S4 Physics Practice Paper: Temperature, Specific Heat Capacity And Latent Heat (12 Questions, 100 Marks, 75 Minutes, With Answers)

S4 Physics Practice Paper: Temperature, Specific Heat Capacity and Latent Heat (12 Questions, 100 Marks, 75 Minutes, with Answers)

2026-11-09

HKDSE Physics: Temperature, Specific Heat Capacity and Latent Heat

12 questions, 100 marks, 75 minutes on specific heat capacity, latent heat, heating curves, thermal equilibrium and the three modes of heat transfer, with marking points.

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Instructions: This paper has three sections, A, B and C. Answer all questions. Total marks: 100. Time allowed: 75 minutes.

Show all working. Answers without working earn no method marks. Write down a formula before substituting numbers.

Give numerical answers to three significant figures unless stated otherwise, and state the correct units.

Spend about 35 minutes on Section A, then use the remaining time for the specific heat and latent heat calculations in Section B and the heat transfer applications in Section C.

Section A: Temperature, Internal Energy and Specific Heat Capacity

Temperature and internal energy, heat capacity and specific heat capacity

#	Question	Answer and marking notes
1	Explain the difference between temperature and internal energy, and distinguish heat capacity from specific heat capacity.	(a) Temperature measures how hot or cold a body is and reflects the average kinetic energy of its molecules (1 mark). (b) Internal energy is the total kinetic and potential energy of all molecules in the body (1 mark). (c) Heat capacity is the energy needed to raise the temperature of the whole body by 1 K, and depends on its mass and material (2 marks). (d) Specific heat capacity is the energy needed to raise the temperature of 1 kg of a substance by 1 K, and is a property of the material (2 marks).
2	0.5 kg of water is heated from 20 °C to 60 °C. Given that the specific heat capacity of water is 4200 J kg⁻¹ K⁻¹, find the energy absorbed and express it in kJ.	(a) Temperature rise = 60 – 20 = 40 K (2 marks). (b) Q = mcΔT = 0.5 × 4200 × 40 (2 marks). (c) Q = 84000 J (3 marks). (d) That is 84 kJ, about 84000 ÷ 3600 = 23.3 Wh (3 marks).
3	A 2 kg aluminium block absorbs 36000 J of heat. Given that the specific heat capacity of aluminium is 900 J kg⁻¹ K⁻¹, find the rise in temperature.	(a) From Q = mcΔT, ΔT = Q ÷ (mc) (2 marks). (b) ΔT = 36000 ÷ (2 × 900) (3 marks). (c) ΔT = 20 K, so the temperature rises by 20 °C (3 marks).
4	0.2 kg of water at 80 °C is mixed with 0.3 kg of water at 20 °C. Assuming no heat loss, find the final temperature.	(a) Let the final temperature be T; the energy lost by the hot water equals the energy gained by the cold water (2 marks). (b) 0.2 × 4200 × (80 – T) = 0.3 × 4200 × (T – 20) (3 marks). (c) This simplifies to 16 – 0.2T = 0.3T – 6, that is 0.5T = 22 (2 marks). (d) T = 44 °C (3 marks).

Section B: Specific Latent Heat, Heating Curves and Thermal Equilibrium

Latent heat of fusion and vaporisation, mixture and electrical methods

#	Question	Answer and marking notes
5	A 350 W heater is placed in 0.5 kg of water for 60 seconds. Assuming no heat loss, find the rise in water temperature.	(a) Energy supplied by the heater E = Pt = 350 × 60 = 21000 J (3 marks). (b) ΔT = E ÷ (mc) = 21000 ÷ (0.5 × 4200) (3 marks). (c) ΔT = 10 K, so the water temperature rises by 10 °C (2 marks).
6	(a) Find the energy needed to melt 0.25 kg of ice at 0 °C completely into water at 0 °C. (b) Find the energy needed to vaporise 0.05 kg of water at 100 °C completely. The specific latent heat of fusion of ice is 3.34 × 10⁵ J kg⁻¹ and that of vaporisation of water is 2.26 × 10⁶ J kg⁻¹.	(a) Energy to melt = 0.25 × 3.34 × 10⁵ = 83500 J (3 marks). (b) Energy to vaporise = 0.05 × 2.26 × 10⁶ = 113000 J (3 marks). (c) During melting the temperature stays at 0 °C and the energy breaks down the crystal structure of ice (2 marks). (d) Vaporisation needs much more energy because the attractive forces between liquid molecules must also be overcome (2 marks).
7	Compare the energy needed to melt 0.1 kg of ice with the energy needed to raise the temperature of the same mass of water by 80 K. Given that the specific heat capacity of water is 4200 J kg⁻¹ K⁻¹, calculate both and explain the result.	(a) Energy to melt 0.1 kg of ice = 0.1 × 3.34 × 10⁵ = 33400 J (3 marks). (b) Energy to raise the same mass of water by 80 K = 0.1 × 4200 × 80 = 33600 J (3 marks). (c) The two values are close, showing that melting 0.1 kg of ice takes about as much energy as heating the same mass of water by 80 °C (2 marks).
8	State the main mode of heat transfer in each case: (a) one end of a metal rod is heated and the other end becomes hot; (b) water at the bottom of a beaker is heated and the whole body of water becomes hot; (c) the Earth receives heat from the Sun.	(a) The metal rod transfers energy from the hot end to the cold end by molecular vibration and free electrons, which is conduction (2 marks). (b) Water at the bottom rises and cooler water sinks, forming a convection current, which is convection (2 marks). (c) Heat from the Sun crosses empty space to reach the Earth, which is radiation (2 marks). (d) Radiation needs no medium and travels as infrared radiation (2 marks).

Section C: Heat Transfer and Insulation

Conduction, convection, radiation and the vacuum flask

#	Question	Answer and marking notes
9	An electrical heater is used to find the specific heat capacity of a metal. The heater supplies 3600 J and raises the temperature of 0.2 kg of the metal by 20 K. Find the specific heat capacity and state whether the actual value would be larger or smaller, with a reason.	(a) Neglecting heat loss, Q = 3600 J (2 marks). (b) c = Q ÷ (mΔT) = 3600 ÷ (0.2 × 20) (3 marks). (c) c = 900 J kg⁻¹ K⁻¹, so the metal is probably aluminium (3 marks). (d) The measured value would be larger, because some energy is lost to the surroundings and the measured temperature rise is smaller (2 marks).
10	Explain why the day-night temperature difference is smaller in coastal areas than in inland areas.	(a) The specific heat capacity of water, about 4200 J kg⁻¹ K⁻¹, is much larger than that of land, about 800 to 1000 (2 marks). (b) In the daytime the sea warms up only slightly, and the sea breeze brings cooler air to the land (2 marks). (c) At night the sea cools down slowly and releases heat to the land (2 marks). (d) Hence the day-night temperature difference is smaller in coastal areas than inland (2 marks).
11	Explain how the vacuum between the double glass walls, the silvered inner surface and the cork stopper of a vacuum flask each reduce heat loss.	(a) The vacuum between the double glass walls reduces conduction and convection (3 marks). (b) The silvered inner surface reflects radiation back into the flask, reducing radiation transfer (3 marks). (c) The cork stopper seals the mouth and reduces conduction and convection (2 marks).
12	State two differences between evaporation and boiling, and explain why evaporation produces a cooling effect.	(a) Evaporation occurs only at the liquid surface and at any temperature (2 marks). (b) Boiling occurs both inside the liquid and at its surface, and only at the boiling point (2 marks). (c) In evaporation the molecules that escape are those with the highest energy, so the average kinetic energy of the remaining liquid falls, producing a cooling effect (2 marks).

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