These questions match real Cambridge IGCSE difficulty. Scoring 50%+ is a solid B, 60%+ is an A, 70%+ is A*. Don't worry if this feels harder -- that's the point!
Instructions
Answer all questions in the spaces provided.
Show all working for calculations. Answers without working may not receive full marks.
You may use a calculator.
Take g = 10 N/kg where needed.
Answer all questions first. When finished, click "Submit Exam" at the bottom of Question 7.
Your answers will be automatically graded when you submit the exam. The model answers will be shown for review.
Question Navigation
Question 1 -- Filling the Dive Cylinder
Total: 12 marks
A dive shop in Plymouth fills scuba cylinders with compressed air. A cylinder of internal volume 12 litres is filled until the air inside is at 200 times atmospheric pressure. The temperature of the air is the same at the start and the end of the comparison in part (b).
(a)[3]
Explain, in terms of particles, how the air inside the cylinder exerts a pressure on its walls.
Model Answer -- 1(a)
The air molecules are in constant, random motion [1]
They collide with the cylinder walls and rebound; each collision exerts a (tiny) force on the wall (change of momentum) [1]
The enormous number of collisions per second over the wall area produces a steady force per unit area — the pressure [1]
⚠ If you missed marks here: Pressure is BOMBARDMENT: molecules moving randomly, colliding with the walls, each impact pushing. All three steps — motion, collisions exerting force, many collisions giving force per unit area — are separately credited. "The air pushes out" restates the question.
Mark 1 -- molecules in constant random motion (1 mark)
Mark 2 -- wall collisions each exert force (1 mark)
Mark 3 -- many collisions per area give steady pressure (1 mark)
(b)[3]
Calculate the volume this air would occupy at atmospheric pressure (at the same temperature).
Model Answer -- 1(b)
p₁V₁ = p₂V₂ (constant temperature) [1]
200 × 12 = 1 × V₂ [1]
V₂ = 2400 litres [1]
⚠ If you missed marks here: Boyle’s law: pV is constant at fixed temperature, so 200 atm × 12 L = 1 atm × V, giving 2400 litres — a small room’s worth of air squeezed into a cylinder. Pressure UP means volume DOWN; if your answer was 0.06 L, the ratio was inverted.
Mark 1 -- p1 V1 equals p2 V2 stated (1 mark)
Mark 2 -- substitution 200 times 12 (1 mark)
Mark 3 -- volume 2400 litres calculated (1 mark)
(c)[3]
While being filled quickly, the cylinder becomes noticeably hot. Explain why, in terms of the air molecules.
Model Answer -- 1(c)
Work is done ON the air as it is compressed (by the pump) [1]
This energy increases the kinetic energy of the air molecules [1]
Average molecular kinetic energy determines temperature, so the air (and the cylinder it touches) gets hotter [1]
⚠ If you missed marks here: Compression does WORK on the gas, and that work becomes molecular kinetic energy — higher average KE literally IS higher temperature. "Friction between molecules" is the standard wrong answer: ideal-gas molecules do not rub.
Mark 1 -- work done on air during compression (1 mark)
Mark 2 -- molecular kinetic energy increases (1 mark)
Mark 3 -- higher average KE means higher temperature (1 mark)
(d)[3]
A filled cylinder left in strong sunshine warms from 20 °C to 45 °C. Its volume does not change. Explain, in terms of particles, why the pressure inside rises.
Model Answer -- 1(d)
At higher temperature the molecules move faster (greater average kinetic energy) [1]
They hit the walls harder (bigger momentum change per collision) [1]
...and more often (more collisions per second) — both effects raise the force on the walls, so the pressure rises at constant volume [1]
⚠ If you missed marks here: Two distinct effects, both needed: each impact is HARDER (faster molecules) and impacts are MORE FREQUENT. This is why cylinders carry sun-exposure warnings — the gas cannot expand, so all the extra molecular energy appears as pressure.
Mark 1 -- faster molecules at higher temperature (1 mark)
Mark 2 -- collisions harder with greater force (1 mark)
Mark 3 -- collisions more frequent raising pressure (1 mark)
Question 2 -- The Storage Heater
Total: 11 marks
A flat in Aberdeen uses night storage heaters. Each heater contains 120 kg of ceramic bricks (specific heat capacity 840 J/(kg °C)). Overnight, cheap-rate electricity warms the bricks from 20 °C to 65 °C; during the day the bricks release the stored energy slowly into the room.
(a)[2]
Define specific heat capacity.
Model Answer -- 2(a)
The energy required per kilogram [1]
...to raise the temperature (of the substance) by 1 °C (per unit temperature rise) [1]
⚠ If you missed marks here: Both "per kilogram" and "per degree" must appear — specific heat capacity is energy per unit mass per unit temperature rise. Leaving out either ratio turns a definition into a vague description.
Mark 1 -- energy per kilogram stated (1 mark)
Mark 2 -- per one degree temperature rise stated (1 mark)
(b)[3]
Calculate the energy stored in the bricks of one heater when they warm from 20 °C to 65 °C.
Model Answer -- 2(b)
E = m c ΔT [1]
E = 120 × 840 × 45 [1]
E = 4 536 000 J ≈ 4.5 MJ [1]
⚠ If you missed marks here: ΔT is the CHANGE: 65 − 20 = 45 °C, not 65. Then 120 × 840 × 45 = 4.5 MJ. Using 65 gives 6.6 MJ — the single most common m c ΔT error at every level.
Mark 1 -- E equals m c delta T stated (1 mark)
Mark 2 -- substitution with 45 degree change (1 mark)
Mark 3 -- energy about 4.5 MJ calculated (1 mark)
(c)[3]
The heating element is rated at 2.0 kW. Calculate the time needed to store this energy, in seconds and in minutes.
Model Answer -- 2(c)
t = E / P = 4 536 000 / 2000 [1]
t = 2268 s [1]
≈ 38 minutes [1]
⚠ If you missed marks here: Time = energy ÷ power with power in WATTS: 4 536 000 J ÷ 2000 W = 2268 s ≈ 38 min. Dividing by 2.0 (kW left unconverted) gives 2.3 million seconds — nearly a month, which the sanity check catches instantly.
Mark 1 -- time equals energy over power used (1 mark)
Mark 2 -- time about 2268 seconds (1 mark)
Mark 3 -- converted to about 38 minutes (1 mark)
(d)[3]
Explain why ceramic bricks are a good choice for storing thermal energy in this way, and why the heater is charged at night.
Model Answer -- 2(d)
The bricks’ large mass and fairly high specific heat capacity mean they store a large amount of energy for a modest temperature rise [1]
They release the energy SLOWLY (over many hours), warming the room through the day [1]
Night-rate electricity is cheaper (demand is low overnight), so storing then and releasing later saves money [1]
⚠ If you missed marks here: Three separate ideas: capacity (large m × c stores a lot), release rate (slow — a brick at 65 °C cannot dump its energy quickly, which is exactly what you want), and economics (cheap off-peak charging). This is thermal storage playing the same game as Dinorwig plays with water.
Mark 1 -- large mass and heat capacity store much energy (1 mark)
Mark 2 -- slow release warms room through day (1 mark)
Mark 3 -- night electricity cheaper so charging then saves money (1 mark)
Question 3 -- De-icing at the Airport
Total: 12 marks
On a freezing morning at Manchester Airport, ground crew must melt 15 kg of ice (already at 0 °C) from an aircraft. The specific latent heat of fusion of ice is 3.3 × 10⁵ J/kg. One de-icing rig delivers thermal energy to the ice at a rate of 25 kW.
(a)[2]
Define specific latent heat of fusion.
Model Answer -- 3(a)
The energy required per kilogram [1]
...to change a substance from solid to liquid WITHOUT any change in temperature [1]
⚠ If you missed marks here: Two essentials: per kilogram, and melting at CONSTANT temperature. Latent means hidden — the energy changes the state, not the thermometer reading, and the definition must say so.
Mark 1 -- energy per kilogram stated (1 mark)
Mark 2 -- solid to liquid with no temperature change (1 mark)
(b)[3]
Calculate the energy needed to melt all 15 kg of ice.
Model Answer -- 3(b)
E = m L [1]
E = 15 × 3.3 × 10⁵ [1]
E = 4.95 × 10⁶ J ≈ 5.0 MJ [1]
⚠ If you missed marks here: E = mL = 15 × 3.3 × 10⁵ = 4.95 × 10⁶ J. No ΔT appears anywhere — the ice is already at its melting point, so every joule goes into the change of state, none into warming.
Mark 1 -- E equals m L stated (1 mark)
Mark 2 -- substitution 15 times 3.3e5 (1 mark)
Mark 3 -- energy about 5.0 MJ calculated (1 mark)
(c)[2]
Calculate the minimum time for the rig to melt all the ice.
Model Answer -- 3(c)
t = E / P = 4.95 × 10⁶ / 25 000 [1]
t = 198 s (≈ 3.3 minutes) [1]
⚠ If you missed marks here: 4.95 × 10⁶ J ÷ 25 000 W = 198 s. "Minimum" because the calculation assumes every joule reaches the ice — in the wind on a real apron, plenty is lost to the air, so the real job takes longer.
Mark 1 -- time energy over power substituted (1 mark)
Mark 2 -- minimum time 198 seconds calculated (1 mark)
(d)[3]
A thermometer in the melting ice reads 0 °C throughout, even while energy pours in. Explain, in terms of molecules, where the energy is going.
Model Answer -- 3(d)
During melting the energy is used to break/loosen the bonds between molecules / pull molecules apart from their fixed lattice positions [1]
So the POTENTIAL energy of the molecules increases... [1]
...but their average KINETIC energy (which determines temperature) does not change — hence a constant 0 °C until all the ice has melted [1]
⚠ If you missed marks here: Temperature tracks average KINETIC energy only. Melting spends the incoming energy on separating molecules (raising POTENTIAL energy) — the molecules do not speed up until the last crystal is gone. That is what "latent" is hiding.
Mark 1 -- energy breaks intermolecular bonds (1 mark)
Mark 2 -- molecular potential energy increases (1 mark)
Mark 3 -- kinetic energy and temperature unchanged during melting (1 mark)
(e)[2]
State two differences between boiling and evaporation.
Model Answer -- 3(e)
Boiling happens at one fixed temperature (the boiling point); evaporation happens at ANY temperature (below it) [1]
Boiling occurs throughout the liquid (bubbles form in the body of the liquid); evaporation occurs only at the SURFACE [1]
⚠ If you missed marks here: Two clean contrasts: WHERE (throughout the bulk with bubbles vs surface only) and WHEN (fixed boiling point vs any temperature). A puddle drying on a cold day is the everyday proof that evaporation needs no boiling point.
Mark 1 -- boiling at fixed temperature evaporation at any (1 mark)
Mark 2 -- boiling throughout liquid evaporation at surface (1 mark)
Question 4 -- Insulating the Terrace
Total: 11 marks
A Victorian terraced house in Leeds is being retrofitted: loft insulation (fibreglass wool), reflective foil panels behind the radiators, and double glazing. Each measure targets a different mechanism of thermal energy transfer.
(a)[3]
Explain how thermal energy is conducted through a solid, and why metals conduct much better than brick or glass.
Model Answer -- 4(a)
Particles in the hot region vibrate with greater amplitude/energy... [1]
...and pass energy to neighbouring particles through the bonds between them (lattice vibration) [1]
Metals ALSO have free electrons which move through the metal, transferring energy quickly — brick and glass have no free electrons, so they conduct poorly [1]
⚠ If you missed marks here: Non-metals conduct only by vibrations passed particle-to-particle — slow. Metals add the express service: free electrons that carry energy through the whole lattice. Naming the free electrons is what separates the third mark from the first two.
Mark 1 -- hot region particles vibrate more (1 mark)
Mark 2 -- energy passed to neighbouring particles (1 mark)
Mark 3 -- free electrons make metals fast conductors (1 mark)
(b)[3]
Explain how a thick layer of fibreglass wool in the loft reduces energy loss through the roof.
Model Answer -- 4(b)
The wool traps air in millions of small pockets [1]
Air is a very poor conductor, so conduction through the layer is small [1]
The trapped air cannot circulate, so convection currents cannot form (and it is convection that would otherwise carry energy to the roof) [1]
⚠ If you missed marks here: The fibreglass itself is not the insulator — the trapped AIR is. Air conducts badly, but only if it is stopped from MOVING: the pockets prevent the convection currents that free air would form. Trapping + poor conduction + no convection = three marks.
Mark 1 -- wool traps pockets of air (1 mark)
Mark 2 -- still air is poor conductor (1 mark)
Mark 3 -- trapped air prevents convection currents (1 mark)
(c)[2]
Explain how a shiny foil panel behind a radiator reduces energy loss into the wall.
Model Answer -- 4(c)
The radiator emits (infrared) radiation towards the wall [1]
A shiny/silvered surface is a poor absorber and good REFLECTOR of infrared, so the radiation is reflected back into the room instead of being absorbed by the wall [1]
⚠ If you missed marks here: This one is about RADIATION — the only transfer mechanism that shininess affects. Shiny surfaces reflect infrared (poor absorbers), bouncing the radiator’s output back into the room rather than warming the brickwork.
Mark 2 -- shiny foil reflects infrared back into room (1 mark)
(d)[3]
Double glazing consists of two panes of glass separated by a narrow gap of air (or a partial vacuum). Explain how the design reduces energy transfer, and why a vacuum gap is even better than an air gap.
Model Answer -- 4(d)
The (still) air in the gap is a poor conductor, cutting conduction between the panes [1]
The gap is kept NARROW so the air cannot form effective convection currents [1]
A vacuum contains no particles at all, so BOTH conduction and convection become impossible across it (only radiation can cross) [1]
⚠ If you missed marks here: Why narrow? A wide air gap would let convection loops ferry energy from warm pane to cold pane. And the vacuum answer must say WHY: conduction and convection both need particles; with none present, only radiation survives.
Mark 1 -- air gap poor conductor (1 mark)
Mark 2 -- narrow gap prevents convection loops (1 mark)
Mark 3 -- vacuum has no particles stopping conduction and convection (1 mark)
Question 5 -- The Cooling Cup of Tea
Total: 12 marks
A food-science student in Birmingham records the temperature of a mug of tea as it cools in a room at 20 °C. Her results are plotted below.
(a)[2]
Use the graph to find the temperature of the tea at 10 minutes, and the starting temperature.
Model Answer -- 5(a)
Start temperature = 90 °C [1]
At 10 minutes: 55 °C [1]
⚠ If you missed marks here: Straight read-offs: the curve starts at 90 °C and passes through 55 °C at t = 10 min. Follow the gridline with a ruler — free marks are lost to sloppy read-offs more often than to physics.
Mark 1 -- starting temperature 90 degrees read (1 mark)
Mark 2 -- temperature 55 degrees at 10 minutes read (1 mark)
(b)[4]
Calculate the average rate of cooling (i) over the first 10 minutes, and (ii) between 30 and 40 minutes.
Model Answer -- 5(b)
(i) fall = 90 − 55 = 35 °C in 10 min [1]
rate = 3.5 °C per minute [1]
(ii) fall = 28.8 − 24.4 ≈ 4.4 °C in 10 min [1]
rate ≈ 0.44 °C per minute — roughly 8 times slower [1]
⚠ If you missed marks here: Rate of cooling = temperature fall ÷ time taken, from the GRAPH: 35 °C in the first 10 min (3.5 °C/min) against about 4.4 °C in the last 10 (0.44 °C/min). Answers quoting just temperatures, without dividing by the time, are not rates.
Mark 1 -- first interval fall 35 degrees identified (1 mark)
Mark 2 -- initial rate 3.5 degrees per minute (1 mark)
Mark 3 -- late interval fall about 4.4 degrees read (1 mark)
Mark 4 -- late rate about 0.44 degrees per minute (1 mark)
(c)[3]
Explain why the tea cools quickly at first and ever more slowly as time passes.
Model Answer -- 5(c)
The rate of energy transfer to the surroundings depends on the temperature DIFFERENCE between the tea and the room [1]
At first the difference is large (70 °C), so energy leaves rapidly [1]
As the tea cools the difference shrinks, so the rate of loss falls — the curve flattens and approaches (but never quite reaches) room temperature at 20 °C [1]
⚠ If you missed marks here: The driving quantity is the temperature DIFFERENCE, not the temperature: 90 °C tea in a 20 °C room loses energy fast; 25 °C tea barely at all. That is why the curve levels off toward 20 °C — and why it can never cross it.
Mark 1 -- cooling rate depends on temperature difference (1 mark)
Mark 2 -- large initial difference gives fast loss (1 mark)
Mark 3 -- shrinking difference flattens curve toward room temperature (1 mark)
(d)[3]
Blowing across the surface of hot tea cools it faster. Explain, in terms of molecules, how evaporation cools the tea, and why blowing helps.
Model Answer -- 5(d)
The MOST energetic molecules at the surface escape as vapour (evaporation) [1]
The average kinetic energy of the molecules left behind falls, so the liquid’s temperature falls [1]
Blowing sweeps the vapour away (and brings drier, moving air), increasing the rate of evaporation — so the cooling is faster [1]
⚠ If you missed marks here: Evaporation is selective: only the FASTEST molecules escape, dragging the average KE of the remainder down — cooling by losing your best players. Blowing helps by removing the humid layer above the surface so escapees are not replaced.
Mark 1 -- most energetic surface molecules escape (1 mark)
Mark 2 -- lower average KE means lower temperature (1 mark)
Mark 3 -- blowing removes vapour speeding evaporation (1 mark)
Question 6 -- The Engineering Works
Total: 12 marks
An engineering works in Sheffield fits a bronze bearing ring onto a steel pump shaft by shrink fitting: at room temperature the hole in the ring is very slightly SMALLER than the shaft. Meanwhile, outside the works, National Grid engineers are checking the sag of the overhead power lines on the hottest day of the summer.
(a)[3]
Explain, in terms of particles, why a solid expands when heated.
Model Answer -- 6(a)
At higher temperature the particles vibrate with greater amplitude/energy (about their positions) [1]
The average separation between particles increases [1]
So the solid as a whole takes up more space; the particles themselves do NOT get bigger [1]
⚠ If you missed marks here: The particles vibrate FURTHER, so their average spacing grows — the expansion lives in the gaps, not in the particles. "The particles expand" is the answer examiners are waiting to punish.
Mark 1 -- particles vibrate with greater amplitude (1 mark)
Mark 2 -- average separation increases (1 mark)
Mark 3 -- particles themselves unchanged in size (1 mark)
(b)[4]
Describe the steps of the shrink-fitting process, and explain why the ring grips the shaft permanently once fitted.
Model Answer -- 6(b)
The ring is heated strongly; it expands — and the HOLE in it expands too (everything scales up, like enlarging a photograph) [1]
While hot, the enlarged hole slips easily over the shaft [1]
As the ring cools it contracts, and the hole tries to return to its original (smaller-than-the-shaft) size [1]
It cannot — the shaft is in the way — so the ring squeezes the shaft with a large force, gripping it without bolts or welds [1]
⚠ If you missed marks here: The make-or-break idea: a heated ring’s HOLE gets bigger, not smaller — expansion enlarges every dimension, gap included, exactly like enlarging a photograph. Then the sequence: heat → hole bigger than shaft → slip on → cool → attempted contraction squeezes the shaft permanently.
Mark 1 -- heating expands ring including its hole (1 mark)
Mark 2 -- enlarged hole slips over shaft while hot (1 mark)
Mark 3 -- cooling ring contracts toward original size (1 mark)
Mark 4 -- blocked contraction grips shaft with large force (1 mark)
(c)[3]
The power lines between pylons are deliberately hung with plenty of slack. Explain why the lines sag more in summer than in winter, and what could happen in winter if they were pulled taut in summer.
Model Answer -- 6(c)
In summer heat the metal lines expand and become longer, so they hang lower (sag more) between the pylons [1]
In winter the lines contract and become shorter, taking up the slack [1]
If they had been strung taut in summer, the winter contraction would put the lines under enormous tension — they could snap or pull the pylons/insulators apart, so the slack is a deliberate safety allowance [1]
⚠ If you missed marks here: Summer sag is the visible half; the dangerous half is winter CONTRACTION. A line taut in July has nowhere to go in January — contraction against a fixed length builds huge tension. The slack is not sloppiness; it is the engineers’ expansion allowance, the same job the gaps in a bridge do.
Mark 1 -- summer expansion lengthens lines increasing sag (1 mark)
Mark 2 -- winter contraction shortens lines reducing sag (1 mark)
Mark 3 -- taut lines would snap under winter contraction tension (1 mark)
(d)[2]
For the same rise in temperature, compare the expansion of solids, liquids and gases, and give the reason in terms of particle bonding.
Model Answer -- 6(d)
Gases expand most, liquids next, solids least [1]
In solids the particles are held by strong bonds that limit their separation; in liquids the bonds are weaker; in gases the particles are essentially free, so the same temperature rise produces the biggest volume change [1]
⚠ If you missed marks here: Order: gas most, liquid, solid least. The reason is bond strength — strong solid bonds resist any increase in separation, while gas particles have nothing holding them together at all.
Mark 1 -- order gases most solids least stated (1 mark)
Mark 2 -- bond strength explanation given (1 mark)
Question 7 -- The Smoke Cell
Total: 10 marks
A physics class in Cardiff observes a smoke cell under a microscope: a tiny air-filled box containing smoke, brightly lit from the side. The smoke particles appear as bright specks that jiggle about ceaselessly and randomly. This is Brownian motion.
(a)[2]
Describe exactly what is observed through the microscope.
Model Answer -- 7(a)
Bright specks (the smoke particles) moving constantly [1]
...in short, random, zig-zag paths — changing direction abruptly with no pattern [1]
⚠ If you missed marks here: Report what the EYE sees: specks (not molecules — those are invisible) in perpetual, random, jerky motion. "The particles vibrate" undersells it; they wander erratically, never settling.
Mark 1 -- bright specks in constant motion seen (1 mark)
Mark 2 -- random zig-zag direction changes described (1 mark)
(b)[3]
Explain what causes the motion of the smoke specks, and what this observation demonstrates about air.
Model Answer -- 7(b)
Air molecules (far too small to see) are moving randomly at high speed and collide with the smoke specks [1]
The collisions are UNEVEN — at any instant more molecules happen to hit one side than the other — so the speck is knocked in a random direction, again and again [1]
This is direct evidence that air consists of molecules in constant random motion (the specks are massive compared with the fast, light molecules battering them) [1]
⚠ If you missed marks here: The specks are being kicked by invisible players: fast, light air molecules striking unevenly from moment to moment. The observation matters because it is EVIDENCE — the jiggling speck is the visible footprint of invisible molecular motion.
Mark 1 -- invisible air molecules bombard the specks (1 mark)
Mark 2 -- uneven collisions knock specks randomly (1 mark)
Mark 3 -- demonstrates air molecules in constant random motion (1 mark)
(c)[3]
The cell is warmed gently and the specks move more violently. Explain why, and state what is special about the temperature −273 °C.
Model Answer -- 7(c)
At higher temperature the air molecules move faster (greater average kinetic energy) [1]
Faster molecules strike the specks harder, so the Brownian motion becomes more vigorous [1]
−273 °C is ABSOLUTE ZERO: the lowest possible temperature, where the particles have their minimum (kinetic) energy [1]
⚠ If you missed marks here: Temperature is a molecular speedometer: warm the cell and the bombardment strengthens. Absolute zero (−273 °C = 0 K) is where molecular kinetic energy reaches its minimum — not where matter vanishes, and not merely "very cold".
Mark 1 -- warmer molecules move faster (1 mark)
Mark 2 -- harder impacts give more vigorous jiggling (1 mark)
Mark 3 -- minus 273 is absolute zero minimum particle energy (1 mark)
(d)[2]
The smoke cell is sealed, so its volume is fixed. State and explain what happens to the pressure of the air inside as it is warmed.
Model Answer -- 7(d)
The pressure increases [1]
The faster molecules collide with the walls harder and more frequently, increasing the average force per unit area [1]
⚠ If you missed marks here: Fixed volume + faster molecules = harder, more frequent wall collisions = higher pressure. The explanation must be in terms of COLLISIONS — "heat makes pressure" is a slogan, not physics.
Mark 1 -- pressure increases stated (1 mark)
Mark 2 -- harder more frequent wall collisions explain it (1 mark)
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