The Physics Question Bank

10 GCSE Physics questions about the thermal camera, from Foundation to Higher, each with a full worked solution.

Every object gives off infrared radiation – and the hotter it is, the more it emits. A thermal camera has a sensor that detects this infrared instead of visible light, then colours each part of the picture by how much infrared it receives. That’s how firefighters find people through smoke and how builders spot heat escaping from a house.

Course: Combined Science + Separate Physics  |  Tier: Foundation + Higher  |  Topics: Waves · Energy · Electricity

Equations you’ll use

Quantity	Symbol	Unit
wave speed	v	metres per second (m/s)
frequency	f	hertz (Hz)
wavelength	λ	metres (m)
power	P	watts (W)
potential difference	V	volts (V)
current	I	amps (A)
charge	Q	coulombs (C)
energy transferred	E	joules (J)
time	t	seconds (s)
mass	m	kilograms (kg)
specific heat capacity	c	joules per kilogram per degree Celsius (J/kg °C)
temperature change	Δθ	degrees Celsius (°C)

Equations: v = f λ  |  P = V I  |  E = P t  |  Q = I t  |  ΔE = m c Δθ  |  1 µm = 1 × 10⁻⁶ m

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.

Foundation: aim for Q1–8.   Higher: aim for Q4–10.

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Questions

Q1 (F) A thermal camera detects infrared with a wavelength of 10 µm. Infrared travels at 3.0 × 10⁸ m/s. Calculate its frequency.

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Convert first: 10 µm = 1.0 × 10⁻⁵ m
F: v = f λ
I: 3.0 × 10⁸ = f × 1.0 × 10⁻⁵
F: f = 3.0 × 10⁸ ÷ 1.0 × 10⁻⁵
A: 3.0 × 10¹³ Hz

Q2 (F) A firefighter uses a thermal camera in a dark, smoky room. Explain why a person shows up brightly on the screen. [2 marks]

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• All objects emit infrared radiation, and the camera detects it (it does not need visible light).
• The person is hotter than the room, so they emit more infrared in a given time and look brighter.

Q3 (F) The camera’s battery supplies a current of 0.50 A at a potential difference of 3.6 V. Calculate the power of the camera.

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F: P = V I
I: P = 3.6 × 0.50
F: P is already the subject
A: 1.8 W

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Q4 (F) The 1.8 W camera is used for 2 hours. Calculate the energy transferred, in kJ.

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Convert first: 2 hours = 2 × 3600 = 7200 s
F: E = P t
I: E = 1.8 × 7200
F: E is already the subject
A: 12 960 J = 13 kJ (2 s.f.)

Q5 (F/H) When fully charged, the camera’s battery can supply a current of 2.6 A for 1.0 hour. Calculate (a) the charge it stores in coulombs and (b) how many hours it lasts if the camera draws 0.50 A.

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Convert first: 1.0 hour = 3600 s
(a) F: Q = I t
I: Q = 2.6 × 3600
F: Q is already the subject
A: 9360 C
(b) F: Q = I t
I: 9360 = 0.50 × t
F: t = 9360 ÷ 0.50 = 18 720 s
A: 18 720 ÷ 3600 = 5.2 hours

Q6 (F/H) A thermal camera shows a badly insulated patch of wall losing heat at 150 W. Calculate the energy lost in one day in kWh, and the cost if electricity costs 28p per kWh.

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Convert first: 150 W = 0.150 kW; 1 day = 24 h
F: E = P t
I: E = 0.150 × 24
F: E is already the subject
A: 3.6 kWh per day; cost = 3.6 × 28 = 100.8p ≈ £1.01 per day

Q7 (F/H) A thermal camera shows a mug containing 0.25 kg of tea cooling from 80 °C to 60 °C in 10 minutes. Calculate (a) the energy lost by the tea and (b) the average rate at which it loses energy. (specific heat capacity of tea = 4200 J/kg °C)

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(a) F: ΔE = m c Δθ
I: ΔE = 0.25 × 4200 × (80 − 60)
F: ΔE is already the subject
A: 21 000 J
(b) Convert first: 10 minutes = 600 s
F: P = E ÷ t
I: P = 21 000 ÷ 600
A: 35 W

Q8 (F/H · S only) What is a perfect black body? Explain why it is also the best possible emitter of radiation. [2 marks]

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• A perfect black body absorbs all of the radiation that hits it – it does not reflect or transmit any.
• A good absorber is also a good emitter, so a perfect black body is the best possible emitter.

Q9 (H · S only) A kettle of water at 100 °C and a person at 37 °C both appear on a thermal camera. Describe how the radiation emitted by the kettle is different from that emitted by the person. [2 marks]

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• The hotter kettle emits a greater intensity of radiation at every wavelength (more energy per second).
• Its peak intensity is at a shorter wavelength than the person’s.

Q10 (H · S only) A person stands in a room. Their body temperature stays at 37 °C. Explain this using ideas about the absorption and emission of infrared radiation, and say what would happen if they moved into a much colder room. [4 marks]

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• The person both absorbs infrared from their surroundings and emits infrared.
• Their temperature stays constant when the rate of absorbing energy equals the rate of emitting energy (with body heat from respiration making up the difference).
• In a colder room, they absorb less infrared from the surroundings but still emit about the same, so they emit more than they absorb.
• Their temperature would fall unless their body transfers more energy (e.g. by shivering) or they reduce emission with clothing.

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⚠️ Where students lose marks

• Saying a thermal camera detects “heat” or “temperature” – it detects infrared radiation.
• Thinking only hot objects emit infrared – all objects emit infrared; hotter ones emit more.
• Forgetting to convert µm to m, or hours to seconds (multiply by 3600) before using Q = I t or E = P t.
• For a hotter body, saying the peak moves to a longer wavelength – it moves to a shorter wavelength.

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