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IGCSE Physics Paper 4 (Theory/Extended) - Challenge

Topic 2: Thermal Physics (2.1 - 2.3) | Harder Difficulty
75 minutes
80
7
75:00

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Question Navigation Score: 0 / 80

Question 1: Solar Water Heater
Total: 12 marks
A rooftop solar water heater in southern England uses a flat black panel to absorb sunlight and heat water for household use. Cold water at 22 °C enters the panel and is heated to 70 °C. The panel holds 40 kg of water at a time.
Black absorbing panel Storage Cold in 22 °C Hot out 70 °C
(a) 2 marks
Model Answer - Q1(a)
Black surfaces are good absorbers of (infrared/thermal) radiation [1]
Matt/rough surfaces absorb more radiation than shiny/smooth surfaces (less radiation is reflected) [1]
⚠ If you missed marks here: "Black attracts heat" or "black gets hotter" scores nothing — the physics word is ABSORBER: black surfaces are good absorbers of (infrared) radiation. The second mark is specifically for the MATT part: matt reflects less than shiny, so more radiation is absorbed — most students explain "black" but forget to explain "matt".
Mark 1: Black surfaces are good absorbers of infrared radiation
Mark 2: Matt surfaces absorb more than shiny surfaces / less reflection
(b) 3 marks
Model Answer - Q1(b)
ΔT = 70 - 22 = 48 °C [1]
E = mcΔT = 40 × 4200 × 48 [1]
E = 8 064 000 J = 8.064 × 106 J (accept 8.1 MJ) [1]
⚠ If you missed marks here: The classic slip is putting 70 (or 22) into E = mcΔT instead of the temperature CHANGE, ΔT = 70 − 22 = 48 °C — using 70 gives 11.8 MJ, too big. Also check you used c = 4200 J/(kg °C) for water; forgetting it (or using 4.2) shifts the answer by a factor of 1000.
Mark 1: Correct temperature change 48 degrees
Mark 2: Correct substitution into E equals mc delta T
Mark 3: Correct answer 8064000 J or 8.064 MJ
(c) 4 marks
(c) (i) 2 marks
Model Answer - Q1(c)(i)
Total power = 1200 × 3.0 = 3600 W [1]
Useful power = 0.56 × 3600 = 2016 W (accept 2000 W) [1]
⚠ If you missed marks here: Two half-steps get skipped: the 1200 W is PER SQUARE METRE, so first multiply by the 3.0 m2 area (3600 W), THEN take 56% of it (2016 W). If your answer was 672 W you applied 56% to 1200 without the area; if it was 3600 W you forgot the efficiency.
Mark 1: Total power 3600 W calculated
Mark 2: Useful power 2016 W or 2000 W
(c) (ii) 2 marks
Model Answer - Q1(c)(ii)
t = E / P = 8 064 000 / 2016 = 4000 s [1]
4000 / 60 = 66.7 minutes (accept 67 minutes) [1]
⚠ If you missed marks here: t = E/P = 8 064 000 / 2016 = 4000 s — the most common loss is leaving the answer in seconds when the question demands MINUTES (divide by 60 → 67 min). Also make sure you divided by the USEFUL power 2016 W, not the total 3600 W (that gives 37 min).
Mark 1: Correct use of time equals energy divided by power, 4000 seconds
Mark 2: Conversion to approximately 67 minutes
(d) 2 marks
Model Answer - Q1(d)
Heat/energy is lost from the panel to the surroundings by conduction/convection/radiation [1]
The intensity of sunlight varies (clouds, angle of Sun changes during the day) / not all radiation is absorbed / some is reflected [1]
⚠ If you missed marks here: "The panel isn't 100% efficient" doesn't score — the 56% is already built into the calculation, so you need reasons BEYOND it. Give two different physical reasons: heat escapes from the panel/water to the surroundings, and the sunlight isn't a steady 1200 W/m2 (clouds, Sun's angle changing).
Mark 1: Heat lost from panel to surroundings by conduction or convection or radiation
Mark 2: Sunlight intensity varies or some radiation reflected or clouds
(e) 1 mark
Model Answer - Q1(e)
Trapped air acts as an insulator / reduces convection losses because air cannot circulate away from the panel [1]
⚠ If you missed marks here: "The glass keeps the heat in" is too vague — the mark is for the TRAPPED AIR: it cannot circulate, so convection currents can't carry warm air away (and still air is a poor conductor). Name the mechanism, not just the outcome.
Mark 1: Trapped air insulator reduces convection losses
Question 2: Thermal Expansion and Railway Engineering
Total: 11 marks
Railway engineers leave small gaps between steel rails when they are laid in cold weather. In Rajasthan, temperatures can range from 5 °C in winter to 50 °C in summer.
Gap expansion
(a) 3 marks
Model Answer - Q2(a)
When temperature increases, particles gain kinetic energy [1]
Particles vibrate more / with greater amplitude [1]
Particles push neighbouring particles further apart, so the material expands / average separation increases [1]
⚠ If you missed marks here: The number one wrong answer is "the particles expand" or "get bigger" — particles stay the SAME SIZE; it's their vibration amplitude and average SEPARATION that increase. For all 3 marks you need the chain: particles gain kinetic energy → vibrate more/with bigger amplitude → average separation increases so the rail gets longer.
Mark 1: Particles gain kinetic energy when temperature increases
Mark 2: Particles vibrate more or greater amplitude
Mark 3: Average separation increases so material expands
(b) 2 marks
Model Answer - Q2(b)
The rails would expand and push against each other / have no room to expand [1]
The rails would buckle / bend / distort, which is dangerous for trains [1]
⚠ If you missed marks here: "The rails would break" or "crack" isn't the expected answer — expanding rails pushed against each other BUCKLE (bend sideways). Two marks means two steps: the rails expand with nowhere to go and push on each other [1], THEN they buckle/distort, which could derail a train [1].
Mark 1: Rails expand and push against each other with no room
Mark 2: Rails buckle or bend or distort creating danger
(c) 2 marks
Model Answer - Q2(c)
27 °C is approximately the midpoint / average of the temperature range (5 °C to 50 °C) [1]
So the rail is under roughly equal compression (in summer) and tension (in winter), preventing extreme stress in either direction [1]
⚠ If you missed marks here: Spot the clue in the numbers: 27 °C is roughly halfway between 5 °C and 50 °C — saying "it's a comfortable temperature" misses the point entirely. The second mark needs the consequence: the rail then experiences roughly equal compression in summer and tension in winter, so neither stress gets extreme.
Mark 1: 27 degrees is midpoint or average of temperature range
Mark 2: Equal compression and tension prevents extreme stress
(d) 4 marks
Cold (straight) Brass Iron Hot (curved) Brass (outside - expands more) Iron (inside) Contact
(d) (i) 2 marks
Model Answer - Q2(d)(i)
Brass expands more than iron for the same temperature rise [1]
Because they are bonded together, the brass side becomes longer, forcing the strip to curve with brass on the outside [1]
⚠ If you missed marks here: Saying "brass expands more" alone is only 1 mark — the second mark needs the BONDED idea: because the two metals cannot slide past each other, the longer brass side is forced to the OUTSIDE of the curve. If you drew or described iron on the outside, that's the classic reversal.
Mark 1: Brass expands more than iron
Mark 2: Bonded together so brass on outside forces strip to curve
(d) (ii) 2 marks
Model Answer - Q2(d)(ii)
Named application, e.g. liquid-in-glass thermometer / rivets / shrink-fitting metal components [1]
Brief explanation showing understanding, e.g. liquid expands up the tube when temperature rises, giving a reading [1]
⚠ If you missed marks here: Naming an application (thermometer, rivets, shrink-fitting) is only half the marks — the second mark is for explaining HOW expansion is used in it (e.g. the liquid expands up the narrow tube as temperature rises). Also avoid re-using the bimetallic strip or railway gaps: the question says "one OTHER application".
Mark 1: Named application such as thermometer or rivets or shrink fitting
Mark 2: Correct explanation of how expansion is used in that application
Question 3: Heating Curve Analysis
Total: 12 marks
A 0.50 kg block of a pure solid substance X is heated steadily at a rate of 500 W. The temperature is recorded every 30 seconds. The graph below shows the heating curve obtained.
-20 0 20 40 60 80 100 Temperature / °C 0 60 120 180 240 300 Time / s A B C D
(a) 4 marks
Model Answer - Q3(a)
Region A: Solid [1]
Region B: Solid and liquid (melting) [1]
Region C: Liquid [1]
Region D: Liquid and gas (boiling) [1]
⚠ If you missed marks here: The marks usually lost are B and D: during a plateau TWO states exist together — B is solid AND liquid (melting), D is liquid AND gas (boiling). Writing just "liquid" for B or "gas" for D describes what it's becoming, not what is actually there.
Mark 1: Region A is solid
Mark 2: Region B is solid and liquid or melting
Mark 3: Region C is liquid
Mark 4: Region D is liquid and gas or boiling
(b) 2 marks
Model Answer - Q3(b)
Melting point = 40 °C [1]
Boiling point = 80 °C [1]
⚠ If you missed marks here: The melting and boiling points are the temperatures of the FLAT sections (B at 40 °C, D at 80 °C) — don't assume water's familiar 0 °C and 100 °C; substance X is not water. If you gave 100 °C for boiling, you answered from memory instead of reading the graph.
Mark 1: Melting point is 40 degrees C
Mark 2: Boiling point is 80 degrees C
(c) 3 marks
Model Answer - Q3(c)
Energy is being supplied / the substance is absorbing energy [1]
The energy is used to break/overcome the bonds/forces between particles (not to increase kinetic energy) [1]
Since temperature is related to the average kinetic energy of particles, and KE is not increasing, the temperature stays constant / energy goes to increasing potential energy [1]
⚠ If you missed marks here: "The heat is used for melting" restates the question and scores little — the marks are for WHERE the energy goes: into breaking the bonds/forces BETWEEN particles, NOT into their kinetic energy. The final mark links KE to temperature: since average KE doesn't rise, temperature can't rise (the energy raises potential energy instead).
Mark 1: Energy is being supplied or absorbed
Mark 2: Energy used to break bonds or overcome forces between particles
Mark 3: Kinetic energy not increasing so temperature constant or potential energy increases
(d) 3 marks
Hint: Read off the temperature change and the time for region A from the graph. Use P = E/t and E = mcΔT.
Model Answer - Q3(d)
From graph: ΔT = 40 - (-20) = 60 °C, time = 60 s. Energy supplied E = 500 × 60 = 30 000 J [1]
c = E / (m × ΔT) = 30 000 / (0.50 × 60) [1]
c = 1000 J/(kg °C) [1]
⚠ If you missed marks here: The graph trap: region A starts at −20 °C, so ΔT = 40 − (−20) = 60 °C — students who read ΔT as 40 get c = 1500 J/(kg °C). Energy first (E = 500 W × 60 s = 30 000 J), then c = E/(mΔT) = 30 000/(0.50 × 60) = 1000 J/(kg °C).
Mark 1: Correct energy calculation 30000 J from power times time
Mark 2: Correct substitution into c equals E divided by m times delta T
Mark 3: Correct answer 1000 J per kg per degree C
Question 4: Designing a Cooking Pot
Total: 11 marks
A manufacturer designs a cooking pot. The pot has a thick copper base, thin stainless steel walls, and a plastic handle. It has a glass lid with a small steam vent hole.
Plastic knob Steam vent Copper base Steel walls Plastic handle Glass lid Heat source
(a) 2 marks
Model Answer - Q4(a)
Copper is a better conductor of heat / thermal energy than stainless steel [1]
So heat transfers quickly and evenly from the flame through the base to the food, reducing hot spots / cooking more evenly [1]
⚠ If you missed marks here: This needs a COMPARISON — "copper is a good conductor" only earns the mark when you say it's a BETTER conductor than stainless steel. The second mark is the consequence: heat passes through the base quickly and evenly to the food, avoiding hot spots.
Mark 1: Copper is better conductor of heat than stainless steel
Mark 2: Heat transfers quickly and evenly reducing hot spots
(b) 2 marks
Model Answer - Q4(b)
Plastic is a poor conductor / good insulator of heat [1]
So very little heat is conducted from the pot to the hand / the handle stays cool enough to hold safely [1]
⚠ If you missed marks here: "So you don't burn your hand" alone is only the second half — the first mark is the physics property: plastic is a POOR CONDUCTOR (good insulator) of heat. State the property, then the consequence: little heat is conducted along the handle to your hand.
Mark 1: Plastic is poor conductor or good insulator of heat
Mark 2: Little heat conducted to hand so handle stays cool or safe to hold
(c) 4 marks
Model Answer - Q4(c)
The process is convection [1]
Water near the base is heated, expands, and becomes less dense [1]
The warmer, less dense water rises and cooler, denser water sinks to take its place [1]
This sets up a convection current that circulates the water, distributing heat throughout the pot [1]
⚠ If you missed marks here: The banned phrase is "heat rises" — heat doesn't rise, HOT WATER rises, and only because heating makes it expand and become LESS DENSE. Full marks needs the whole loop: name convection, hot water expands/less dense and rises, cooler denser water sinks to replace it, forming a circulating convection current.
Mark 1: Process is convection
Mark 2: Water heated near base expands becomes less dense
Mark 3: Warm water rises cooler denser water sinks to replace it
Mark 4: Convection current circulates distributing heat throughout
(d) 2 marks
Model Answer - Q4(d)
The lid traps steam / water vapour inside the pot, reducing energy lost by evaporation [1]
More of the thermal energy from the heater goes into raising the water temperature rather than being carried away by escaping steam [1]
⚠ If you missed marks here: "The lid keeps the heat in" is too vague for both marks. The key mechanism is EVAPORATION: escaping water vapour carries away large amounts of energy (latent heat), so trapping the steam under the lid keeps that energy in the water — name evaporation/steam for the first mark, the energy staying in the water for the second.
Mark 1: Lid traps steam reduces energy lost by evaporation
Mark 2: More thermal energy raises water temperature instead of escaping
(e) 1 mark
Model Answer - Q4(e)
Eheat = mcΔT = 1.5 × 4200 × 76 = 478 800 J; Eboil = mL = 0.20 × 2.26 × 106 = 452 000 J; Etotal = 478 800 + 452 000 = 930 800 J ≈ 931 kJ [1]
⚠ If you missed marks here: This needs TWO terms added together, and each has its own mass: heat ALL 1.5 kg through ΔT = 100 − 24 = 76 °C (mcΔT), but only 0.20 kg boils away (mL). The classic errors: using 100 instead of 76, or putting 1.5 kg into the mL term (that gives 3.39 MJ — way too big).
Mark 1: Total energy approximately 930000 J or 931 kJ combining heating and boiling
Question 5: Spacecraft Re-entry and Thermal Protection
Total: 12 marks
When a spacecraft re-enters Earth's atmosphere at high speed, friction with air molecules generates extreme heating. The heat shield on the spacecraft's underside is designed to protect the crew. Some heat shields use an ablative material that melts and then vaporises, carrying heat away from the spacecraft.
(a) 4 marks
Model Answer - Q5(a)
When the material melts, energy is absorbed as latent heat of fusion to break bonds between particles [1]
When the liquid vaporises, further energy is absorbed as latent heat of vaporisation to overcome the remaining intermolecular forces [1]
This energy is taken from the surface/surroundings, preventing the temperature from rising further / absorbing heat that would otherwise reach the crew cabin [1]
The vapour/gas carries the absorbed energy away from the spacecraft as it is released into the atmosphere [1]
⚠ If you missed marks here: "It melts so it absorbs heat" compresses four marks into one sentence — the scheme wants each stage: latent heat of FUSION absorbed on melting, latent heat of VAPORISATION absorbed on vaporising, this energy taken WITHOUT the temperature rising (that's what "latent" means), and the vapour physically carrying the energy away. Use the term "latent heat" — "absorbs heat" alone is too weak.
Mark 1: Melting absorbs latent heat of fusion breaking bonds
Mark 2: Vaporisation absorbs latent heat of vaporisation overcoming intermolecular forces
Mark 3: Energy absorbed prevents temperature rising protects crew
Mark 4: Vapour carries absorbed energy away from spacecraft
(b) 2 marks
Model Answer - Q5(b)
During vaporisation, particles must completely separate from each other / all bonds/attractive forces must be overcome [1]
During melting, particles only need to partly overcome the forces / move from fixed positions but remain close together, so less energy is needed [1]
⚠ If you missed marks here: "Vaporising needs more energy" just restates the question — the marks are for WHY in particle terms: vaporisation must COMPLETELY separate the particles (breaking all attractions), while melting only loosens them from fixed positions and they stay close together. You need both halves of the comparison.
Mark 1: Vaporisation requires completely separating particles overcoming all bonds
Mark 2: Melting only partly overcomes forces particles remain close needing less energy
(c) During re-entry, 12 kg of the shield material vaporises. The specific latent heat of vaporisation of the material is 5.0 × 106 J/kg. 3 marks
(c) (i) 2 marks
Model Answer - Q5(c)(i)
E = mL = 12 × 5.0 × 106 [1]
E = 6.0 × 107 J (or 60 MJ or 60 000 000 J) [1]
⚠ If you missed marks here: E = mL = 12 × 5.0 × 106 — the usual slip is the power of ten (getting 6.0 × 106 instead of 6.0 × 107). No temperature change appears in this formula: if you wrote mcΔT anywhere, you mixed up latent heat with specific heat capacity.
Mark 1: Correct substitution E equals mL
Mark 2: Correct answer 60000000 J or 60 MJ
(c) (ii) 1 mark
Model Answer - Q5(c)(ii)
m = E / Lv = 6.0 × 107 / 2.26 × 106 = 26.5 kg (accept 26 to 27 kg) [1]
⚠ If you missed marks here: Two traps: you must switch to WATER's latent heat (2.26 × 106 J/kg), not the shield material's 5.0 × 106 (that gives 12 kg back); and the 200 kg is a red herring — the water is already at 100 °C, so no mcΔT term is needed, just m = E/L.
Mark 1: Correct answer approximately 26 to 27 kg of water converted to steam
(d) 3 marks
Model Answer - Q5(d)
Evaporation occurs at any temperature; boiling occurs only at the boiling point [1]
Evaporation occurs only at the surface; boiling occurs throughout the liquid (bubbles form inside) [1]
Evaporation is a slow, gradual process; boiling is rapid / vigorous (any one from: no bubbles vs bubbles, evaporation causes cooling of remaining liquid) [1]
⚠ If you missed marks here: Each mark needs a PAIRED comparison — "evaporation is slow" alone doesn't score; you must complete it with "...but boiling is rapid". The two most forgotten pairs: evaporation happens at ANY temperature vs boiling only at the boiling point, and evaporation happens only at the SURFACE vs boiling throughout the liquid (bubbles form inside).
Mark 1: Evaporation at any temperature boiling only at boiling point
Mark 2: Evaporation at surface only boiling throughout liquid with bubbles
Mark 3: Evaporation slow and gradual boiling rapid or evaporation causes cooling
Question 6: Desert Animals and Thermal Energy Transfer
Total: 10 marks
The Thar Desert in Rajasthan experiences extreme temperatures: over 50 °C during the day and below 5 °C at night. Animals have evolved adaptations to survive these conditions.
(a) 3 marks
Model Answer - Q6(a)
Large ears have a large surface area for heat loss [1]
Blood flowing near the surface loses heat by radiation to the cooler surroundings / by convection to the air [1]
The cooled blood then returns to the body, reducing the fox's core body temperature [1]
⚠ If you missed marks here: "Big ears lose more heat" is only the first mark — you also need to NAME the transfer processes (radiation to the surroundings and/or convection to the air) and close the loop: the cooled blood flows back into the body, lowering the core temperature. Three marks = surface area + named mechanism + returning blood.
Mark 1: Large surface area for heat loss
Mark 2: Blood near surface loses heat by radiation or convection
Mark 3: Cooled blood returns to body reducing core temperature
(b) 3 marks
Model Answer - Q6(b)
Sweat is water on the skin surface; the most energetic / fastest-moving particles escape from the liquid surface (evaporate) [1]
These escaping particles take energy / latent heat of vaporisation away from the skin [1]
The average kinetic energy of the remaining particles decreases, so the skin temperature drops / the skin is cooled [1]
⚠ If you missed marks here: "Sweat is cold / sweat cools you down" scores nothing — the marks are for the particle story: it's the MOST ENERGETIC particles that escape (evaporate), taking latent heat from the skin, so the AVERAGE kinetic energy of the particles left behind falls — and lower average KE means lower temperature.
Mark 1: Most energetic fastest particles escape from liquid surface evaporate
Mark 2: Escaping particles take energy latent heat away from skin
Mark 3: Average kinetic energy decreases so skin temperature drops cooled
(c) 2 marks
Model Answer - Q6(c)
Soil/earth/sand is a poor conductor of heat, so it acts as an insulator and reduces heat loss from the burrow [1]
The surface loses heat rapidly by radiation to the cold night sky, but the burrow is shielded from this radiative loss / the ground retains thermal energy absorbed during the day [1]
⚠ If you missed marks here: "It's warmer underground" restates the question — the question says "using ideas about thermal energy transfer", so you must name the mechanisms: soil is a poor CONDUCTOR (insulates the burrow), and the open surface loses heat fast by RADIATION to the night sky while the burrow is shielded from that.
Mark 1: Soil is poor conductor insulator reduces heat loss from burrow
Mark 2: Surface loses heat by radiation but burrow shielded or ground retains thermal energy
(d) 2 marks
Outer wall Vacuum Silvered surfaces Insulating lid Hot liquid
Explain how two features of the Thermos flask reduce thermal energy loss. For each feature, state which method of heat transfer is reduced.
Model Answer - Q6(d)
Vacuum between walls: prevents heat loss by conduction and convection (since there are no particles to transfer energy) [1]
Silvered/shiny surfaces: reduce heat loss by radiation (shiny surfaces are poor emitters and good reflectors of infrared radiation) [1]
⚠ If you missed marks here: Each feature must be matched to the RIGHT transfer method: the vacuum stops conduction AND convection (no particles to carry energy) but does NOT stop radiation — that's what the silvered surfaces are for (poor emitters / good reflectors of infrared). Mixing up which feature blocks which method is the classic error.
Mark 1: Vacuum prevents conduction and convection no particles
Mark 2: Silvered surfaces reduce radiation poor emitters good reflectors
Question 7: Experimental Design - Measuring Specific Heat Capacity
Total: 12 marks
A student wants to determine the specific heat capacity of aluminium by the method of mixtures. She plans to heat a block of aluminium and then place it into cold water in an insulated container, measuring the temperature changes.
(a) 5 marks
Model Answer - Q7(a)
Measure the mass of the aluminium block using a balance (mAl) [1]
Heat the aluminium block in boiling water for several minutes so it reaches 100 °C; measure its initial temperature Thot with a thermometer [1]
Measure a known mass of cold water (mw) using a measuring cylinder or balance, and record its initial temperature Tcold with a thermometer, placing it in an insulated (e.g. polystyrene/calorimeter) container [1]
Quickly transfer the hot block into the cold water, stir gently, and record the highest/final temperature Tfinal reached by the water [1]
Use the equation: energy lost by aluminium = energy gained by water, i.e. mAl × cAl × (Thot - Tfinal) = mw × cw × (Tfinal - Tcold), rearrange for cAl [1]
⚠ If you missed marks here: The marks most often dropped: forgetting to measure the MASSES (both block and water), not saying the container is INSULATED, and never stating the key physics — energy lost by aluminium = energy gained by water. A method that lists apparatus but never says which measurements to record, or how c is calculated from them, caps out at 2-3 marks.
Mark 1: Measure mass of aluminium block using balance
Mark 2: Heat block in boiling water to 100 degrees measure initial hot temperature
Mark 3: Measure mass of cold water record initial cold temperature insulated container
Mark 4: Transfer hot block into cold water stir record final temperature
Mark 5: Energy lost by aluminium equals energy gained by water rearrange for specific heat capacity
(b) 3 marks
Quantity Value
Mass of aluminium block 0.40 kg
Initial temperature of block 100 °C
Mass of water 0.25 kg
Initial temperature of water 18 °C
Final temperature (water + block) 28 °C
Calculate the specific heat capacity of aluminium using these results.
Model Answer - Q7(b)
Energy gained by water = mwcwΔTw = 0.25 × 4200 × (28 - 18) = 0.25 × 4200 × 10 = 10 500 J [1]
Energy lost by aluminium = mAl × cAl × ΔTAl = 0.40 × cAl × (100 - 28) = 0.40 × cAl × 72 [1]
Setting equal: 10 500 = 0.40 × cAl × 72 = 28.8 × cAl; cAl = 10 500 / 28.8 = 365 J/(kg °C) (accept 360 - 370) [1]
⚠ If you missed marks here: The two objects have DIFFERENT temperature changes: the water rises 28 − 18 = 10 °C but the block falls 100 − 28 = 72 °C — using 10 for the aluminium (or 82 = 100 − 18) is the classic slip. Then equate: energy gained by water (10 500 J) = energy lost by block, and solve for c.
Mark 1: Energy gained by water correctly calculated as 10500 J
Mark 2: Correct expression for energy lost by aluminium with temperature change 72
Mark 3: Correct answer approximately 365 J per kg per degree C
(c) 4 marks
Model Answer - Q7(c)
Heat is lost from the hot block during transfer from the boiling water to the cold water, so the block enters at a temperature lower than 100 °C. This means less energy is actually transferred to the water than assumed, giving a calculated cAl that is too low. [1+1]
Heat is lost from the water to the surroundings (even with insulation), so the final temperature is lower than it should be. The measured temperature rise of the water is smaller, meaning less energy appears to be gained, giving a cAl that is too low. [1+1]
Also accept: Some heat is absorbed by the container/calorimeter itself, not just the water, so the energy gained by water alone underestimates total energy transferred, giving cAl too low.
⚠ If you missed marks here: "Human error" or "inaccurate thermometer" scores zero — you need physical heat-loss errors (block cools during transfer, heat escapes to surroundings/container). And each error is only HALF the marks: the other half is stating the direction of the effect and why — here both errors make the calculated value TOO LOW, which is exactly what the question tells you happened (365 vs 900).
Mark 1: Source of error - heat lost during transfer of block or block cools before entering water
Mark 2: This makes calculated specific heat capacity too low
Mark 3: Source of error - heat lost to surroundings or absorbed by container
Mark 4: This makes calculated specific heat capacity too low or measured temperature rise smaller
Score Summary
Topic 2: Thermal Physics - Cambridge Challenge

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Question Breakdown

IGCSE Physics 0625 | Paper 4 Cambridge Challenge | Topic 2: Thermal Physics

Cambridge Assessment International Education

Total: 80 marks | Q1: 12 | Q2: 11 | Q3: 12 | Q4: 11 | Q5: 12 | Q6: 10 | Q7: 12