IB DP Physics: Molecular model, temperature, internal energy and heating calculations (DP Physics) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover the first half of topic B.1 (Thermal energy transfers) in the IB Physics guide, first assessment 2025, for Standard Level and Higher Level; none of B.1 is marked additional higher level. Constants are supplied in the questions where needed. Start with the study guide first if a topic is new, and see the revision notes for formulas. Course page: IB DP Physics. Checklist: IB DP Physics checklist.

Questions

1. Compare the arrangement and motion of the particles in a solid and in a gas. [4]

2. A rock sample of mass 0.84 kg is lowered into a measuring cylinder and displaces 310 cm³ of water. Calculate the density of the rock in kg m⁻³. [3]

3. The temperature of a gas rises from 290 K to 335 K.

(a) State the temperature change in degrees Celsius. [1] (b) Convert 25 °C to kelvin, and 90 K to degrees Celsius. Use 273 for the conversion. [2] (c) A student says that the average kinetic energy of the particles doubles when the temperature rises from 10 °C to 20 °C. Explain why this is incorrect. [2]

4. Argon atoms of mass 6.63 × 10⁻²⁶ kg are in a gas at 540 K. The Boltzmann constant is 1.38 × 10⁻²³ J K⁻¹.

(a) Calculate the average kinetic energy of an atom. [2] (b) Determine the temperature at which the average kinetic energy is half this value. [1] (c) Calculate the speed corresponding to the average kinetic energy at 540 K. [2]

5. Define the internal energy of a block of metal, and explain why a hot block at 80 °C of mass 0.05 kg and a large water bath at 45 °C can be in contact while energy flows from the block to the bath. [4]

6. An electric heater of power 150 W warms a 2.4 kg aluminium block from 18 °C to 46 °C. Take c for aluminium as 900 J kg⁻¹ K⁻¹ and assume all the energy goes into the block. Calculate the time taken. [4]

7. A freezer removes energy from 0.75 kg of water that is already at 0 °C, at a steady rate of 420 W. The specific latent heat of fusion of water is 3.34 × 10⁵ J kg⁻¹. Calculate the time taken for all the water to freeze, in seconds. [4]

8. Distinguish between boiling and evaporation, and explain why evaporation cools the liquid that remains. [4]

9. A 0.50 kg sample of ice is heated by a steady 800 W heater. The table shows the readings.

Time / s	0	26	235	496
Temperature / °C	−20	0	0	100

The temperature is constant between 26 s and 235 s.

(a) Explain, using the molecular model, why the temperature is constant during this interval. [2] (b) Determine the specific latent heat of fusion of ice from the data. [3] (c) Show that the specific heat capacity of the liquid is about 4180 J kg⁻¹ K⁻¹. [2]

10. An aluminium block of mass 0.35 kg is heated to 160 °C and dropped into 0.90 kg of water at 24 °C in a polystyrene cup. Take c as 900 J kg⁻¹ K⁻¹ for aluminium and 4180 J kg⁻¹ K⁻¹ for water. Calculate the final temperature, stating one assumption. [5]

11. Steam of mass 0.040 kg at 100 °C is passed into 0.30 kg of water at 20 °C in an insulated container. The steam condenses and mixes with the water. Use L_v = 2.26 × 10⁶ J kg⁻¹ and c = 4180 J kg⁻¹ K⁻¹.

(a) Show that the water alone would need 1.0 × 10⁵ J to reach 100 °C, and deduce that the final temperature is below 100 °C. [3] (b) Calculate the final temperature. [4] (c) Suggest why steam at 100 °C causes a worse burn than the same mass of water at 100 °C. [2]

Answers

1. In a solid, particles are close together and fixed in position, vibrating about those positions [1]; in a gas they are far apart [1] and move rapidly and randomly in straight lines between collisions [1]. Forces between solid particles are strong and hold them in place, but in a gas they are negligible except during collisions [1]. [4] Examiner insight: Comparison answers earn credit when both states are described against the same feature, such as spacing or motion, rather than describing each state in a separate paragraph.

2. V = 310 cm³ = 3.10 × 10⁻⁴ m³ [1]; ρ = m/V = 0.84 / 3.10 × 10⁻⁴ [1]; ρ = 2.71 × 10³ kg m⁻³ [1]. [3] Examiner insight: The unit conversion is where this kind of question is usually lost, so show the volume in m³ as a separate line.

3. (a) 335 − 290 = 45 °C [1] (b) 25 + 273 = 298 K [1]; 90 − 273 = −183 °C [1] (c) Average kinetic energy depends on the kelvin temperature in direct proportion [1]; 10 °C to 20 °C is 283 K to 293 K, a rise of about 4 percent, not a doubling [1]. Examiner insight: Show the kelvin values in the explanation, because a bare statement that Celsius is the wrong scale is less convincing than the numbers.

4. (a) Ek = 3/2 kB T = 1.5 × 1.38 × 10⁻²³ × 540 [1]; 1.12 × 10⁻²⁰ J [1] (b) Ek is proportional to T, so T = 270 K [1] (c) ½mv² = Ek, so v = √(2 × 1.118 × 10⁻²⁰ / 6.63 × 10⁻²⁶) [1]; v = 581 m s⁻¹ [1] Examiner insight: Keep the unrounded Ek for part (c); round once, at the end.

5. Internal energy is the sum of the intermolecular potential energy arising from forces between the molecules and the random kinetic energy of those molecules [1]. Energy flows because of the temperature difference, from higher to lower temperature [1]; the block is at 80 °C and the bath is at 45 °C so the net flow is from the block to the bath [1], even though the bath has more total internal energy because it is much larger [1]. [4] Examiner insight: A strong answer says that direction depends on temperature, not on the amount of internal energy, and uses the two temperatures given.

6. Q = mcΔT = 2.4 × 900 × 28 [1]; Q = 6.05 × 10⁴ J [1]; t = Q/P = 60 480 / 150 [1]; t = 403 s [1]. [4] Examiner insight: A change of 28 °C equals 28 K, so no conversion is needed, and stating that explicitly shows you understand the scales.

7. Q = mL = 0.75 × 3.34 × 10⁵ [1]; Q = 2.51 × 10⁵ J [1]; t = Q/P = 250 500 / 420 [1]; t = 596 s [1]. [4] Examiner insight: No temperature change is involved, so Q = mcΔT is the wrong tool here, and the state of the water is the clue.

8. Boiling happens throughout the liquid at the boiling point, at a fixed temperature [1]; evaporation happens only at the surface and can occur at any temperature below the boiling point [1]. In evaporation it is the fastest molecules that leave through the surface [1]; the molecules that stay have a lower average kinetic energy, so the liquid cools [1]. [4] Examiner insight: Naming the escape of the highest-energy molecules is what turns a description of cooling into an explanation.

9. (a) The supplied energy raises the potential energy as particles move apart [1], not their average kinetic energy, so the temperature does not change [1]. (b) Energy supplied = Pt = 800 × (235 − 26) = 1.672 × 10⁵ J [1]; L = Q/m = 167 200 / 0.50 [1]; L = 3.34 × 10⁵ J kg⁻¹ [1] (c) Time to heat the liquid = 496 − 235 = 261 s, so Q = 800 × 261 = 2.088 × 10⁵ J [1]; c = Q/(mΔT) = 208 800 / (0.50 × 100) = 4176 J kg⁻¹ K⁻¹, about 4180 [1] Examiner insight: A show-that answer should state the unrounded value before the rounded one and must use the time interval for the liquid only, not the whole run.

10. Assume no energy is lost to the surroundings, and none of the water boils [1]. Energy lost by aluminium = energy gained by water [1]: 0.35 × 900 × (160 − T) = 0.90 × 4180 × (T − 24) [1]; 315(160 − T) = 3762(T − 24), so 50 400 + 90 288 = 4077T [1]; T = 34.5 °C [1]. [5] Examiner insight: A final temperature outside the range of the two starting values points to a sign error, so test your answer against that range before moving on.

11. (a) Q = mcΔT = 0.30 × 4180 × 80 = 1.00 × 10⁵ J [1]; condensing steam releases 0.040 × 2.26 × 10⁶ = 9.04 × 10⁴ J [1], which is less than 1.0 × 10⁵ J, so the final temperature is below 100 °C [1] (b) Energy released = energy gained: 9.04 × 10⁴ + 0.040 × 4180 × (100 − T) = 0.30 × 4180 × (T − 20) [1]; 90 400 + 16 720 − 167.2T = 1254T − 25 080 [1]; 132 200 = 1421.2T [1]; T = 93.0 °C [1] (c) Condensing on the skin gives out the latent heat of vaporization as well [1], on top of what the water gives out when it cools, so more energy reaches the skin for the same mass [1]. Examiner insight: Test the limiting case in part (a) first, because the working in part (b) is only valid if all the steam condenses and the final temperature is below 100 °C.

Where marks are usually lost

• Substituting cm³ or g without first converting to m³ or kg.
• Using Celsius temperatures in Ek = 3/2 kB T.
• Rounding part-way through, so the third significant figure of the answer drifts.
• Applying Q = mcΔT across a phase change, or forgetting the latent heat stage.
• Writing “heat rises” or “temperature rises” in a phase change explanation rather than describing potential energy.
• Describing the direction of transfer in terms of which body has more energy.
• Leaving out the assumption of no energy loss in a mixing question.
• Leaving units off the final answer.

Next steps

• Review the formulas in the revision notes.
• Work back through the examples in the study guide.
• Return to the IB DP Physics course page and tick off the checklist.
• Take the free 10-minute diagnostics to find your next weak area.
• Book a free trial class.

Official syllabus

Source: the Physics guide for the IB Diploma Programme (first assessment 2025), International Baccalaureate Organization. Topic B.1, Thermal energy transfers, first half.