Free B.1 Thermal energy transfers questions for IB Physics (SL & HL), part of B The particulate nature of matter. Each question has the answer and the marking points an examiner would look for.
Generate a full B.1 Thermal Energy Transfers worksheet →All B The particulate nature of matter questionsA mass of 0.05 kg of ice at 0 °C is added to 0.3 kg of water at 40 °C. Specific heat capacity of water = 4200 J kg−1 K−1; specific latent heat of fusion of ice = 3.34 × 105 J kg−1.
What is the final temperature?
Answer: A
A solid is heated at constant power. Its temperature rises, stays constant while it melts, then rises again. What happens to the mean kinetic energy and the potential energy of its particles while it melts?
| mean kinetic energy | potential energy | |
|---|---|---|
| A | constant | constant |
| B | increases | increases |
| C | constant | increases |
| D | increases | constant |
Answer: C
A wall conducts thermal energy at rate P. The wall is replaced by one of the same material with 2 times the thickness and 2 times the area. The temperatures on each side stay the same. Which statements are correct?
I. The new rate of conduction is 2P/2.
II. The rate of conduction depends on the thermal conductivity of the material.
III. The new rate of conduction is 4P.
Answer: A
A glass window is 1.2 m × 0.80 m and 4.0 mm thick. The thermal conductivity of glass is 0.80 W m−1 K−1. The inside temperature is 20 °C and the outside temperature is 5 °C.
Calculate the rate of thermal energy transfer through the glass.
Explain why double glazing with an air gap reduces this rate.
A radiator of surface area 1.5 m2 and emissivity 0.90 is at 330 K. Calculate the power it radiates.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | 2900 W ΔQ/Δt = kAΔT/Δx ✓ = 0.80 × 0.96 × 15/0.004 = 2.9 × 103 W ✓ | 2 |
| (b) | air poor conductor air has a much lower thermal conductivity ✓ the narrow gap limits convection ✓ | 2 |
| (c) | 910 W P = eσAT4 ✓ = 0.90 × 5.67 × 10−8 × 1.5 × 3304 = 910 W ✓ | 2 |
An electric kettle of power 2200 W contains 1 kg of water at 15 °C. Specific heat capacity of water = 4200 J kg−1 K−1; specific latent heat of vaporization = 2.26 × 106 J kg−1.
Calculate the minimum time for the water to reach 100 °C.
The kettle fails to switch off and the water boils for 90 s. Calculate the mass of water that boils away.
Distinguish, in molecular terms, between the energy supplied while the water heats up and the energy supplied while it boils.
In practice the water takes 179 s to reach 100 °C. Calculate the average rate of energy loss.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | 162 s Q = mcΔθ = 357000 J ✓ t = QP = 162 s ✓ | 2 |
| (b) | 0.0876 kg Q = Pt = 198000 J ✓ m = QL = 0.0876 kg ✓ | 2 |
| (c) | KE vs PE while heating, the mean kinetic energy of the molecules increases (temperature rises) ✓ while boiling, the energy increases the potential energy of the molecules ✓ by breaking intermolecular bonds; the temperature stays constant ✓ | 3 |
| (d) | 206 W useful power = 357000179 = 1994 W ✓ loss = 206 W ✓ | 2 |
A student heats a 0.5 kg block of aluminium with an electrical heater. The potential difference across the heater is 12.0 ± 0.1 V and the current is 4.6 ± 0.1 A. A thermometer in the block is read every 60 s.
| t / s | 60 | 120 | 180 | 240 | 300 | 360 |
|---|---|---|---|---|---|---|
| θ / °C ± 0.5 | 24.1 | 30.7 | 37.4 | 44.3 | 51.1 | 57.8 |
Calculate the power of the heater and its percentage uncertainty.
Use the data to determine the specific heat capacity of the block.
The accepted value for aluminium is 900 J kg−1 K−1. Explain why the experimental value is larger.
Suggest why the temperature does not rise during the first part of the heating.
Suggest two improvements to the experiment.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | 55.2 W ± 3.01% P = VI = 55.2 W ✓ percentage uncertainties add ✓ 0.112.0 + 0.14.6 → 3.01% ✓ | 3 |
| (b) | 974 J kg−1 K−1 rate of temperature rise = 0.113 K s−1 from the linear region ✓ c = Pm × (Δθ/Δt) ✓ c ≈ 974 J kg−1 K−1 (accept ±8%) ✓ | 3 |
| (c) | thermal energy lost some energy is transferred to the surroundings / not all goes into the block ✓ so the temperature rise is smaller than expected, giving a larger calculated c ✓ | 2 |
| (d) | heater warms up first the heater and thermometer take time to warm up / thermal energy takes time to conduct to the thermometer ✓ | 1 |
| (e) | insulate; oil in holes insulate the block (lagging) ✓ put oil/thermal paste in the holes for good thermal contact (or start below room temperature and finish the same amount above) ✓ | 2 |
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Meet Riyaz →Original IB-style practice questions written for mentorriyaz.in. These are not IB past paper questions, and this site is not affiliated with or endorsed by the International Baccalaureate Organization. Updated 2026-11-04. Paper generator · mentorriyaz.in