10 GCSE Physics questions about the heat pump that get harder as you go, from Foundation to Higher, each with a full worked answer. An air source heat pump works like a fridge in reverse: it uses electrical energy to move energy from the cold air outside into your house. For every 1 J of electrical energy it uses, it can deliver about 3 J into the home – the extra comes from the thermal energy store of the outside air.
Course: Combined Science + Separate Physics | Tier: Foundation + Higher | Topics: Energy · Particle model
📌 KEY DEFINITIONS – learn them word for word
Specific heat capacity: the amount of energy needed to raise the temperature of 1 kg of a substance by 1 °C.
Power: the rate at which energy is transferred, or the rate at which work is done.
Efficiency: useful output energy transfer ÷ total input energy transfer.
Equations you’ll use
| Quantity | Symbol | Unit |
|---|---|---|
| energy transferred | E | joules (J) or kilowatt-hours (kWh) |
| power | P | watts (W) or kilowatts (kW) |
| time | t | seconds (s) or hours (h) |
| mass | m | kilograms (kg) |
| specific heat capacity | c | joules per kilogram per degree Celsius (J/kg °C) |
| temperature change | Δθ | degrees Celsius (°C) |
Use FIFA for every calculation
- 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.
Questions
Q1 (F) A heat pump uses 2.0 kW of electrical power for 5 hours. Calculate the electrical energy it uses in kilowatt-hours (kWh).
Show solution
F: E = P t
I: E = 2.0 kW × 5 h
F: already the subject
A: 10 kWh
Q2 (F) Electricity costs 25p per kWh. Calculate the cost of the 10 kWh used by the heat pump.
Show solution
cost = energy × price per kWh = 10 × 25p = 250p = £2.50
Q3 (F) Calculate the same energy, 2.0 kW for 5 hours, in joules.
Show solution
Convert first: 2.0 kW = 2000 W; 5 h = 5 × 3600 = 18 000 s
F: E = P t
I: E = 2000 × 18 000
F: already the subject
A: 36 000 000 J (3.6 × 107 J)
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Q4 (F) A hot water tank holds 150 kg of water. The heat pump heats it from 15 °C to 55 °C. Calculate the energy transferred to the water. (c = 4200 J/kg °C)
Show solution
F: ΔE = m c Δθ
I: ΔE = 150 × 4200 × (55 − 15)
F: already the subject
A: 150 × 4200 × 40 = 25 200 000 J (2.52 × 107 J)
Q5 (F/H) The heat pump delivers 3 J of energy to the water for every 1 J of electrical energy it uses. Calculate the electrical energy needed to heat the tank in Q4, in kWh. (1 kWh = 3 600 000 J)
Show solution
Step 1: electrical energy = 25 200 000 ÷ 3 = 8 400 000 J
Step 2: 8 400 000 ÷ 3 600 000 = 2.3 kWh
Q6 (F/H) A gas boiler is 90% efficient. It gives a useful output power of 18 kW to a house. Calculate the total input power from the gas.
Show solution
F: efficiency = useful power output ÷ total power input
I: 0.90 = 18 ÷ input
F: input = 18 ÷ 0.90
A: 20 kW
Q7 (F/H) Explain why replacing a gas boiler with an electric heat pump can reduce carbon dioxide emissions. [4 marks]
Show answer
• A gas boiler burns natural gas, a fossil fuel, which releases carbon dioxide.
• A heat pump burns no fuel in the home.
• Its electricity can come from renewable resources such as wind and solar, which release no CO₂ when generating.
• It delivers more energy than the electrical energy it uses, so less energy needs to be generated in total.
Q8 (H) A heat pump uses 3.0 kW for 6 hours a day for 150 days of winter. Electricity costs 25p per kWh. Calculate the total cost.
Show solution
Step 1 – energy per day: E = P t = 3.0 × 6 = 18 kWh
Step 2 – energy for winter: 18 × 150 = 2700 kWh
Step 3 – cost: 2700 × £0.25 = £675
Q9 (H) A radiator contains 12 kg of water. It cools from 45 °C to 35 °C in 10 minutes as it heats a room. Calculate the average power transferred to the room. (c = 4200 J/kg °C)
Show solution
Step 1: ΔE = m c Δθ = 12 × 4200 × 10 = 504 000 J
Step 2: 10 minutes = 600 s
F: P = E ÷ t
I: P = 504 000 ÷ 600
A: 840 W
Q10 (H) Heat pumps work best in well-insulated homes. Explain how insulation reduces the energy a home needs to stay warm. Refer to thermal conductivity in your answer. [6 marks]
Show answer
Marking points (any 6):
• Energy is transferred from the warm house to the colder surroundings, mainly by conduction through walls, roof, windows and floor.
• The rate of energy transfer depends on the thermal conductivity of the material.
• Insulating materials (loft wool, cavity foam) have low thermal conductivity…
• …so the rate of energy transfer out of the house is lower.
• Thicker walls or insulation also reduce the rate of transfer.
• Double glazing traps a layer of air or gas between the panes, which reduces conduction.
• So less energy is dissipated to the surroundings, and the heat pump needs to transfer less energy (and costs less to run) to keep the house at the same temperature.
⚠️ Where students lose marks
- Mixing up kW and W, or hours and seconds – use kW and h for kWh, and W and s for joules.
- Forgetting to find the temperature change (55 − 15) before using ΔE = m c Δθ.
- Thinking more energy comes out than goes in breaks conservation of energy. It doesn’t – the extra energy comes from the outside air.
- Writing that insulation “stops heat getting in”. Insulation reduces the rate of energy transfer out of the house.
More devices: all 40 everyday devices · Energy · Particle model
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