COP_ref = T_c / (T_h - T_c) and COP_hp = T_h / (T_h - T_c).
• →Energy conservation requires that the heat delivered to the hot reservoir equals the heat extracted from the cold reservoir plus the work input: Q_h = Q_c + W.
• →COP is a dimensionless quantity and can be greater than 1, unlike efficiency which is always less than 1.

• State that a heat pump or refrigerator uses work input to transfer thermal energy from a cold region to a hot region.
• Apply the energy balance Q_H = Q_C + W to a reversed heat engine.
• Calculate the coefficient of performance of a refrigerator using COP = Q_C/W.
• Calculate the coefficient of performance of a heat pump using COP = Q_H/W.
• Describe practical uses such as domestic refrigerators, air conditioners and heat pumps for space heating.

• 💡Draw and label an energy flow diagram showing Q_C, W and Q_H before substituting into equations.
• 💡Check whether the question asks for refrigerator COP or heat pump COP and select the matching expression.
• 💡Use energy conservation Q_H = Q_C + W to find any missing energy value before calculating COP.
• 💡Always write down the formula you are using before substituting numbers. This helps you gain method marks even if you make an arithmetic error.
• 💡When calculating COP, ensure temperatures are in kelvin. A common mistake is to use degrees Celsius, which leads to incorrect answers.
• 💡In explanations, clearly distinguish between a refrigerator and a heat pump, and state which COP definition you are using. Use the terms 'cold reservoir' and 'hot reservoir' precisely.

• Treating a refrigerator as a device that creates coldness; correct this by explaining that it removes thermal energy from the cold interior and rejects it to warmer surroundings.
• Using the heat pump COP expression for a refrigerator or vice versa; correct this by matching the useful energy transfer to the device: Q_C for a refrigerator and Q_H for a heat pump.
• Assuming COP must be less than 1; correct this by noting that COP can exceed 1 because work input moves additional thermal energy rather than creating it.
• Students often think that a reversed heat engine violates the second law of thermodynamics because heat flows from cold to hot. In fact, this is allowed because work is done to drive the process, and the total entropy of the universe still increases.
• Many students confuse the COP of a refrigerator with that of a heat pump, using the wrong heat value in the numerator. Remember: refrigerator COP uses heat extracted from the cold reservoir (Q_c), while heat pump COP uses heat delivered to the hot reservoir (Q_h).
• Some students believe that COP cannot exceed 1, similar to efficiency. However, COP can be much greater than 1 because it compares heat transfer to work input, not energy output to energy input.

1. 1Day 1-2: Review the basics of heat engines and the second law of thermodynamics. Ensure you understand the direction of natural heat flow and the role of work in reversed processes.
2. 2Day 3-4: Learn the definitions and formulas for COP of refrigerators and heat pumps. Practice calculating COP using both energy values and temperatures.
3. 3Day 5-6: Work through past exam questions on reversed heat engines, focusing on calculations and explanations. Check your answers against mark schemes.
4. 4Day 7-8: Create a summary sheet with key formulas, definitions, and common pitfalls. Test yourself on active recall prompts.
5. 5Day 9-10: Attempt a full past paper section on thermodynamics, including reversed heat engines, under timed conditions. Review any mistakes and revisit weak areas.

• 📋Calculation of COP from given heat transfers and work input. Advice: Identify whether it is a refrigerator or heat pump, then use the correct formula.
• 📋Calculation of maximum theoretical COP using reservoir temperatures. Advice: Convert temperatures to kelvin and use the Carnot COP expressions.
• 📋Explanation of why a reversed heat engine does not violate the second law. Advice: Emphasise the role of work input and the overall increase in entropy.
• 📋Comparison of real and ideal COP values, with reasons for differences. Advice: Discuss irreversibilities such as friction and heat losses.