AP Physics · course AP 2
AP Physics 2: Algebra-Based
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.
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.
A gas cycle with a sign-convention trap
A fixed amount of gas follows the rectangular cycle A→B→C→D→A on the supplied pressure–volume data: A=(1 litre,100 kPa), B=(3 litre,100 kPa), C=(3 litre,300 kPa), D=(1 litre,300 kPa). Straight graph segments join the points. Use ΔU = Q−Wby, with work by the gas positive in expansion. For A→B the internal-energy increase is 300 J. Supplied data (Volume / litre; Pressure / kPa): (1, 100), (3, 100), (3, 300), (1, 300), (1, 100). (a) Find work and heat transferred on A→B, and determine net work by the gas over the complete cycle. [3 marks] (b) Find net heat over the full cycle and assess whether this direction of operation is a heat engine. [3 marks]
Rectangular gas cycle
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Pressure / kPa
Horizontal axis (x): Volume / litre
Read the exact data as a table
| Volume / litre | Pressure / kPa |
|---|---|
| 1 | 100 |
| 3 | 100 |
| 3 | 300 |
| 1 | 300 |
| 1 | 100 |
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): Find work and heat transferred on A→B, and determine net work by the gas over the complete cycle.
Reveal the worked solution
- At 100 kPa, expansion by 2 litres gives Wby = pΔV = 100000(0.002) = 200 J. Note that 1 kPa·litre = 1 J.
- The first law gives Q = ΔU+Wby = 300+200 = 500 J into the gas.
- On C→D, Wby = 300000(−0.002) = −600 J; vertical legs do no boundary work. Net work is 200−600 = −400 J, so 400 J is done on the gas.
- Internal energy is a state function, so returning to A gives ΔUcycle = 0.
- The first law then gives Qnet = Wby,net = −400 J: net energy leaves as heat while net work is supplied.
- A net-output heat engine would require the opposite orientation, but its efficiency cannot be found from the enclosed area alone; heat input on the relevant legs must also be known.
Answer (a) A→B: Wby = +200 J and Q = +500 J; net cycle Wby = −400 J. (b) Net Q = −400 J; this direction consumes net work rather than delivering heat-engine work.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
An engine's advertised efficiency exceeds its thermal limit
A cyclic engine absorbs 1200 J per cycle from a 600 K reservoir and rejects heat to a 300 K reservoir. The manufacturer claims 800 J of work per cycle. The supplied graph shows the maximum reversible efficiency 1−Tc/Th as cold-reservoir temperature varies for Th = 600 K. Supplied model: For heat Q received reversibly by a reservoir at absolute temperature T, its entropy change is ΔS = Q/T, with Q negative for energy leaving that reservoir. Supplied data (Cold reservoir / K; Maximum efficiency): (150, 0.75), (300, 0.5), (450, 0.25), (600, 0). (a) Check the claim using both energy conservation and the second law. Find maximum permitted work and minimum rejected heat per cycle. [3 marks] (b) A real engine instead delivers 450 J per cycle. Find its efficiency and total reservoir entropy change; explain why waste heat cannot all be recycled into work in the same cycle. [3 marks]
Reversible efficiency limit
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Maximum efficiency
Horizontal axis (x): Cold reservoir / K
Read the exact data as a table
| Cold reservoir / K | Maximum efficiency |
|---|---|
| 150 | 0.75 |
| 300 | 0.5 |
| 450 | 0.25 |
| 600 | 0 |
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): Check the claim using both energy conservation and the second law. Find maximum permitted work and minimum rejected heat per cycle.
Reveal the worked solution
- Energy conservation alone allows Qout = 1200−800 = 400 J, so a first-law calculation does not expose the problem.
- The reversible limit is 1−300/600 = 0.50. Maximum work is 0.50(1200) = 600 J, and at least 600 J must be rejected.
- At the claimed output the reservoir entropy change would be −1200/600+400/300 = −2/3 J K⁻¹ per cycle, violating the nondecrease of total entropy.
- The actual rejected heat is 1200−450 = 750 J. Efficiency is 450/1200 = 0.375.
- Reservoir entropy change is −1200/600+750/300 = −2+2.5 = +0.50 J K⁻¹ per cycle, consistent with irreversibility.
- The working substance returns to its initial state each cycle. Converting all heat from a single reservoir into cyclic work would violate the second law; reducing losses cannot remove the need for heat rejection at these temperatures.
Answer (a) Claimed efficiency = 66.7% exceeds the 50% limit; Wmax = 600 J and Qrejected,min = 600 J. (b) Efficiency = 37.5%; reservoir entropy increase = 0.50 J K⁻¹ per cycle.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.