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Specification: AQA A-level Physics (7408) 3.11.2.6 Engineering physics option  |  Paper: 3 Section B  |  Data and formulae booklet: COP_ref = Q_C/W = Q_C/(Q_H − Q_C) = T_C/(T_H − T_C) and COP_hp = Q_H/W = Q_H/(Q_H − Q_C) = T_H/(T_H − T_C) are given in the exam

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The equation

refrigerator: COP_ref = Q_C ÷ W = Q_C ÷ (Q_H − Q_C), maximum T_C ÷ (T_H − T_C)    heat pump: COP_hp = Q_H ÷ W = Q_H ÷ (Q_H − Q_C), maximum T_H ÷ (T_H − T_C)

Quantity	Symbol	Unit
heat taken from the cold space	QC	joules (J)
heat delivered to the hot space	QH	joules (J)
work done on the working fluid	W	joules (J)
temperature of the cold space	TC	kelvin (K)
temperature of the hot space	TH	kelvin (K)
coefficient of performance	COP	no unit

Rearranged: W = Q_C ÷ COP_ref = Q_H ÷ COP_hp  |  Q_H = Q_C + W  |  COP_hp = COP_ref + 1 for the same device  |  Powers (W) can replace energies. Temperatures must be in kelvin, and the useful heat is Q_C for a refrigerator but Q_H for a heat pump.

Use FIFA for every answer

• F – Formula: write the equation as it appears on the sheet.
• I – Insert: put in the numbers (convert to standard units first).
• F – Fix: rearrange to make the unknown the subject.
• A – Answer: calculate and always give the unit.

Core: Q1–4.   Exam standard: Q5–8.   Stretch: Q9–10.

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Questions

Q1 (Core) A refrigerator removes 900 J of heat from its interior while 300 J of work is done on the working fluid. Calculate its coefficient of performance.

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F: COP_ref = Q_C ÷ W
I: COP_ref = 900 ÷ 300
F: COP_ref is already the subject
A: 3.0 (no unit)

Q2 (Core) A heat pump takes heat from outside air at 278 K and delivers it to a house at 308 K. Calculate its maximum possible coefficient of performance, to 3 significant figures.

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F: maximum COP_hp = T_H ÷ (T_H − T_C)
I: maximum COP_hp = 308 ÷ (308 − 278) = 308 ÷ 30
F: maximum COP_hp is already the subject
A: 10.26… = 10.3 (3 s.f.)

Q3 (Core) The inside of a freezer is at −18 °C and the kitchen is at 22 °C. Calculate the maximum possible coefficient of performance of the freezer, to 2 significant figures.

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Convert first: T_C = −18 + 273 = 255 K; T_H = 22 + 273 = 295 K
F: maximum COP_ref = T_C ÷ (T_H − T_C)
I: maximum COP_ref = 255 ÷ (295 − 255) = 255 ÷ 40
F: maximum COP_ref is already the subject
A: 6.375 = 6.4 (2 s.f.)
Watch out: a refrigerator uses T_C on top, a heat pump uses T_H. Using T_H here gives 7.4, which is wrong.

Q4 (Core) A heat pump with a coefficient of performance of 3.5 delivers heat to a house at a rate of 7.0 kW. Calculate the electrical power input and the rate at which heat is taken from the outside air.

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Convert first: 7.0 kW = 7000 W
F: COP_hp = Q_H ÷ W
I: 3.5 = 7000 ÷ W
F: W = 7000 ÷ 3.5 = 2000 W
A: power input = 2.0 kW; heat from outside Q_C = Q_H − W = 7000 − 2000 = 5000 W = 5.0 kW

Q5 (Exam) In each cycle a refrigerator delivers 1.25 kJ of heat to the kitchen and removes 0.95 kJ from its interior. Calculate, to 2 significant figures, (a) its COP as a refrigerator, (b) its COP if it were used as a heat pump. Show that COP_hp = COP_ref + 1.

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Convert first: Q_H = 1250 J; Q_C = 950 J
Step 1 – (a) refrigerator
F: COP_ref = Q_C ÷ (Q_H − Q_C)
I: COP_ref = 950 ÷ (1250 − 950) = 950 ÷ 300 = 3.166…
A: 3.2 (2 s.f.)
Step 2 – (b) heat pump
COP_hp = Q_H ÷ (Q_H − Q_C) = 1250 ÷ 300 = 4.166…, so 4.2 (2 s.f.)
Step 3 – the link
COP_hp = (Q_C + W) ÷ W = Q_C ÷ W + 1 = COP_ref + 1, and 3.166… + 1 = 4.166…

Q6 (Exam) A ground-source heat pump takes heat from the ground at 10 °C and heats water for underfloor heating to 35 °C. Its actual COP is 4.0. Calculate its maximum possible COP, and its actual COP as a percentage of the maximum, to 2 significant figures. Suggest why the actual COP is lower.

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Convert first: T_C = 10 + 273 = 283 K; T_H = 35 + 273 = 308 K
Step 1 – maximum COP
F: maximum COP_hp = T_H ÷ (T_H − T_C)
I: maximum COP_hp = 308 ÷ 25
A: 12.32 = 12 (2 s.f.)
Step 2 – comparison
4.0 ÷ 12.32 = 0.324…, so 32% of the maximum (2 s.f.)
Why lower: friction in the compressor, heat losses, and the working fluid must be colder than the ground and hotter than the water for heat to flow, so the real temperature difference is larger.

Q7 (Exam) A house needs 45 MJ of heat in one day. It is heated by a heat pump with a COP of 3.0. Calculate the electrical energy used, in kWh, to 2 significant figures, and compare it with a direct electric heater. (1 kWh = 3.6 × 10⁶ J)

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Convert first: 45 MJ = 45 × 10⁶ J
F: COP_hp = Q_H ÷ W
I: 3.0 = 45 × 10⁶ ÷ W
F: W = 45 × 10⁶ ÷ 3.0 = 15 × 10⁶ J
A: 15 × 10⁶ ÷ 3.6 × 10⁶ = 4.166… kWh = 4.2 kWh (2 s.f.). A direct heater would use all 45 MJ = 12.5 kWh, three times as much.

Q8 (Exam) A freezer has a coefficient of performance of 2.5 and its compressor uses 80 W. Calculate how long it takes to freeze 0.50 kg of water that is already at 0 °C, to 2 significant figures. (Specific latent heat of fusion of water = 3.34 × 10⁵ J kg⁻¹)

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Step 1 – heat to be removed
Q_C = ml = 0.50 × 3.34 × 10⁵ = 167 000 J
Step 2 – rate of heat removal
F: COP_ref = Q_C ÷ W, so rate of heat removal = COP_ref × power input
I: rate = 2.5 × 80 = 200 W
Step 3 – time
F: t = Q_C ÷ rate = 167 000 ÷ 200 = 835 s
A: 840 s (2 s.f.), about 14 minutes
Watch out: the 80 W is the work input, not the rate of cooling. Multiply by the COP to get the rate at which heat is removed.

Q9 (Stretch) An air-source heat pump keeps a house at 21 °C. Its actual COP is 30% of the maximum possible value, and the rate of heat loss from the house is proportional to the temperature difference between inside and outside. When it is 1 °C outside, the house loses heat at 6.0 kW. Calculate the electrical power needed, to 2 significant figures, when it is (a) 1 °C outside, (b) −9 °C outside.

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Convert first: T_H = 21 + 273 = 294 K; 1 °C = 274 K; −9 °C = 264 K
Step 1 – (a) COP at 1 °C
maximum COP_hp = 294 ÷ (294 − 274) = 294 ÷ 20 = 14.7; actual = 0.30 × 14.7 = 4.41
Step 2 – (a) power
W = Q_H ÷ COP_hp = 6000 ÷ 4.41 = 1360.5… W
A: 1.4 kW (2 s.f.)
Step 3 – (b) new heat loss
the temperature difference rises from 20 K to 30 K, so heat loss = 6.0 × 30 ÷ 20 = 9.0 kW
Step 4 – (b) COP and power at −9 °C
maximum COP_hp = 294 ÷ 30 = 9.8; actual = 0.30 × 9.8 = 2.94; W = 9000 ÷ 2.94 = 3061.2… W
A: 3.1 kW (2 s.f.)
Watch out: on colder days the heat pump is hit twice. The house needs more heat and the COP falls, so the power more than doubles.

Q10 (Stretch) A refrigerator’s interior is kept at 3 °C and it releases heat to a kitchen at 35 °C. Its actual COP is 2.8 and its compressor uses 95 W when running. Heat leaks into the refrigerator at an average rate of 120 W. Calculate (a) the maximum possible COP, (b) the rate of heat removal while the compressor runs, (c) the fraction of the time the compressor runs, to 2 significant figures, (d) the electrical energy used per day, to 2 significant figures.

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Convert first: T_C = 3 + 273 = 276 K; T_H = 35 + 273 = 308 K
Step 1 – (a) maximum COP
maximum COP_ref = 276 ÷ (308 − 276) = 276 ÷ 32 = 8.625 (8.6 to 2 s.f.)
Step 2 – (b) rate of heat removal
Q_C per second = COP_ref × W per second = 2.8 × 95 = 266 W
Step 3 – (c) fraction of time running
on average the fridge must remove 120 W, so fraction = 120 ÷ 266 = 0.451…
A: 45% of the time (2 s.f.)
Step 4 – (d) energy per day
average electrical power = 0.451… × 95 = 42.85… W; energy = 42.85… × 24 × 3600 = 3.702… × 10⁶ J
A: 3.7 MJ (2 s.f.), about 1.0 kWh
Watch out: the heat released to the kitchen is Q_C + W = 266 + 95 = 361 W while running, more than is removed from inside. A fridge with its door open warms the kitchen.

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