10 GCSE Physics questions about the 3D printer, from Foundation to Higher, each with a full worked answer. A 3D printer pushes a thin plastic thread, called filament, into a hot nozzle at about 210 °C. The plastic melts, squeezes out as a fine line and quickly solidifies again, building the object up layer by layer on a heated bed that stops the first layers from cooling too fast and peeling off.
Course: Combined Science + Separate Physics | Tier: Foundation + Higher | Topics: Particle model of matter, Energy
📌 KEY DEFINITIONS – learn them word for word
Specific heat capacity: the amount of energy required to raise the temperature of one kilogram of the substance by one degree Celsius.
Specific latent heat: the amount of energy required to change the state of one kilogram of the substance with no change in temperature. The specific latent heat of fusion is for changing between solid and liquid.
Internal energy: the total kinetic energy and potential energy of all the particles (atoms and molecules) that make up a system.
Equations you’ll use
ΔE = m c Δθ | E = m L | E = P t | efficiency = useful output energy transfer ÷ total input energy transfer
| Quantity | Symbol | Unit |
|---|---|---|
| change in thermal energy | ΔE | joules (J) |
| mass | m | kilograms (kg) |
| specific heat capacity | c | joules per kilogram per degree Celsius (J/kg °C) |
| temperature change | Δθ | degrees Celsius (°C) |
| energy for a change of state | E | joules (J) |
| specific latent heat | L | joules per kilogram (J/kg) |
| power | P | watts (W) |
| time | t | seconds (s) |
Data for this page: PLA plastic: c = 1800 J/kg °C, specific latent heat of fusion L = 40 000 J/kg (a simplified value). Aluminium: c = 900 J/kg °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 print uses 50 g of PLA filament. Calculate the energy needed to raise its temperature from 20 °C to 210 °C. (Ignore the change of state.)
Show solution
Convert first: 50 g = 0.050 kg; Δθ = 210 − 20 = 190 °C
F: ΔE = m c Δθ
I: ΔE = 0.050 × 1800 × 190
F: ΔE is already the subject
A: 17 100 J (17.1 kJ)
Q2 (F) Calculate the energy needed to melt the 50 g of PLA. Specific latent heat of fusion of PLA = 40 000 J/kg.
Show solution
Convert first: 50 g = 0.050 kg
F: E = m L
I: E = 0.050 × 40 000
F: E is already the subject
A: 2000 J
Q3 (F) The printer’s heated bed is an aluminium plate with a mass of 0.45 kg. Calculate the energy needed to warm it from 20 °C to 60 °C.
Show solution
F: ΔE = m c Δθ
I: ΔE = 0.45 × 900 × (60 − 20) = 0.45 × 900 × 40
F: ΔE is already the subject
A: 16 200 J (16.2 kJ)
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Q4 (F) The bed heater has a power of 220 W. Calculate the shortest time it could take to supply the 16 200 J needed, to 2 significant figures.
Show solution
F: E = P t
I: 16 200 = 220 × t
F: t = 16 200 ÷ 220 = 73.6 s
A: 74 s (2 s.f.)
Q5 (F/H) In fact, the 220 W bed takes 2.5 minutes to reach 60 °C. Calculate the efficiency of heating the bed, as a percentage to 2 significant figures.
Show solution
Convert first: 2.5 minutes = 2.5 × 60 = 150 s
Step 1 – total energy in: E = P t = 220 × 150 = 33 000 J
F: efficiency = useful output energy ÷ total input energy
I: efficiency = 16 200 ÷ 33 000
F: efficiency is already the subject
A: 0.49 = 49% (2 s.f.)
Q6 (F/H) A large print takes 6 hours. The printer’s average power is 120 W. Calculate the energy used in kWh and the cost if electricity costs 27p per kWh.
Show solution
Convert first: 120 W = 0.12 kW
F: E = P t
I: E = 0.12 × 6
F: E is already the subject
A: 0.72 kWh
Cost: 0.72 × 27 = 19.44p ≈ 19p
Q7 (F/H) As the filament melts in the nozzle, energy is still being supplied but its temperature stays the same. Explain why, in terms of particles. [3 marks]
Show answer
• The energy supplied increases the internal energy of the plastic. (1)
• During melting it is used to break / weaken the forces (bonds) between the particles, which increases their potential energy. (1)
• The kinetic energy of the particles does not increase, so the temperature stays the same until all the plastic has melted. (1)
Q8 (H) The nozzle pushes out 0.90 g of PLA every minute. Each gram must be heated from 20 °C to 210 °C and melted. Calculate the minimum power needed, to 2 significant figures. The nozzle’s heater is rated at 40 W – comment on your answer.
Show solution
Convert first: 0.90 g = 0.000 90 kg; 1 minute = 60 s
Step 1 – heating: ΔE = m c Δθ = 0.000 90 × 1800 × 190 = 307.8 J
Step 2 – melting: E = m L = 0.000 90 × 40 000 = 36 J
Step 3 – total energy per minute: 307.8 + 36 = 343.8 J
F: E = P t
I: 343.8 = P × 60
F: P = 343.8 ÷ 60 = 5.73 W
A: 5.7 W (2 s.f.)
Comment: this is much less than 40 W – most of the heater’s energy is dissipated to the surroundings (the metal nozzle and cooling fans), and extra power lets it heat up quickly.
Q9 (H) A finished 30 g part cools from 210 °C to 20 °C, solidifying on the way. Calculate the total energy it transfers to the surroundings, in kJ to 2 significant figures.
Show solution
Convert first: 30 g = 0.030 kg; Δθ = 210 − 20 = 190 °C
Step 1 – cooling: ΔE = m c Δθ = 0.030 × 1800 × 190 = 10 260 J
Step 2 – solidifying: E = m L = 0.030 × 40 000 = 1200 J
Step 3 – total: 10 260 + 1200 = 11 460 J
A: 11 kJ (2 s.f.)
The same energy is released when it solidifies as was needed to melt it.
Q10 (H) In Q4 and Q5 the heated bed took twice as long to warm up as the calculation predicted. Explain why, and describe how the printer could be changed to reduce the energy wasted while it prints. [6 marks]
Show answer
Model answer – marking points:
• Not all of the energy supplied goes into the thermal store of the aluminium plate.
• As soon as the bed is warmer than the room, energy is transferred to the thermal store of the surroundings – by heating the air above it (which rises and carries energy away) and by infrared radiation.
• Energy is also conducted into the frame and parts touching the bed.
• The hotter the bed gets, the faster it loses energy, so the efficiency is low (about 49%).
• To reduce this: put a layer of thermal insulation (material with low thermal conductivity, e.g. cork or foam) under the bed so less energy is conducted away.
• Put the printer inside an enclosure so warm air is trapped and less energy is carried away by moving air; switch the bed off as soon as the print finishes.
⚠️ Where students lose marks
- Leaving the mass in grams – c and L are “per kilogram”, so 50 g must become 0.050 kg.
- Using the final temperature instead of the temperature change Δθ in ΔE = m c Δθ.
- Mixing up the two equations: use m c Δθ when the temperature changes and m L when the state changes. If both happen, calculate each part and add them.
- Saying the particles “speed up” during melting – the temperature is constant, so their kinetic energy is not increasing; their potential energy increases.
More devices: all 40 everyday devices · Particle model questions · Energy questions
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