S4 Physics Paper: Thermal Physics - Temperature, Heat Capacity, Latent Heat And Heat Transfer (100 Marks)

S4 Physics Paper: Thermal Physics – Temperature, Heat Capacity, Latent Heat and Heat Transfer (100 marks)

2026-11-11

S4 Physics – thermal physics and heat transfer

S4 physics thermal practice paper: temperature and internal energy, specific heat capacity, latent heat, conduction, convection and radiation, with full answers and mark.

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Paper details　S4 physics thermal physics practice paper: temperature and internal energy, specific heat capacity, latent heat of fusion, conduction, convection, radiation and the method of mixtures.

Section A – Temperature, internal energy and thermometers

Answer in full sentences; show the formula, substitution and unit in calculations.

#	Question	Answer and marking notes
1	Distinguish between temperature and internal energy, and state what decides the direction of heat flow.	Temperature measures the average kinetic energy of the molecules and indicates hotness, in K or °C. Internal energy is the total kinetic plus potential energy of all the molecules, in J, and depends on temperature, mass and state. Heat flows from the hotter body to the cooler one until both reach the same temperature. (4 marks: temperature 1; internal energy 1; factors 1; direction 1)
2	Convert 20 °C to K, and 300 K to °C.	T = 20 + 273 = 293 K. t = 300 – 273 = 27 °C. (4 marks: 2 each, units required)
3	What property does a liquid-in-glass thermometer use, and why is the bore made very narrow?	It uses the expansion of a liquid (mercury or alcohol) whose volume changes with temperature; in a narrow bore a small volume change gives a long column, so finer scale divisions and a more sensitive reading are possible. (4 marks: expansion 2; narrow bore improves sensitivity 2)
4	Why is an alcohol thermometer used near dry ice at -10 °C rather than a mercury thermometer?	Mercury freezes at about -39 °C while alcohol freezes near -115 °C, so at 263 K (-10 °C) alcohol is still liquid and expands normally; alcohol is also non-toxic if the tube breaks. (4 marks: freezing points 2; still liquid at 263 K 1; safety 1)

Section B – Specific heat capacity and heating by a heater

Take the specific heat capacity of water as 4200 J/(kg K) and ignore heat loss unless stated otherwise.

#	Question	Answer and marking notes
5	Define specific heat capacity and write the formula for heat energy.	Specific heat capacity is the energy needed to raise the temperature of 1 kg of a substance by 1 K (or 1 °C), in J/(kg K). Formula: E = m c dT where m is mass, c specific heat capacity and dT the temperature change. (4 marks: definition 2; unit 1; formula 1)
6	How much energy is needed to heat 0.5 kg of water from 20 °C to 60 °C?	dT = 60 – 20 = 40 K. E = m c dT = 0.5 x 4200 x 40 = 84000 J (about 84.0 kJ). (4 marks: dT 1; substitution 1; answer 1; unit 1)
7	If a 350 W heater supplies all its energy to the water, how long does it take?	E = P t, so t = E / P = 84000 / 350 = 240 s (4.0 minutes). (5 marks: formula 1; substitution 2; answer 1; unit 1)
8	In practice only 80% of the heater energy reaches the water. Find the actual time and suggest one way to reduce the loss.	Useful power = 350 x 0.80 = 280 W; t = 84000 / 280 = 300 s. Wrap the beaker with insulation, fit a lid, and keep the heater close to the base to cut conduction, convection and radiation losses. (5 marks: useful power 1; time 2; method 2)

Section C – Melting, latent heat and change of state

Take the latent heat of fusion of ice as 334000 J/kg and the specific heat capacity of water as 4200 J/(kg K).

#	Question	Answer and marking notes
9	Explain latent heat of fusion and why the temperature stays constant while melting.	Latent heat of fusion is the energy needed to change 1 kg of a solid completely into liquid at its melting point; for ice it is 334000 J/kg. The energy absorbed breaks the ordered arrangement of particles (raising potential energy) rather than raising average kinetic energy, so the temperature stays constant. (4 marks: definition 2; energy raises potential energy 1; constant temperature 1)
10	How much energy is needed to melt 0.2 kg of ice at 0 °C completely into water at 0 °C?	E = m L = 0.2 x 334000 = 66800 J (about 66.8 kJ). (5 marks: formula 1; substitution 2; answer 1; unit 1)
11	The melted water is then heated from 0 °C to 25 °C. Find the energy needed and compare it with the melting energy.	E = m c dT = 0.2 x 4200 x 25 = 21000 J. The melting energy of 66800 J is about 3.2 times this value, so a change of state needs far more energy than warming the same mass. (5 marks: calculation 3; comparison 1; conclusion 1)
12	Find the total energy absorbed in changing 0 °C ice into water at 25 °C, and state where the energy comes from.	Total = 66800 + 21000 = 87800 J. The energy comes from the warmer surroundings and container by conduction, convection and radiation, which is why ice melts by itself at room temperature. (4 marks: total 2; transfer modes 1; conclusion 1)

Section D – Conduction, convection and radiation

Name the heat transfer process involved and explain the reason; a name alone gains no marks.

#	Question	Answer and marking notes
13	Describe conduction, convection and radiation, giving one example of each.	Conduction: particles or free electrons pass energy to neighbours, best in metals, e.g. a metal spoon. Convection: a fluid expands on heating, becomes less dense and rises while cooler fluid sinks, e.g. water circulating in a heated pot. Radiation: energy carried by electromagnetic waves without a medium, e.g. sunlight warming the ground. (4 marks: three mechanisms 3; example 1)
14	In the same cup of hot water a metal spoon feels much hotter than a wooden one. Explain in terms of particles.	Metals contain many free electrons that carry energy quickly along the spoon, so their thermal conductivity is high; wood has no free electrons and relies on particle vibration, so its conductivity is low. The metal spoon therefore delivers more heat to the hand in a short time. (4 marks: free electrons 2; wood poor conductor 1; comparison 1)
15	In a heated room should the heater be placed high or low? What about an air conditioner? Explain.	A heater should be low: warm air is less dense and rises while cool air sinks to replace it, giving a convection current that heats the whole room. An air conditioner should be high: the cold air it blows out is denser and sinks while warm air rises to be cooled, so the room cools evenly. (4 marks: heater low 2; air conditioner high 2, with convection reasoning)
16	A vacuum flask uses a vacuum between double walls, silvered surfaces and a cork stopper. State which heat transfer process each part reduces.	The vacuum layer has no matter, so it stops conduction and convection. The silvered walls reflect radiation, cutting radiative transfer. The cork is a poor conductor and seals the neck, reducing conduction and convection losses. (4 marks: vacuum 2; silvering 1; cork 1)

Section E – Experiment: specific heat capacity by the method of mixtures

Take the specific heat capacity of water as 4200 J/(kg K) and ignore heat loss.

#	Question	Answer and marking notes
17	0.2 kg of metal is heated to 100 °C and quickly put into 0.1 kg of water at 25 °C; the final common temperature is 30 °C. State the energy principle used.	If heat loss is neglected, the energy lost by the metal equals the energy gained by the water: m_m c_m dT_m = m_w c_w dT_w. This follows from conservation of energy and allows the unknown specific heat capacity to be found. (5 marks: conservation 1; loss equals gain 2; formula 2)
18	Calculate the specific heat capacity of the metal.	Energy gained by water: E = 0.1 x 4200 x (30 – 25) = 3150 J. Metal temperature drop dT = 100 – 30 = 70 K. c = E / (m dT) = 3150 / (0.2 x 70) = 225 J/(kg K). (6 marks: water energy 2; metal dT 1; formula 1; answer 1; unit 1)
19	State two sources of error that make the result too low, with one improvement for each.	The metal loses heat to the air while being transferred, so the energy released is overestimated as smaller and c comes out low; move it quickly with tongs and pre-read the water temperature. The calorimeter loses heat and water evaporates, so the water gains more than calculated and c is again low; use a polystyrene cup with a lid. Incomplete mixing and thermometer lag add error; stir continuously and read once the temperature is steady. (5 marks: 2 per error, 1 per improvement, max 5)

Section F – Integrated application: efficiency of an electric kettle

Take the specific heat capacity of water as 4200 J/(kg K) and give energy in J and kJ.

#	Question	Answer and marking notes
20	An electric kettle heats 1.5 kg of water from 20 °C to boiling at 100 °C. Find the energy needed.	dT = 100 – 20 = 80 K. E = m c dT = 1.5 x 4200 x 80 = 504000 J (504.0 kJ). (6 marks: dT 1; substitution 2; answer 2; unit 1)
21	The kettle is rated 2000 W with an efficiency of 90%; find the heating time.	Useful power = 2000 x 0.90 = 1800 W. t = E / useful power = 504000 / 1800 = 280 s (about 4.7 minutes). (5 marks: useful power 1; formula 1; calculation 2; answer and unit 1)
22	Why is the actual heating time longer than the calculated value? Suggest one way to save energy.	The body and lid lose heat to the air, the kettle itself absorbs energy, and evaporating water carries away latent heat, so only part of the electrical energy stays in the water and the real time is longer. To save energy, boil only the amount needed, store the water in a vacuum flask instead of reboiling, and descale the element to keep heat transfer efficient. (5 marks: explanation 3; suggestion 2)

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