10 GCSE Physics questions about the air fryer, from Foundation to Higher, each with a full worked solution. An air fryer is a small electric oven. A heating element gets very hot, and a fan blows the hot air quickly around the food. Because the cooking space is small, less energy is wasted heating air and metal, so it usually costs less to run than a big oven.
Course: Combined Science + Separate Physics | Tier: Foundation + Higher | Topics: Energy · Electricity
📌 KEY DEFINITIONS – learn them word for word
Power: the rate at which energy is transferred. 1 watt (W) = 1 joule transferred per second.
Specific heat capacity: the amount of energy needed to raise the temperature of 1 kg of a substance by 1 °C.
Efficiency: useful output energy transfer ÷ total input energy transfer.
Dissipated energy: energy that is transferred to the surroundings in a way that is not useful, usually heating them. We say it is “wasted”.
Equations you’ll use
E = P t | P = V I | V = I R | ΔE = m c Δθ (given on the equation sheet) | efficiency = useful output ÷ total input
| Quantity | Symbol | Unit |
|---|---|---|
| energy transferred | E | joules (J) |
| power | P | watts (W) |
| time | t | seconds (s) |
| potential difference | V | volts (V) |
| current | I | amps (A) |
| resistance | R | ohms (Ω) |
| mass | m | kilograms (kg) |
| specific heat capacity | c | J/kg °C |
| temperature change | Δθ | °C |
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.
Questions
Q1 (F) A 1500 W air fryer is switched on for 15 minutes. Calculate the energy it transfers.
Show solution
Convert first: 15 minutes = 900 s
F: E = P t
I: E = 1500 × (15 × 60) = 1500 × 900
F: E is already the subject
A: 1 350 000 J (1.35 MJ)
Q2 (F) The 1500 W air fryer is plugged into the 230 V mains. Calculate the current in it.
Show solution
F: P = V I
I: 1500 = 230 × I
F: I = 1500 ÷ 230
A: 6.5 A (2 s.f.)
Q3 (F) Electricity costs 25p per kilowatt-hour (kWh). Calculate the cost of using the 1.5 kW air fryer for 15 minutes.
Show solution
Step 1 – energy in kWh: E = P t = 1.5 kW × 0.25 h = 0.375 kWh
Step 2 – cost: 0.375 × 25p = 9.375p
A: 9.4p (about 9p)
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Q4 (F) A family cooks chips in a 2.5 kW oven for 40 minutes, or in a 1.5 kW air fryer for 20 minutes. Calculate how much energy, in kWh, the air fryer saves.
Show solution
Oven: E = P t = 2.5 kW × (40 ÷ 60) h = 1.67 kWh
Air fryer: E = 1.5 kW × (20 ÷ 60) h = 0.50 kWh
A: saving = 1.67 − 0.50 = 1.17 kWh (about 1.2 kWh)
Q5 (F/H) The aluminium basket has a mass of 0.30 kg. It heats up from 20 °C to 200 °C. The specific heat capacity of aluminium is 900 J/kg °C. Calculate the energy needed to heat the basket.
Show solution
F: ΔE = m c Δθ
I: ΔE = 0.30 × 900 × (200 − 20) = 0.30 × 900 × 180
F: ΔE is already the subject
A: 48 600 J
Q6 (F/H) In 10 minutes the air fryer is supplied with 900 000 J. Only 540 000 J goes into heating the food. Calculate the efficiency of the air fryer.
Show solution
F: efficiency = useful output energy ÷ total input energy
I: efficiency = 540 000 ÷ 900 000
F: already the subject
A: 0.60 (or 60%)
Q7 (F/H) Explain why an air fryer usually uses less energy than a large oven to cook the same food. [4 marks]
Show answer
Marking points (any four):
• the air fryer has a lower power rating
• it cooks for a shorter time, so E = P t is smaller
• the cooking space is smaller, so less energy is needed to heat the air and metal walls
• the fan moves hot air over the food quickly, so energy reaches the food faster
• less energy is dissipated (wasted) heating the kitchen, so it is more efficient
Q8 (F/H) The heating element is a long thin wire. Explain why the wire gets hot when there is a current in it. [2 marks]
Show answer
• the current (moving charges) does work against the resistance of the wire
• this transfers energy to the thermal energy store of the wire, so its temperature rises
Q9 (H) 0.50 kg of potato chips (specific heat capacity 3500 J/kg °C) is heated from 20 °C to 100 °C. The air fryer is 1500 W and 60% efficient. Calculate the minimum time needed. Give your answer in minutes.
Show solution
Step 1 – energy needed: ΔE = m c Δθ = 0.50 × 3500 × 80 = 140 000 J
Step 2 – useful power: 0.60 × 1500 = 900 W
Step 3 – time: t = E ÷ P = 140 000 ÷ 900 = 156 s
A: 156 ÷ 60 = 2.6 minutes
In real life it takes longer, because energy is also needed to evaporate water from the chips.
Q10 (H) Use your answer to Q2 to calculate the resistance of the heating element when it is working.
Show solution
Use unrounded current: I = 1500 ÷ 230 = 6.52 A
F: V = I R
I: 230 = 6.52 × R
F: R = 230 ÷ 6.52
A: 35 Ω (2 s.f.)
⚠️ Where students lose marks
- Forgetting to change minutes into seconds when using E = P t in joules.
- Using watts instead of kilowatts (or seconds instead of hours) when calculating energy in kWh.
- Using the final temperature instead of the temperature change (Δθ) in ΔE = m c Δθ.
- Saying energy is “used up” or “lost”. Say it is dissipated to the surroundings.
More devices: all 40 everyday devices · Energy · Electricity
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