Refrigerators and Heat Pumps
For a cyclic refrigerator, W = Q_H − Q_C, COP_R = Q_C/W, and COP_HP = Q_H/W = COP_R + 1; reversible values are T_C/(T_H−T_C) and T_H/(T_H−T_C), respectively, in kelvin.
Why this shows up in the exam
Ice making and refrigeration · Heat-pump heating · Engine-driven refrigerator combinations
Learn the idea
Refrigerators use work to move heat from cold to hot; heat pumps value the heat delivered to the hot side. The same reversed device has two performance measures depending on whether cooling or heating is the desired output.
🧠 Memory hook: Fridge counts cold heat; heat pump counts hot heat.
Get this one clearly and it pays off every single time it shows up in the paper. 🎯
Formulas & facts to keep ready
- COP_R = Q_C/W — refrigerator performance based on heat removed from the cold region
- COP_HP = Q_H/W = COP_R + 1 — heat-pump performance based on heat delivered
- COP_R,Carnot = T_C/(T_H-T_C) — maximum refrigerator COP between two kelvin temperatures
How to approach it
- 1Identify the desired heat transfer
- 2Write W = Q_H − Q_C
- 3Use the proper COP and apply the Carnot value only for reversible minimum-work operation
Common slip-ups that cost marks
- •Calling COP an efficiency that must be below one
- •Using Q_H in refrigerator COP
- •Using Celsius directly in reversible COP formulas
🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.
Original chapter practice
Original questions for this chapter, not past-paper questions or an exact mapping to this individual concept.
A gas absorbs 500 J of heat and does 200 J of work. What is the change in its internal energy?
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