MYP Physics · unit 07
Thermal physics
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
Optional enrichment: this challenge may extend beyond the usual syllabus depth and assumes confidence with algebra and graph interpretation. 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.
A gas law that fails on a Celsius axis
Optional enrichment: this challenge may extend beyond the usual syllabus depth and assumes confidence with algebra and graph interpretation. 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.
Selecting insulation from cooling evidence
Optional enrichment: this challenge may extend beyond the usual syllabus depth and assumes confidence with algebra and graph interpretation. 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.