S4 Physics Practice Paper: Heat Capacity, Specific Heat Capacity And Changes Of State (12 Questions, 100 Marks, 75 Minutes)

S4 Physics Practice Paper: Heat Capacity, Specific Heat Capacity and Changes of State (12 questions, 100 marks, 75 minutes)

2026-11-09

HKDSE Physics: Heat Capacity, Specific Heat Capacity and Changes of State

12 questions, 100 marks, 75 minutes on specific heat capacity, heat capacity, latent heat of fusion and vaporisation, heating curves, evaporation and boiling, and insulation at home.

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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 30 minutes on Section A, then work through latent heat and heating curves in Section B, and finish with heat transfer in Section C.

Section A　Specific Heat Capacity and Heat Capacity

Section A: Show the formula and working for each calculation, with units.

#	Question	Answer and marking notes
1	Data 1: the specific heat capacity of water is 4200 J kg⁻¹ K⁻¹. (a) Define specific heat capacity and state what J, kg and K stand for in the unit. (b) Explain why car radiators use water as a coolant rather than the same mass of metal. (c) Calculate the energy needed to heat 2 kg of water from 20 °C to 60 °C.	(a) Specific heat capacity is the energy absorbed by 1 kg of a substance when its temperature rises by 1 K (3 marks); in the unit J kg⁻¹ K⁻¹, J stands for energy (joules), kg for mass and K for temperature rise (1 mark). (b) Water has a large specific heat capacity, so it absorbs a lot of heat with only a small temperature rise and can carry heat away from the engine without becoming very hot itself (2 marks). (c) Q = mcΔT = 2 × 4200 × 40 = 336000 J (3 marks).
2	(a) Distinguish between heat capacity and specific heat capacity, and state what each symbol means in C = mc. (b) A 0.5 kg metal block at 100 °C is placed in 0.4 kg of water at 20 °C, and the final temperature is 30 °C. Assuming no heat loss, find the specific heat capacity of the metal. (c) State one assumption made in the calculation.	(a) Heat capacity is the energy needed to raise the temperature of a whole object by 1 K (unit J K⁻¹); specific heat capacity is the energy needed to raise the temperature of unit mass by 1 K (unit J kg⁻¹ K⁻¹); in C = mc, C is the heat capacity, m the mass and c the specific heat capacity (3 marks). (b) Heat gained by water = 0.4 × 4200 × (30 − 20) = 16800 J; heat lost by the metal = 0.5 × c × (100 − 30) = 35c; setting 35c = 16800 gives c = 480 J kg⁻¹ K⁻¹ (4 marks). (c) No heat is lost to the surroundings and the container absorbs no heat (2 marks).
3	(a) 0.5 kg of water at 80 °C is mixed with 0.5 kg of water at 20 °C. Find the final temperature. (b) The temperature measured by the thermometer is slightly lower than the calculated value; explain why. (c) If 0.3 kg of water at 20 °C is mixed with 0.2 kg of water at 90 °C instead, find the final temperature.	(a) Let the final temperature be T: 0.5 × 4200 × (80 − T) = 0.5 × 4200 × (T − 20), giving T = 50 °C (3 marks). (b) Some heat is lost to the air and to the container, so less energy is actually absorbed by the water and the measured temperature is lower (2 marks). (c) 0.3 × 4200 × (T − 20) = 0.2 × 4200 × (90 − T), i.e. 0.3T − 6 = 18 − 0.2T, giving T = 48 °C (3 marks).
4	(a) Define the specific latent heat of fusion of ice. (b) The specific latent heat of fusion of ice is 3.34 × 10⁵ J kg⁻¹. Calculate the energy needed to melt 0.2 kg of ice at 0 °C completely. (c) Explain why ice at 0 °C keeps absorbing energy while melting but its temperature does not rise.	(a) The specific latent heat of fusion is the energy needed to change 1 kg of a solid completely into liquid at its melting point without changing temperature (3 marks). (b) Q = mL = 0.2 × 3.34 × 10⁵ = 66800 J (3 marks). (c) The energy absorbed is used to break the bonds between particles and increase their potential energy; the average kinetic energy stays the same, so the temperature stays at 0 °C (3 marks).

Section B　Changes of State and Latent Heat

Section B: Use Q = mcΔT, Q = mL and Q = Pt, and keep three significant figures.

#	Question	Answer and marking notes
5	Data 2: a 0.5 kg block of ice at −10 °C is heated uniformly by a 500 W heater. The specific heat capacity of ice is 2100 J kg⁻¹ K⁻¹ and its specific latent heat of fusion is 3.34 × 10⁵ J kg⁻¹. (a) Calculate the time needed to raise the ice from −10 °C to 0 °C. (b) Calculate the time needed to melt the whole block. (c) Describe how the thermometer reading changes during melting.	(a) Q = mcΔT = 0.5 × 2100 × 10 = 10500 J; t = Q ÷ P = 10500 ÷ 500 = 21 s (3 marks). (b) Q = mL = 0.5 × 3.34 × 10⁵ = 167000 J; t = 167000 ÷ 500 = 334 s (3 marks). (c) During melting the thermometer reading stays at 0 °C and only rises after all the ice has melted (2 marks).
6	(a) Distinguish between boiling and evaporation, giving one difference. (b) The specific latent heat of vaporisation of water is 2.26 × 10⁶ J kg⁻¹. Calculate the energy needed to vaporise 0.05 kg of water at 100 °C completely. (c) Explain why steam at 100 °C causes more serious scalds than boiling water at the same temperature.	(a) Boiling occurs at a fixed temperature (the boiling point) with vaporisation throughout the liquid; evaporation occurs at any temperature and only at the surface of the liquid (3 marks). (b) Q = mL = 0.05 × 2.26 × 10⁶ = 113000 J (3 marks). (c) Steam condenses on the skin and releases its specific latent heat of vaporisation, and further energy is released as it cools from 100 °C to body temperature, so the scald is far worse than with boiling water (2 marks).
7	(a) State three factors that affect the rate of evaporation. (b) Explain why sweating helps to cool the body. (c) Explain why a pressure cooker cooks food faster than an ordinary pot.	(a) The temperature of the liquid, the surface area of the liquid, the speed of air movement (wind) and the humidity of the air, any three (3 marks). (b) As sweat evaporates it absorbs the specific latent heat of vaporisation from the skin, removing heat from the body and cooling it (3 marks). (c) The pressure inside a pressure cooker is higher, so the boiling point of water is raised; food is cooked at a higher temperature and the chemical reactions proceed faster (2 marks).
8	Data 3: a kettle rated at 2000 W heats 1.5 kg of water from 20 °C to boiling point (100 °C). (a) Calculate the energy absorbed by the water. (b) Assuming no heat loss, calculate the heating time needed. (c) In practice the time needed is longer; give two reasons.	(a) Q = mcΔT = 1.5 × 4200 × 80 = 504000 J (3 marks). (b) t = Q ÷ P = 504000 ÷ 2000 = 252 s, about 4.2 minutes (3 marks). (c) Heat is lost to the surrounding air, the body of the kettle and the heating element absorb heat, and not all the electrical energy becomes internal energy of the water (any two, 3 marks).

Section C　Heat Transfer and Applications

Section C: Apply the concepts of conduction, convection and radiation, and answer in full sentences.

#	Question	Answer and marking notes
9	With reference to Data 3, the kettle is used continuously for 1.5 hours and electricity costs $1.2 per kWh. (a) Calculate the electrical energy consumed in kWh. (b) Calculate the cost of the electricity. (c) Suggest two ways to reduce the electricity used for boiling water.	(a) Energy = power × time = 2 kW × 1.5 h = 3 kWh (3 marks). (b) Cost = 3 × 1.2 = $3.6 (2 marks). (c) Boil only the amount of water needed, use a vacuum flask to avoid re-boiling, and descale the kettle regularly to keep the heating efficiency high (any two, 3 marks).
10	(a) Explain how the sealed air gap in a double-glazed window reduces heat loss. (b) Explain why a dark-coloured roof makes a house hotter in summer. (c) State two building measures that can reduce heat loss from a home.	(a) The sealed air is a poor conductor, so conduction is reduced; convection is suppressed in the narrow air layer; the glass and any low-emissivity coating reduce radiative transfer (3 marks). (b) Dark surfaces absorb solar radiation better; the energy absorbed is then transferred indoors by infrared radiation and conduction, making the house hotter (2 marks). (c) Add insulation to the roof and walls, use double-glazed windows, seal gaps around doors and windows, or use light-coloured external walls (any two, 3 marks).
11	(a) Describe the steps of an experiment using the electrical method (heater coil) to measure the specific heat capacity of a metal block. (b) Explain why the metal block should be insulated and lagged in the experiment. (c) State two sources of experimental error and say whether each makes the measured value too large or too small.	(a) Wrap the metal block in insulation, insert a thermometer and a heater coil, and connect them to a power supply and timer; measure the current, voltage, heating time and temperature rise; then find the specific heat capacity from Q = VIt and c = Q ÷ (mΔT) (3 marks). (b) To reduce heat loss to the surroundings so that the measured temperature rise is close to the theoretical value; otherwise the measured specific heat capacity is too large (2 marks). (c) Heat loss to the surroundings (makes c too large), uneven temperature inside the block (makes ΔT too small and c too large), and errors in measuring mass or time (3 marks).
12	Data 4: a 300 W heater warms a 1 kg aluminium block for 5 minutes. The specific heat capacity of aluminium is 900 J kg⁻¹ K⁻¹. (a) Calculate the temperature rise of the block. (b) The measured temperature rise is smaller than the calculated value; suggest one explanation. (c) Give one everyday application of the concept of specific heat capacity and explain it.	(a) Q = Pt = 300 × 300 = 90000 J; ΔT = Q ÷ (mc) = 90000 ÷ (1 × 900) = 100 °C (3 marks). (b) Some heat is lost to the air or the block is not perfectly insulated, so less energy is used to raise its temperature and the measured rise is smaller (2 marks). (c) Water is used as a coolant because of its large specific heat capacity; coastal areas have smaller temperature changes than inland areas because the sea has a large specific heat capacity (any one, explained, 3 marks).

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