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IGCSE Physics Paper 4 (Theory / Extended)

Topic 2: Thermal Physics -- Mock Exam 1
1 hour 15 minutes
80
7
75:00
0625 / 0972

Instructions

Question Navigation

Question 1 -- States of Matter and Particle Model
Total: 12 marks
(a) [3]
The diagrams below represent the arrangement of particles in three states of matter.

Diagram A: Particles closely packed in a regular, ordered pattern, vibrating about fixed positions.
Diagram B: Particles close together but randomly arranged, able to slide over each other.
Diagram C: Particles far apart, moving randomly at high speed in all directions.

For each diagram, state the state of matter it represents.
Diagram A Diagram B Diagram C
Model Answer -- 1(a)
A = solid [1]
B = liquid [1]
C = gas [1]
Mark 1 -- A = solid
Mark 2 -- B = liquid
Mark 3 -- C = gas
(b) [4]
Compare the motion and spacing of particles in a solid and in a gas.
Model Answer -- 1(b)
In a solid, particles vibrate about fixed positions [1]
In a gas, particles move randomly at high speed in all directions [1]
In a solid, particles are closely packed / touching [1]
In a gas, particles are far apart / large spaces between them [1]
Mark 1 -- solid particles vibrate about fixed positions
Mark 2 -- gas particles move randomly at high speed
Mark 3 -- solid particles closely packed
Mark 4 -- gas particles far apart
(c) [2]
A solid is heated and melts to form a liquid. Describe what happens to the particles during melting.
Model Answer -- 1(c)
Particles gain (kinetic) energy and vibrate more / faster [1]
Particles break free from their fixed positions and are able to move/slide over each other (but remain close together) [1]
Mark 1 -- particles gain energy / vibrate more
Mark 2 -- break free from fixed positions / slide over each other
(d) [3]
Explain, in terms of particles, why a gas fills its container and exerts a pressure on the walls.
Model Answer -- 1(d)
Gas particles move randomly in all directions at high speed, so they spread out to fill the entire container [1]
The particles collide with the walls of the container [1]
Each collision exerts a force on the wall; the total force of many collisions per unit area is the gas pressure [1]
Mark 1 -- particles move randomly, fill container
Mark 2 -- particles collide with walls
Mark 3 -- collisions exert force / force per unit area = pressure
Question 2 -- Brownian Motion and Temperature Scales
Total: 10 marks
(a) [3]
A student uses a microscope to observe smoke particles in a glass cell. The smoke particles are seen to move in a random, jerky, zigzag path.

(i) State the name given to this type of motion. [1]
(ii) Explain why the smoke particles move in this way. [2]
Model Answer -- 2(a)
(i) Brownian motion [1]
(ii) The smoke particles are bombarded / hit by air molecules (which are too small to see) [1]
(ii) The air molecules move randomly, so the collisions are uneven / unequal from different directions, causing the smoke particles to change direction randomly [1]
Mark 1 -- Brownian motion named
Mark 2 -- hit/bombarded by air molecules
Mark 3 -- uneven collisions cause random direction changes
(b) [2]
(i) State what is meant by absolute zero. [1]
(ii) State the value of absolute zero in degrees Celsius. [1]
Model Answer -- 2(b)
(i) Absolute zero is the lowest possible temperature, at which particles have minimum / zero kinetic energy and stop moving (cease all motion) [1]
(ii) -273 °C (accept -273.15 °C) [1]
Mark 1 -- correct definition of absolute zero
Mark 2 -- -273 °C
(c) [2]
Convert the following temperatures:
(i) 100 °C to kelvin [1]
(ii) 350 K to degrees Celsius [1]
Model Answer -- 2(c)
T(K) = T(°C) + 273
(i) 100 + 273 = 373 K [1]
(ii) 350 - 273 = 77 °C [1]
Mark 1 -- 373 K
Mark 2 -- 77 °C
(d) [3]
A sealed container holds a fixed mass of gas at 27 °C. The gas is then heated to 127 °C at constant volume.

Explain, in terms of particles, why the pressure of the gas increases.
Model Answer -- 2(d)
When temperature increases, the particles gain kinetic energy and move faster [1]
The particles hit the walls of the container more frequently / more often [1]
Each collision exerts a greater force (because particles have more momentum), so the total pressure increases [1]
Mark 1 -- particles gain KE / move faster
Mark 2 -- collisions more frequent
Mark 3 -- greater force per collision / pressure increases
Question 3 -- Boyle's Law
Total: 12 marks
(a) [2]
State Boyle's law.
Model Answer -- 3(a)
For a fixed mass of gas at constant temperature [1]
the pressure is inversely proportional to the volume (or pV = constant) [1]
Mark 1 -- fixed mass, constant temperature stated
Mark 2 -- p inversely proportional to V / pV = constant
(b) [3]
Explain Boyle's law in terms of the particle model of a gas.
Model Answer -- 3(b)
When the volume of the container decreases, the same number of particles occupies a smaller space [1]
The particles hit the walls more frequently (more collisions per second with the walls) [1]
The rate of change of momentum at the walls increases, so the force per unit area (pressure) increases [1]
Mark 1 -- same particles in smaller space
Mark 2 -- more frequent collisions with walls
Mark 3 -- greater force per unit area / pressure increases
(c) [4]
A diver at the surface of a lake fills her lungs with 6.0 litres of air at atmospheric pressure 100 kPa. She then dives to a depth where the total pressure on her body is 300 kPa.

Assuming the temperature remains constant, calculate the volume of air in her lungs at this depth.
Model Answer -- 3(c)
p1V1 = p2V2
Correct formula stated [1]
100 × 6.0 = 300 × V2
Correct substitution [1]
V2 = (100 × 6.0) / 300 = 600 / 300
Correct rearrangement and calculation [1]
V2 = 2.0 litres
Correct answer with unit: 2.0 litres [1]
Mark 1 -- correct formula (p1V1 = p2V2)
Mark 2 -- correct substitution
Mark 3 -- correct rearrangement
Mark 4 -- correct answer: 2.0 litres
(d) [3]
A student performs an experiment to verify Boyle's law. She measures the pressure and volume of a trapped gas sample and records the following data:

Pressure / kPa Volume / cm3
100 60
150 40
200 30
300 20
(i) Use the data to show that these results are consistent with Boyle's law. [2]
(ii) What graph would give a straight line through the origin if Boyle's law is obeyed? [1]
Model Answer -- 3(d)
(i) Calculate pV for each pair of readings: 100 × 60 = 6000, 150 × 40 = 6000, 200 × 30 = 6000, 300 × 20 = 6000 [1]
(i) pV is constant (= 6000 kPa·cm3) for all readings, which is consistent with Boyle's law (pV = constant) [1]
(ii) A graph of pressure (p) against 1/volume (1/V) would give a straight line through the origin [1]
Mark 1 -- pV calculated for readings
Mark 2 -- pV constant, consistent with law
Mark 3 -- p vs 1/V gives straight line through origin
Question 4 -- Thermal Expansion
Total: 10 marks
(a) [2]
Explain, in terms of particles, why a metal bar expands when it is heated.
Model Answer -- 4(a)
When heated, the particles gain kinetic energy and vibrate with greater amplitude / more vigorously [1]
The particles push their neighbours further apart, so the average separation between particles increases, causing the bar to expand [1]
Mark 1 -- particles gain KE / vibrate more
Mark 2 -- average separation increases / expansion
(b) [3]
Railway tracks in India are laid with small gaps between successive rails.

(i) Explain why these gaps are necessary. [2]
(ii) In which season would these gaps be the smallest? Give a reason. [1]
Model Answer -- 4(b)
(i) When the temperature rises (e.g. during summer), the metal rails expand [1]
(i) Without gaps, the expanding rails would push against each other, creating enormous forces that could buckle/bend the track [1]
(ii) Summer, because the rails expand the most in hot weather, filling the gaps [1]
Mark 1 -- rails expand when heated
Mark 2 -- without gaps, rails buckle/bend
Mark 3 -- summer (gaps smallest due to expansion)
(c) [2]
A metal lid is stuck tightly on a glass jar. Suggest how thermal expansion could be used to remove the lid, and explain why this works.
Model Answer -- 4(c)
Run hot water over the metal lid (or heat the lid) [1]
Metal expands more than glass for the same temperature rise, so the lid becomes looser and can be unscrewed [1]
Mark 1 -- heat the lid (hot water etc.)
Mark 2 -- metal expands more than glass
(d) [3]
The diagram below shows a bimetallic strip made of brass and iron, riveted together. When the strip is heated, it bends.

Diagram: A straight horizontal strip. The top layer is labelled "Brass" and the bottom layer is labelled "Iron". Rivets join the two layers together at regular intervals.

(i) State which direction the strip bends when heated (towards the brass or towards the iron). [1]
(ii) Explain why the strip bends. [2]
BRASS IRON Bimetallic strip (before heating)
Model Answer -- 4(d)
(i) The strip bends towards the iron side (downwards / concave on the iron side) [1]
(ii) Brass expands more than iron for the same temperature rise (brass has a higher coefficient of thermal expansion) [1]
(ii) Because the strips are riveted together and cannot expand independently, the brass side becomes longer, forcing the strip to curve towards the iron side [1]
Mark 1 -- bends towards iron
Mark 2 -- brass expands more than iron
Mark 3 -- joined together, brass longer, forces curve
Question 5 -- Specific Heat Capacity and Changes of State
Total: 12 marks
(a) [1]
Define specific heat capacity.
Model Answer -- 5(a)
Specific heat capacity is the energy required per unit mass per unit temperature rise (or: the energy needed to raise the temperature of 1 kg of a substance by 1 °C / 1 K) [1]
Mark 1 -- correct definition (energy per unit mass per unit temp rise)
(b) [4]
A 2.0 kg aluminium block at 25 °C is heated using a 500 W electric heater for 3 minutes and 32 seconds. The temperature rises to 143 °C.

The specific heat capacity of aluminium is 900 J/(kg °C).

(i) Calculate the energy supplied by the heater. [2]
(ii) Calculate the energy needed to produce this temperature rise using the specific heat capacity equation. [2]
Model Answer -- 5(b)
(i) Energy from heater:
E = P × t = 500 × 212
Correct conversion of time: 3 min 32 s = 212 s, and substitution [1]
E = 106 000 J (= 106 kJ)
Correct answer with unit: 106 000 J [1]
(ii) Energy from SHC equation:
ΔE = mcΔθ = 2.0 × 900 × (143 - 25)
Correct substitution with Δθ = 118 °C [1]
ΔE = 2.0 × 900 × 118 = 212 400 J
Correct answer: 212 400 J (212.4 kJ) [1]
Note: The energy from the heater (106 kJ) is less than the theoretical energy (212.4 kJ). In practice, this discrepancy would be reversed -- the heater supplies more energy than what the block absorbs, because energy is lost to the surroundings. The question is designed so students practise both calculations independently.
Mark 1 -- correct time conversion and substitution (E = Pt)
Mark 2 -- correct heater energy: 106 000 J
Mark 3 -- correct substitution into ΔE = mcΔθ
Mark 4 -- correct answer: 212 400 J
(c) [4]
The graph below shows how the temperature of a pure substance changes as it is heated steadily from solid to gas.

Heating curve description: The temperature rises from -20 °C (section P), then remains constant at 0 °C for a period (section Q), then rises again (section R), then remains constant at 100 °C for a period (section S), then rises again (section T).
Time Temperature / °C -20 0 100 P Q R S T
(i) State the name of the change of state occurring during section Q. [1]
(ii) State the name of the change of state occurring during section S. [1]
(iii) Explain why the temperature remains constant during section Q, even though the substance is being heated. [2]
Model Answer -- 5(c)
(i) Melting (or fusion) [1]
(ii) Boiling (or vaporisation) [1]
(iii) The energy supplied is being used to break the bonds / overcome the forces of attraction between particles [1]
(iii) The energy increases the potential energy of the particles (not their kinetic energy), so the temperature does not change [1]
Mark 1 -- Q = melting
Mark 2 -- S = boiling
Mark 3 -- energy used to break bonds
Mark 4 -- PE increases not KE / temperature constant
(d) [3]
State three differences between boiling and evaporation.
Model Answer -- 5(d)
Boiling occurs at a fixed temperature (the boiling point); evaporation occurs at any temperature below the boiling point [1]
Boiling occurs throughout the liquid (bubbles form inside); evaporation occurs only at the surface of the liquid [1]
Boiling is rapid and vigorous; evaporation is a slow, gradual process [1]
Mark 1 -- boiling at fixed temp vs evaporation at any temp
Mark 2 -- boiling throughout liquid vs evaporation at surface only
Mark 3 -- boiling rapid vs evaporation slow
Question 6 -- Conduction and Convection
Total: 12 marks
(a) [4]
Explain how thermal energy is conducted through a metal rod when one end is placed in a flame. Your answer should refer to both lattice vibrations and free electrons.
Model Answer -- 6(a)
Lattice vibrations: At the hot end, particles vibrate with greater amplitude / gain kinetic energy [1]
These vibrating particles collide with neighbouring particles, passing on kinetic energy along the rod from particle to particle [1]
Free electrons: Metals have free (delocalised) electrons that move throughout the structure [1]
At the hot end, free electrons gain kinetic energy, move rapidly through the metal, and transfer energy to cooler particles by colliding with them. This is much faster than lattice vibrations, which is why metals are good conductors [1]
Mark 1 -- hot end particles vibrate more
Mark 2 -- vibrations passed to neighbours
Mark 3 -- free/delocalised electrons mentioned
Mark 4 -- electrons transfer energy quickly through metal
(b) [2]
Explain why non-metals (such as wood or plastic) are poor conductors of thermal energy compared to metals.
Model Answer -- 6(b)
Non-metals do not have free (delocalised) electrons [1]
They can only conduct by lattice vibrations (particle to particle), which is a much slower process [1]
Mark 1 -- no free electrons
Mark 2 -- only lattice vibrations / slower
(c) [4]
The diagram shows a beaker of water being heated from below by a Bunsen burner. A few crystals of potassium permanganate (purple dye) have been placed at the bottom of the beaker.

Diagram: A beaker of water sits on a tripod above a Bunsen burner. Purple streaks rise from the bottom centre, spread across the top, and come back down at the sides, forming a circulation pattern.
dye crystals Bunsen burner
(i) What is the name of the process that causes the purple dye to circulate through the water? [1]
(ii) Explain, in terms of density, how this process works. [3]
Model Answer -- 6(c)
(i) Convection [1]
(ii) Water at the bottom is heated, expands, and becomes less dense [1]
(ii) The less dense (warmer) water rises and is replaced by cooler, denser water that sinks from the sides [1]
(ii) This sets up a convection current -- a continuous circulation of water carrying the dye around the beaker [1]
Mark 1 -- convection named
Mark 2 -- heated water expands / becomes less dense
Mark 3 -- warm water rises, cooler denser water sinks
Mark 4 -- continuous circulation / convection current
(d) [2]
Explain why convection cannot occur in a solid.
Model Answer -- 6(d)
Convection requires particles to move from one place to another (bulk movement of fluid) [1]
In a solid, particles are held in fixed positions and cannot flow, so convection currents cannot be established [1]
Mark 1 -- convection needs particles to move freely / bulk flow
Mark 2 -- solid particles fixed in position / cannot flow
Question 7 -- Thermal Radiation and Earth's Temperature
Total: 12 marks
(a) [2]
(i) State one way in which thermal radiation (infrared) is different from conduction and convection. [1]
(ii) State one everyday example of thermal energy transfer by radiation. [1]
Model Answer -- 7(a)
(i) Radiation does not require a medium / can travel through a vacuum (conduction and convection both need particles/matter) [1]
(ii) Any valid example, e.g.: feeling warmth from the Sun / heat from a campfire / warmth from a room heater / heat from a hot iron (accept any sensible example) [1]
Mark 1 -- radiation needs no medium / travels through vacuum
Mark 2 -- valid everyday example
(b) [4]
A student investigates how surface colour affects the rate of emission and absorption of thermal radiation. She uses two identical metal cans -- one painted matt black and one painted shiny white. Both are filled with the same volume of hot water at 80 °C and left to cool.

(i) State which can will cool faster. [1]
(ii) Explain your answer to (i). [1]
(iii) The student repeats the experiment, this time placing the empty cans at equal distances from a radiant heater and measuring the temperature rise. State and explain which can heats up faster. [2]
Model Answer -- 7(b)
(i) The matt black can cools faster [1]
(ii) Dark, matt surfaces are better emitters of thermal radiation than light, shiny surfaces, so the black can loses heat more quickly [1]
(iii) The matt black can heats up faster [1]
(iii) Dark, matt surfaces are also better absorbers of thermal radiation than light, shiny surfaces [1]
Mark 1 -- matt black can cools faster
Mark 2 -- dark matt surfaces are better emitters
Mark 3 -- matt black can heats up faster
Mark 4 -- dark matt surfaces are better absorbers
(c) [3]
Explain how the temperature of the Earth is maintained at a roughly constant average value. Your answer should refer to the balance between energy absorbed and energy emitted.
Model Answer -- 7(c)
The Earth absorbs thermal radiation (short-wavelength, visible and UV) from the Sun [1]
The Earth also emits thermal radiation (longer-wavelength infrared) into space [1]
The Earth's average temperature remains roughly constant because the rate of energy absorbed from the Sun equals the rate of energy emitted into space (thermal equilibrium) [1]
Mark 1 -- Earth absorbs radiation from the Sun
Mark 2 -- Earth emits infrared radiation into space
Mark 3 -- constant temperature when absorbed = emitted
(d) [3]
In many Indian cities, buildings are painted white and have thick walls. Using your knowledge of thermal energy transfer, explain how each of these design features helps to keep the inside of a building cool during summer.

(i) White-painted exterior walls [1]
(ii) Thick walls [2]
Model Answer -- 7(d)
(i) White / light-coloured surfaces are poor absorbers (good reflectors) of thermal radiation, so less heat from the Sun is absorbed by the building [1]
(ii) Thick walls contain more material for heat to conduct through, slowing down the rate of heat transfer from outside to inside [1]
(ii) The thick walls act as thermal insulation -- by the time heat conducts through, the outside temperature may have dropped (e.g. at night), keeping the interior temperature more stable [1]
Mark 1 -- white surfaces reflect / poor absorbers of radiation
Mark 2 -- thick walls slow conduction / more material
Mark 3 -- thermal insulation / temperature stability

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IGCSE Physics 0625/0972 -- Paper 4 Theory (Extended) -- Topic 2: Thermal Physics -- Mock Exam 1
For examination preparation only. Not an official Cambridge Assessment paper.