IB Diploma Physics · unit B
The particulate nature of matter
Connect the ideas, make your working visible, and test the result. Use the hints when you need a starting point; open the full solution after your attempt.
Extra challenging questions
A selection from this pathway. Open a topic below for its complete practice and challenge set.
Heating ice while heat escapes
A 0.10 kg sample of ice begins at 0°C. A heater supplies 100 W, while a separate calibration shows a constant 20 W loss to the surroundings during this experiment. The sample-temperature observations are supplied below. Use latent heat of fusion 3.36×10⁵ J kg⁻¹ and water specific heat capacity 4200 J kg⁻¹ K⁻¹. Ignore vessel heat capacity. Supplied data (Time / s; Temperature / °C): (0, 0), (210, 0), (420, 0), (525, 20). (a) Predict the melting duration and then the time to reach 20°C after melting. Check both against the data. [3 marks] (b) Explain the constant-temperature interval with a particle model and evaluate a calculation that ignores losses. [3 marks]
Observed heating curve
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Temperature / °C
Horizontal axis (x): Time / s
Read the exact data as a table
| Time / s | Temperature / °C |
|---|---|
| 0 | 0 |
| 210 | 0 |
| 420 | 0 |
| 525 | 20 |
These are the supplied values, not a worked answer. Lines join the supplied values; they are not a fitted model.
Show a hint
Begin with part (a): Predict the melting duration and then the time to reach 20°C after melting. Check both against the data.
Reveal the worked solution
- Net heating power is 100−20 = 80 W, since the sample does not receive the full heater input.
- Melting needs mL = 0.10(3.36×10⁵) = 33600 J; duration is 33600/80 = 420 s.
- Warming liquid by 20 K needs mcΔT = 0.10(4200)(20) = 8400 J; duration is 8400/80 = 105 s. Both match the supplied curve.
- During melting, transferred energy changes intermolecular potential energy rather than increasing mean kinetic energy, so temperature need not rise.
- The internal energy still increases through the phase change; temperature is not a measure of the total internal energy.
- Ignoring losses gives 33600/100 = 336 s, 84 s too short. Insulation and a lid can reduce unwanted transfers; repeats alone do not remove systematic heat loss.
Answer (a) Melting takes 420 s; subsequent warming takes 105 s. (b) Temperature stays constant while internal energy increases through a change in particle arrangement; ignoring losses underestimates the required time.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Selecting insulation from cooling evidence
Two identical vessels each contain 0.50 kg of water initially at 80°C in a 20°C room. Vessel A has a matte black exposed surface; B has a shiny jacket and an insulating lid. The measured cooling data are supplied. Use water specific heat capacity 4200 J kg⁻¹ K⁻¹; neglect vessel heat capacity. Supplied data (Time / min; Water temperature / °C): (0, 80), (2, 70), (4, 62), (6, 55). (a) Vessel B reaches 70°C after 6 min. Compare average rates of energy loss for A and B over the first 6 min. [3 marks] (b) Explain the role of conduction, convection, evaporation and radiation, and propose a fair test separating the lid from the shiny finish. [3 marks]
Water cooling in vessel A
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Water temperature / °C
Horizontal axis (x): Time / min
Read the exact data as a table
| Time / min | Water temperature / °C |
|---|---|
| 0 | 80 |
| 2 | 70 |
| 4 | 62 |
| 6 | 55 |
These are the supplied values, not a worked answer. Lines join the supplied values; they are not a fitted model.
Show a hint
Begin with part (a): Vessel B reaches 70°C after 6 min. Compare average rates of energy loss for A and B over the first 6 min.
Reveal the worked solution
- For A, energy loss is mcΔT = 0.50(4200)(80−55) = 52500 J over 360 s, giving about 146 W.
- For B, energy loss is 0.50(4200)(80−70) = 21000 J, giving 58.3 W.
- These are time averages. A's decreasing temperature drop per interval supports a cooling rate that falls as its temperature approaches the room temperature.
- Insulation reduces conduction through the walls; a lid restricts warm-air convection and evaporation. A shiny surface lowers thermal radiation exchange compared with a similar matte surface.
- Use a factorial comparison: identical vessels with/without a lid and with shiny/matte surfaces, holding water mass, starting temperature, geometry, room and probe depth constant.
- Repeat each setup and log the room temperature. Comparing A and B alone shows the combined improvement but cannot apportion the effect between lid and finish.
Answer (a) A loses about 146 W on average; B loses about 58.3 W. (b) Change one feature at a time with equal initial conditions; a combined design comparison cannot identify each feature's individual effect.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
A planet's energy balance with an albedo feedback
A planet receives stellar irradiance 1360 W m⁻² on a surface perpendicular to incoming light. Its albedo is initially 0.30. Assume a spherical planet and uniform effective emission σT⁴, where σ = 5.67×10⁻⁸ W m⁻² K⁻⁴. The supplied graph gives emitted flux for several effective temperatures. The actual surface is warmer than this simple emission temperature. Supplied data (Temperature / K; Emitted flux / W m⁻²): (240, 188.117), (255, 239.742), (270, 301.327), (288, 390.079). (a) Calculate globally averaged absorbed stellar flux and equilibrium effective temperature, explaining the factor of four. [3 marks] (b) Albedo drops to 0.25. Find the new absorbed flux and temperature, and distinguish the greenhouse mechanism from the albedo change. [3 marks]
Blackbody emitted flux
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Emitted flux / W m⁻²
Horizontal axis (x): Temperature / K
Read the exact data as a table
| Temperature / K | Emitted flux / W m⁻² |
|---|---|
| 240 | 188.117 |
| 255 | 239.742 |
| 270 | 301.327 |
| 288 | 390.079 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Begin with part (a): Calculate globally averaged absorbed stellar flux and equilibrium effective temperature, explaining the factor of four.
Reveal the worked solution
- The planet intercepts light over a disc πR² but emits across a spherical area 4πR², so averaged incoming irradiance is S/4.
- Absorbed flux is (1−0.30)(1360)/4 = 238 W m⁻².
- Equilibrium requires σT⁴ = 238, giving T = (238/σ)¹/⁴≈254.6 K. This is the effective emission temperature, not automatically the surface temperature.
- With albedo 0.25, absorbed flux is 0.75(1360)/4 = 255 W m⁻², an increase of 17 W m⁻².
- New equilibrium temperature is (255/σ)¹/⁴≈259 K, about 4.4 K above the initial effective temperature.
- Albedo changes reflected incoming light. Greenhouse gases absorb and emit outgoing infrared radiation and alter energy transfer between surface and atmosphere. Melting reflective ice can lower albedo and reinforce warming; this simple model does not quantify the complete coupled climate response.
Answer (a) Absorbed flux = 238 W m⁻²; effective temperature≈254.6 K. (b) New flux = 255 W m⁻²; effective temperature≈259 K; lower albedo gives warming in this model.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
A gas law that fails on a Celsius axis
A fixed amount of dilute gas is sealed in a rigid container. After equilibrium at each bath temperature, its absolute pressure is measured as supplied below. Assume ideal-gas behaviour and no leakage. The gas occupies 2.0×10⁻³ m³; take R = 8.31 J mol⁻¹ K⁻¹. Supplied data (Temperature / °C; Absolute pressure / kPa): (0, 100), (27.3, 110), (54.6, 120), (81.9, 130). (a) Extrapolate the pressure–temperature line to zero pressure and estimate the amount of gas at 0°C. [3 marks] (b) Predict pressure at 136.5°C and explain the microscopic mechanism. What if the container is allowed to expand at constant pressure instead? [3 marks]
Gas thermometer calibration
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Absolute pressure / kPa
Horizontal axis (x): Temperature / °C
Read the exact data as a table
| Temperature / °C | Absolute pressure / kPa |
|---|---|
| 0 | 100 |
| 27.3 | 110 |
| 54.6 | 120 |
| 81.9 | 130 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Begin with part (a): Extrapolate the pressure–temperature line to zero pressure and estimate the amount of gas at 0°C.
Reveal the worked solution
- The slope is 10/27.3 kPa °C⁻¹. Extrapolating 100 kPa down to zero requires a decrease of 273°C.
- Use absolute temperature T = 273 K and SI pressure 100000 Pa in pV = nRT.
- n = 100000(0.002)/(8.31×273) ≈0.0882 mol. The mathematical intercept motivates an absolute scale; a real gas condenses before this extrapolation reaches zero.
- 136.5°C corresponds to 409.5 K, 1.5 times 273 K, so fixed-volume pressure rises by the same factor to 150 kPa.
- Higher absolute temperature means greater mean molecular kinetic energy; more vigorous momentum transfer to the walls raises pressure at fixed volume.
- At fixed pressure and fixed particle number, V/T is constant, so the volume increases by 1.5. Dividing by Celsius temperatures would give an invalid proportionality.
Answer (a) Zero-pressure intercept ≈−273°C; amount ≈0.0882 mol. (b) Rigid-container pressure = 150 kPa; at constant pressure the volume would rise to 3.0×10⁻³ m³.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.