International GCSE Physics · Pearson Edexcel 4PH1 · Topic 5
Solids, liquids and gases
Original exam-style questions with mark schemes. Questions marked “Paper 2 only” use content assessed only on Paper 2. Notes 5(b) →Notes 5(c) →Notes 5(d) →
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Freefull mark schemes
Original questions in the style of each exam boardEvery answer re-solved independently before publishingCreated by Evan Kotronis with AI assistanceUpdated 13 October 2026 · new questions added regularlyHow every question is checked →
1IGCSE-PHY-5-001Converting Celsius to kelvinPapers 1 & 2Grade 4–51 mark
The temperature inside a domestic freezer is −23 °C. What is this temperature on the Kelvin scale?
Show mark scheme
B (250 K);
A: subtracts 273 instead of adding it (−23 − 273)
C: ignores the minus sign and adds 23 to 273
D: drops the minus sign and does not add 273 (thinks the Kelvin and Celsius scales share the same zero)
2IGCSE-PHY-5-002Pressure acts equally in all directionsPapers 1 & 2Grade 4–51 mark
A small pressure sensor is held at a fixed depth in a tank of still water. The flat face of the sensor is turned to point upwards, then sideways, then downwards. Which statement describes the three pressure readings?
Show mark scheme
C (the reading is the same in all three positions);
A: thinks pressure only acts downwards because it is caused by the weight of water above
B: thinks the water pushes upwards more strongly (confuses pressure with upthrust)
D: thinks pressure in a liquid acts only vertically
4IGCSE-PHY-5-004Practical: density of a regular blockPapers 1 & 2Grade 4–56 marks
A student investigates the density of a rectangular block of candle wax. The diagram shows the measurements of the block. The student measures the mass of the block as 72 g.
(a)
The student measures the mass using a top-pan balance. State what the student should do to the balance before placing the block on it.
(1)
(b)
Calculate the volume of the block in m³.
(2)
(c)
Calculate the density of the wax in kg/m³. Use the formula density = mass ÷ volume
(2)
(d)
Suggest how the student could make the length measurements more accurate.
(1)
Show mark scheme
(a)
set the balance to zero / tare the balance (with nothing on the pan);
Allow
check the balance reads 0.00 g
press the 'zero' button
Do not accept
calibrate (unqualified)
weigh it twice
(b)
evaluation of volume in cm³;
(V =) 8.0 × 5.0 × 2.0 = 80 (cm³)
conversion to m³;
8.0 × 10⁻⁵ (m³)
Allow
0.00008 (m³)
conversion of each length to m first, e.g. 0.080 × 0.050 × 0.020
Do not accept
8.0 × 10⁻² m³ (divided by 1000)
(c)
substitution;
e.g. ρ = 0.072 ÷ 8.0 × 10⁻⁵
evaluation;
(ρ =) 900 (kg/m³)
Allow
ecf from (b)
0.9 g/cm³ converted correctly to 900 kg/m³
Do not accept
0.9 kg/m³
900 000 (mass left in g)
(d)
any one from:
use vernier callipers / micrometer / instrument with higher resolution;
measure each side at several places and take a mean;
view the ruler at eye level / perpendicular to avoid parallax;
Allow
repeat and average the measurements
Do not accept
use a longer ruler
repeat the experiment (with no reference to averaging)
5IGCSE-PHY-5-021Why heating a sealed aerosol can raises the gas pressurePapers 1 & 2Grade 4–51 mark
A sealed aerosol can contains gas. The can is left in strong sunlight and the temperature of the gas increases. The volume of the can does not change. Which statement explains why the pressure of the gas increases?
Show mark scheme
C (the gas molecules move faster, so they hit the walls more often and with more force);
A: thinks the molecules themselves expand when heated (molecules do not change size)
B: thinks heating creates more gas; the can is sealed so the number of molecules is fixed
D: invents an attraction to the walls; pressure is due to collisions of moving molecules
6IGCSE-PHY-5-022Comparing liquid pressure: depth and density both matterPapers 1 & 2Grade 4–51 mark
Four containers each hold a different liquid at rest. The water is in a large, wide tank. The other liquids are in narrow tubes. Liquid / name: water, cooking oil, salt solution, mercury Depth of liquid / m: 0.50, 0.65, 0.48, 0.040 Density of liquid / kg/m³: 1000, 800, 1200, 13600 At the bottom of which container is the pressure difference due to the liquid the greatest?
Show mark scheme
C (salt solution);
A: water is in the widest tank with the largest volume, but pressure difference does not depend on the width or volume of liquid (h × ρ = 500)
B: cooking oil is the deepest liquid, but its density is low (h × ρ = 520 < 576)
D: mercury is by far the densest liquid, but its depth is very small (h × ρ = 544 < 576)
9IGCSE-PHY-5-005Kelvin temperature and molecular kinetic energyPapers 1 & 2Grade 6–71 mark
The temperature of a fixed mass of gas is increased from 27 °C to 327 °C. What happens to the average kinetic energy of the gas molecules?
Show mark scheme
D (increases by a factor of 2);
A: uses Celsius temperatures in the ratio (327 ÷ 27 ≈ 12)
B: thinks kinetic energy is proportional to the square of the Kelvin temperature (2² = 4)
C: confuses the change in average speed (√2 ≈ 1.4) with the change in kinetic energy
10IGCSE-PHY-5-006Water tower: pressure difference and force on a tap washerPapers 1 & 2Grade 6–77 marks
A water tower supplies water to a block of flats, as shown in the diagram. Tap A is on the ground floor. The water surface in the tank is 32 m above tap A. Tap B is on the top floor, 18 m above tap A. density of water = 1000 kg/m³ g = 10 N/kg
(a)
Calculate the pressure difference between the water surface in the tank and tap A.
(2)
(b)
When tap A is closed, a rubber washer of area 1.5 × 10⁻⁴ m² blocks the flow. Calculate the force on the washer due to the pressure difference in (a).
(3)
(c)
Tap B has an identical washer. Explain how the force on the washer in tap B compares with the force on the washer in tap A.
(2)
Show mark scheme
(a)
substitution;
e.g. p = 32 × 1000 × 10
evaluation;
(p =) 320 000 (Pa)
Allow
320 kPa
3.2 × 10⁵ Pa
use of g = 9.8 or 9.81 giving 313 600 / 313 920
(b)
rearrangement;
e.g. F = p × A
substitution;
e.g. F = 320 000 × 1.5 × 10⁻⁴
evaluation;
(F =) 48 (N)
Allow
ecf from (a)
47 N from g = 9.8
Do not accept
2.1 × 10⁹ (divided p by A)
(c)
height/depth of water above tap B is smaller (14 m);
so pressure (difference) at B is smaller and so force is smaller;
11IGCSE-PHY-5-007Compressing air in a sealed syringePapers 1 & 2Grade 6–77 marks
A plastic syringe contains 60 cm³ of air at a pressure of 100 kPa. The nozzle of the syringe is sealed. The plunger is pushed in slowly until the volume of the air is 24 cm³. The temperature of the air stays the same.
(a)
Calculate the new pressure of the air in the syringe.
(3)
(b)
Explain, in terms of the air molecules, why the pressure of the air increases when its volume decreases.
(3)
(c)
Suggest why the plunger must be pushed in slowly.
(1)
Show mark scheme
(a)
substitution;
e.g. 100 × 60 = p₂ × 24
rearrangement;
e.g. p₂ = 100 × 60 ÷ 24
evaluation;
(p₂ =) 250 (kPa)
Allow
250 000 Pa
substitution and rearrangement in either order
Do not accept
40 kPa (inverted ratio)
(b)
(same number of) molecules in a smaller volume / molecules closer together;
molecules collide with the walls more often / more collisions per second;
(so) greater (total) force on each unit area of wall;
Allow
more frequent collisions for MP2
greater force per unit area means greater pressure for MP3
Do not accept
molecules move faster
molecules hit the walls harder
molecules get bigger
(c)
(so that) the temperature of the air stays constant / does not rise;
Allow
allows time for energy to be transferred to the surroundings (so temperature stays the same)
12IGCSE-PHY-5-008Car tyre pressure and temperaturePapers 1 & 2Grade 6–78 marks
Early in the morning the air in a car tyre has a pressure of 220 kPa at a temperature of 7 °C. After a long drive the temperature of the air in the tyre is 42 °C. Assume that the volume and mass of air in the tyre stay constant.
(a)
Convert 7 °C to kelvin.
(1)
(b)
Calculate the pressure of the air in the tyre after the drive.
(3)
(c)
A student says: 'The temperature has increased six times, so the pressure will be six times bigger.' Explain why the student is wrong.
(2)
(d)
Explain why there is a lowest possible temperature, called absolute zero.
(2)
Show mark scheme
(a)
280 (K);
Allow
280.15 K
Do not accept
−266 K
(b)
conversion of 42 °C to 315 K;
substitution;
e.g. 220 ÷ 280 = p₂ ÷ 315
evaluation;
(p₂ =) 248 (kPa)
Allow
247.5 (kPa)
250 kPa
ecf from (a)
Do not accept
1320 kPa (Celsius used)
(c)
pressure is proportional to temperature in kelvin (not °C);
(Kelvin) temperature increases from 280 K to 315 K which is only about 1.1 times / increase is about 12.5%;
Allow
0 °C is not zero temperature / Celsius scale does not start at absolute zero for MP1
Do not accept
pressure does not depend on temperature
(d)
as temperature decreases the (average) kinetic energy / speed of the molecules decreases;
at absolute zero the molecules have no / minimum kinetic energy so they cannot slow down / lose any more energy;
13IGCSE-PHY-5-009Practical: density of steel from ball bearings by displacementPapers 1 & 2Grade 6–79 marks
A student determines the density of steel using small, identical steel ball bearings. Each ball bearing has a volume of about 0.3 cm³. The student uses a measuring cylinder that can be read to 0.2 cm³. These are the student's results. mass of 40 ball bearings = 84.8 g water level before adding ball bearings = 42.0 cm³ water level after adding 40 ball bearings = 52.8 cm³
(a)
Explain why the student uses 40 ball bearings rather than one ball bearing.
(2)
(b)
Calculate the density of steel in kg/m³.
(4)
(c)
State one precaution the student should take when reading the water level in the measuring cylinder.
(1)
(d)
Some air bubbles are trapped between the ball bearings under the water. Explain the effect this has on the calculated density.
(2)
Show mark scheme
(a)
(rise in level for) one ball bearing is too small to measure (accurately) / is close to the resolution of the cylinder;
larger volume reduces the (percentage) uncertainty / makes result more accurate;
Allow
the change in water level would be only about 0.3 cm³ for MP1
Do not accept
to make it a fair test
to repeat the experiment
(b)
volume of 40 ball bearings = 10.8 (cm³);
substitution into density = mass ÷ volume;
e.g. 84.8 ÷ 10.8
evaluation;
7.85 (g/cm³)
conversion to kg/m³;
7850 (kg/m³)
Allow
7852 / 7900 (kg/m³)
converting mass and volume to kg and m³ first: 0.0848 ÷ 1.08 × 10⁻⁵
ecf for incorrect volume
Do not accept
1610 / 1606 (kg/m³) from using 52.8 cm³ as volume
(c)
read the bottom of the meniscus with eye level to the scale;
14IGCSE-PHY-5-011Air bubble rising through a lakePapers 1 & 2Grade 6–79 marks
A diver at a depth of 12 m in a freshwater lake breathes out a bubble of air of volume 3.0 cm³. The bubble rises to the surface. atmospheric pressure at the lake surface = 1.0 × 10⁵ Pa density of water = 1000 kg/m³ g = 10 N/kg Assume the temperature and mass of the air in the bubble stay constant.
(a)
Calculate the total pressure on the bubble at a depth of 12 m.
(3)
(b)
Calculate the volume of the bubble when it reaches the surface.
(3)
(c)
Explain, in terms of the air molecules, why the bubble expands as it rises.
(3)
Show mark scheme
(a)
substitution into p = hρg;
e.g. 12 × 1000 × 10
evaluation of pressure difference;
120 000 (Pa)
addition of atmospheric pressure;
(p =) 220 000 (Pa)
Allow
2.2 × 10⁵ Pa
220 kPa
g = 9.8 giving 217 600 Pa
Do not accept
120 000 Pa as final answer (scores MP1 and MP2 only)
(b)
substitution into p₁V₁ = p₂V₂;
e.g. 220 000 × 3.0 = 100 000 × V₂
rearrangement;
e.g. V₂ = 220 000 × 3.0 ÷ 100 000
evaluation;
(V₂ =) 6.6 (cm³)
Allow
ecf from (a): 120 000 Pa gives 3.6 cm³
6.5 cm³ from g = 9.8
Do not accept
1.4 cm³ (inverted ratio)
(c)
pressure (of the water) outside the bubble decreases (as depth decreases);
pressure of the gas inside is greater than the pressure outside so (the force pushes out and) the bubble expands;
(as volume increases) molecules collide with each unit area of the bubble wall less often so gas pressure decreases (until it equals the outside pressure);
15IGCSE-PHY-5-025Pressure under a narrow shoe heel (cm² to m²)Papers 1 & 2Grade 6–71 mark
A person of weight 540 N balances for a moment with all of her weight on the heel of one shoe. The area of the heel in contact with the floor is 1.2 cm². What is the pressure on the floor under the heel?
Show mark scheme
A (4.5 × 10⁶ Pa);
B: 4.5 × 10⁴ Pa — converts cm² to m² by dividing by 100 instead of 10 000
C: 450 Pa — does not convert cm² to m² (540 ÷ 1.2)
16IGCSE-PHY-5-026Practical: density of engine coolant from a mass–volume graphPapers 1 & 2Grade 6–78 marks
A student investigates the density of a car engine coolant. She places an empty measuring cylinder on a top-pan balance. She pours coolant into the measuring cylinder 10 cm³ at a time and records the total mass shown on the balance after each addition. Volume of coolant / cm³: 0, 10, 20, 30, 40, 50 Total mass / g: 84.0, 94.6, 105.3, 115.8, 126.4, 137.0 The graph shows her results with a line of best fit. The coolant is made by mixing equal volumes of water and antifreeze. density of water = 1000 kg/m³
(a)
State what the total mass at a volume of 0 cm³ represents.
(1)
(b)
Determine the gradient of the line of best fit. Give the unit.
(3)
(c)
Use your answer to (b) to calculate the density of the coolant in kg/m³.
(2)
(d)
Assume that the volume of the coolant is equal to the volume of water plus the volume of antifreeze. Use your answer to (c) to determine the density of the antifreeze.
(2)
Show mark scheme
(a)
mass of the (empty) measuring cylinder;
Do not accept
mass of the coolant
zero error of the balance
(b)
values from a large triangle on the line;
e.g. (137.0 − 84.0) ÷ 50
evaluation in range 1.04 to 1.08;
unit g/cm³;
Allow
g cm⁻³
g per cm³
correct value with no or wrong unit scores 2
Do not accept
0.94 (cm³/g) (volume ÷ mass)
(c)
conversion: × 1000 / 1 g/cm³ = 1000 kg/m³;
(ρ =) 1060 (kg/m³);
Allow
ecf from (b)
1040 to 1080 kg/m³
Do not accept
1.06 × 10⁻³ (divides by 1000)
1.06 × 10⁶ (multiplies by 10⁶)
(d)
idea that 1 m³ of coolant contains 0.5 m³ of water (500 kg) and 0.5 m³ of antifreeze / density of coolant is the mean of the two densities;
17IGCSE-PHY-5-027Hospital saline drip and blood pressure at different heightsPapers 1 & 2Grade 6–77 marks
In a hospital, a bag of saline solution is hung on a stand above a patient's arm, as shown in the diagram. The saline flows down a tube and through a needle into a vein in the patient's arm. density of saline solution = 1020 kg/m³ g = 10 N/kg
(a)
For the saline to flow into the vein, the pressure difference between the liquid surface in the bag and the needle must be at least 9.2 kPa. Calculate the minimum height h.
(3)
(b)
When a person stands up, their feet are 1.3 m below their heart. density of blood = 1060 kg/m³ Calculate the difference in blood pressure between the heart and the feet due to this difference in height.
(2)
(c)
A nurse measures blood pressure with an inflatable cuff wrapped around the upper arm. The cuff should be at the same height as the heart. Explain why the reading would be too high if the cuff were placed lower than the heart.
(2)
Show mark scheme
(a)
rearrangement h = p ÷ (ρ × g);
substitution with 9200 Pa;
e.g. h = 9200 ÷ (1020 × 10)
evaluation;
(h =) 0.90 (m)
Allow
0.902 m
90 cm
rearrangement and substitution in either order
Do not accept
9.0 × 10⁻⁴ (m) (kPa not converted)
1.1 (m) (ρg ÷ p)
(b)
substitution;
e.g. p = 1.3 × 1060 × 10
evaluation;
(p =) 13 800 (Pa)
Allow
13 780 Pa
13.8 kPa
13 500 Pa from g = 9.8
Do not accept
1378 (Pa) (g omitted)
(c)
cuff below heart has a (greater) height/column of blood above it (compared with heart level);
pressure increases with depth / extra pressure difference = hρg so pressure at cuff is greater;
18IGCSE-PHY-5-028Constant-volume gas thermometer: temperature of liquid nitrogenPapers 1 & 2Grade 6–78 marks
A constant-volume gas thermometer contains a fixed mass of helium in a sealed metal bulb. A gauge measures the pressure of the helium. When the bulb is in melting ice at 0 °C, the pressure of the helium is 100.0 kPa. When the bulb is in liquid nitrogen, the pressure of the helium is 28.2 kPa.
(a)
Calculate the temperature of the liquid nitrogen in kelvin.
(3)
(b)
Convert your answer to (a) into °C.
(1)
(c)
Explain why the volume of the helium must be kept constant.
(2)
(d)
State the temperature, in °C, at which the pressure of the helium would be zero. Explain, in terms of the helium particles, why the pressure would be zero at this temperature.
(2)
Show mark scheme
(a)
0 °C = 273 K;
substitution into p₁/T₁ = p₂/T₂;
e.g. 100.0 ÷ 273 = 28.2 ÷ T₂
evaluation;
(T₂ =) 77 (K)
Allow
76.986 K
77.0 K
Do not accept
968 (K) (inverted ratio 273 × 100 ÷ 28.2)
(b)
−196 (°C);
Allow
ecf: their (a) − 273
Do not accept
196 (°C) (minus sign missing)
350 (°C) (adds 273)
(c)
pressure (of a fixed mass of gas) also depends on / changes with volume;
so the pressure would not depend on temperature alone / pressure is only proportional to Kelvin temperature at constant volume;
Allow
if the bulb expanded the pressure would be lower than expected for that temperature
Do not accept
so the gas does not escape
(d)
−273 (°C);
particles have no / minimum kinetic energy / stop moving, so no collisions with the walls / no force on the walls;
19IGCSE-PHY-5-029Smoke cell: evidence for random motion of air moleculesPapers 1 & 2Grade 6–79 marks
A student looks through a microscope at a small glass cell containing air and a little smoke. The smoke particles are seen as tiny bright specks of light. The diagram shows the path of one smoke particle.
(a)
Describe the motion of the smoke particle.
(2)
(b)
Explain why the smoke particle moves in this way.
(3)
(c)
The air in the cell is warmed. Describe and explain how the motion of the smoke particles changes.
(2)
(d)
The temperature of the air increases from 290 K to 319 K. Calculate the percentage increase in the average kinetic energy of the air molecules.
(2)
Show mark scheme
(a)
any two from:
random / unpredictable direction;
sudden / frequent changes of direction;
moves in short straight lines / zig-zag path;
changes speed;
Do not accept
moves in circles
vibrates about a fixed position
(b)
air molecules move randomly / in all directions at high speed;
air molecules collide with / bombard the smoke particle;
collisions on different sides are unequal (at any instant) so the resultant force (and direction of movement) keeps changing;
Allow
reference to Brownian motion is not required (ignore the name)
Do not accept
smoke particles are alive
convection currents carry the smoke
smoke particles collide with each other (as the cause)
(c)
smoke particles move faster / more vigorously / further between changes of direction;
air molecules have a greater average (kinetic energy / speed) so hit the smoke particles harder / more often;
Do not accept
smoke particles get bigger
(d)
average kinetic energy ∝ Kelvin temperature / 319 ÷ 290 (= 1.1);
20IGCSE-PHY-5-030Practical data: volume of trapped air at different pressures (oil column)Papers 1 & 2Grade 6–78 marks
A student uses the apparatus shown to investigate how the volume of a fixed mass of air depends on its pressure. Air is trapped in a glass tube above some oil. The student uses a foot pump to increase the pressure on the oil, which compresses the trapped air. The volume of the trapped air is read from the scale. Pressure / kPa: 100, 125, 150, 200, 250, 300 Volume / cm³: 30.0, 24.0, 20.0, 16.5, 12.0, 10.0
(a)
Identify the anomalous result. Use calculations to support your answer.
(2)
(b)
Calculate the volume of the trapped air when the pressure is 400 kPa.
(2)
(c)
The student wants to show that the pressure is inversely proportional to the volume. Suggest what graph the student should plot, and describe the shape of the graph she should expect.
(2)
(d)
State two variables that the student must keep constant.
(2)
Show mark scheme
(a)
p × V calculated for at least two results (3000);
200 kPa / 16.5 cm³ result identified (p × V = 3300);
Allow
correct identification with no supporting calculation scores 1
identification by other valid reasoning, e.g. pressure doubled from 100 to 200 kPa so volume should halve to 15.0 cm³
(b)
substitution into p₁V₁ = p₂V₂;
e.g. 100 × 30.0 = 400 × V₂
evaluation;
(V₂ =) 7.5 (cm³)
Allow
any non-anomalous pair used
8.25 cm³ (anomalous pair used) scores 1
Do not accept
120 (cm³) (V ∝ p)
(c)
pressure against 1 / volume (or volume against 1 / pressure);
21IGCSE-PHY-5-010Data analysis: density of a liquid from a pressure–depth graphPapers 1 & 2Grade 8–99 marks
A scientist lowers a pressure sensor into a tall tank of salt solution and measures the total pressure at different depths. Depth / m: 0.0, 0.5, 1.0, 1.5, 2.0, 2.5 Pressure / kPa: 101.0, 106.5, 112.0, 120.5, 123.0, 128.5 The graph shows the results with a line of best fit. g = 10 N/kg
(a)
State the pressure at a depth of 0.0 m and explain what this pressure is.
(2)
(b)
Identify the anomalous result.
(1)
(c)
Determine the gradient of the line of best fit in Pa/m.
(2)
(d)
Use your answer to (c) to calculate the density of the salt solution.
(3)
(e)
At a depth of 2.0 m the sensor is turned so that it faces sideways. State the pressure reading.
(1)
Show mark scheme
(a)
101 (kPa);
(pressure of the) atmosphere (acting on the surface of the liquid);
Allow
101 000 Pa
air pressure
Do not accept
zero
pressure of the liquid
(b)
(depth) 1.5 m / 120.5 kPa;
Do not accept
any other value
(c)
correct values from line;
e.g. (128.5 − 101.0) kPa ÷ 2.5 m
evaluation in Pa/m;
11 000 (Pa/m)
Allow
10 800 to 11 200 Pa/m
11 kPa/m with no conversion for MP1 only
Do not accept
gradient using the anomalous point alone
(d)
idea that gradient = density × g;
substitution;
e.g. ρ = 11 000 ÷ 10
evaluation;
(ρ =) 1100 (kg/m³)
Allow
ecf from (c)
1080 to 1120 kg/m³
use of one data pair with atmospheric pressure subtracted, e.g. (123 − 101) × 1000 ÷ (2.0 × 10)
22IGCSE-PHY-5-012Evaluating a claim: is the ring pure gold? (displacement can)Papers 1 & 2Grade 8–97 marks
A jeweller is offered a ring that the seller claims is made of pure gold. density of pure gold = 19 300 kg/m³ density of water = 1000 kg/m³ The jeweller measures the mass of the ring as 13.5 g. She lowers the ring into a displacement can filled with water up to the spout and collects the water that overflows in a beaker. The mass of the water collected is 0.90 g.
(a)
Calculate the density of the ring in kg/m³.
(3)
(b)
The jeweller estimates that her value for the mass of water collected could be too high or too low by up to 0.10 g. Evaluate whether her results show that the ring is not pure gold. You should include a calculation in your answer.
(3)
(c)
Suggest one way the jeweller could reduce the percentage uncertainty in her value for the volume of the ring.
(1)
Show mark scheme
(a)
volume of ring = volume of water displaced = 0.90 cm³ / 9.0 × 10⁻⁷ m³;
substitution into density = mass ÷ volume;
e.g. 0.0135 ÷ 9.0 × 10⁻⁷
evaluation in kg/m³;
15 000 (kg/m³)
Allow
13.5 ÷ 0.90 = 15 g/cm³ correctly converted
ecf for incorrect volume
Do not accept
15 kg/m³
0.067 / 67 (kg/m³) from dividing mass of water by mass of ring
(b)
minimum volume = 0.80 cm³ (from 0.80 g of water);
maximum density = 16 900 kg/m³ (16.9 g/cm³);
(this is) less than 19 300 kg/m³ so the ring is not pure gold;
Allow
for a pure gold ring the displaced water would have to be 0.70 g, which is more than 0.10 g below 0.90 g, for MP1 and MP2
density range 13 500 to 16 900 kg/m³ for MP1 and MP2
argument that water left on the spout or ring makes the collected mass too small, so the volume is at least 0.90 cm³ and the density at most 15 000 kg/m³ (< 19 300 kg/m³), for MP1 and MP2
Do not accept
conclusion that the ring is pure gold
conclusion with no comparison to 19 300 kg/m³
(c)
any one from:
test several identical rings/items together (so a larger mass of water is displaced);
use a smaller displacement can / one with a narrow spout so less water is left behind;
wait until dripping stops and shake the drops off the ring into the beaker;
23IGCSE-PHY-5-013Gas pressure, temperature and absolute zeroPapers 1 & 2Grade 8–96 marks
A student investigates how the pressure of a fixed mass of air depends on its temperature. The air is sealed in a glass flask connected to a pressure gauge. The flask is placed in a water bath and the volume of the air stays constant. The graph shows the student's results. The student has extended the line back to the point where the pressure would be zero.
(a)
Explain, in terms of the air molecules, why the pressure changes as the temperature increases, and how the graph supports the idea of an absolute zero of temperature. You should refer to the Kelvin scale in your answer.
(6)
Show mark scheme
(a)
Indicative content:
MP1. molecules move randomly / at high speed in all directions;
MP2. molecules collide with the walls of the flask and exert a force;
MP3. pressure is the total force per unit area on the walls;
MP4. increasing temperature increases the average kinetic energy / speed of the molecules;
MP5. molecules hit the walls more often / more collisions per second;
MP6. molecules hit the walls harder / with greater force per collision;
MP7. volume constant so increased collision rate is not due to molecules being closer together;
MP8. extended line reaches zero pressure at about −273 °C;
MP9. at this temperature molecules would have no / minimum kinetic energy / stop moving / no collisions;
MP10. so temperature cannot be lower than this: absolute zero;
MP11. absolute zero is 0 K / the zero of the Kelvin scale (K = °C + 273);
MP12. pressure is (directly) proportional to Kelvin temperature / Kelvin temperature is proportional to average kinetic energy;
Allow
award 1 mark per MP to a maximum of 6, subject to the level descriptors
−270 °C to −280 °C for MP8
Do not accept
molecules expand when heated
molecules stop moving at 0 °C
pressure proportional to Celsius temperature
Level 0 (0)
No rewardable content.
Level 1 (1–2)
Isolated points about molecular motion or the graph; little or no link between temperature, collisions and pressure.
Level 2 (3–4)
Explains how increased temperature increases pressure via molecular speed and collisions, OR links the graph intercept to absolute zero with some molecular reasoning; reasoning partly linked.
Level 3 (5–6)
Coherent, linked explanation covering both the molecular cause of the pressure increase (speed, frequency and force of collisions) and how extrapolation to zero pressure at −273 °C (0 K) shows absolute zero.
24IGCSE-PHY-5-031Weather balloon: height at which it burstsPapers 1 & 2Grade 8–97 marks
A weather balloon is filled with 4.0 m³ of helium at ground level, where the atmospheric pressure is 100 kPa. The balloon rises. Its thin rubber skin stretches easily, so the pressure of the helium is always equal to the atmospheric pressure outside the balloon. The balloon bursts when its volume reaches 20 m³. The graph shows how atmospheric pressure changes with height above the ground.
(a)
Assume that the temperature of the helium stays constant. Determine the height above the ground at which the balloon bursts.
(3)
(b)
A student says: 'The atmospheric pressure falls by the same amount for every kilometre that the balloon rises.' Evaluate the student's statement. Use data from the graph in your answer.
(4)
Show mark scheme
(a)
substitution into p₁V₁ = p₂V₂;
e.g. 100 × 4.0 = p₂ × 20
p₂ = 20 (kPa);
height read from graph at their p₂: 12 (km);
Allow
11.6 to 12.4 km
ecf from an incorrect p₂ read correctly from the graph
Do not accept
500 (kPa) (inverted ratio)
about 1.7 km (graph read at 80 kPa = 100 − 20)
(b)
fall in pressure over a low interval read from graph, e.g. 0 to 2 km: 100 → 77 kPa (about 23 kPa);
fall in pressure over a high interval read from graph, e.g. 10 to 12 km: 26 → 20 kPa (about 6 kPa);
the fall per km gets smaller as height increases, so the statement is wrong;
pressure falls by about the same fraction for each equal rise / halves about every 5 km (100 → 50 kPa at ≈ 5.2 km, 50 → 25 kPa at ≈ 10.4 km);
Allow
any correctly read pair of equal height intervals for MP1 and MP2 (readings ± 2 kPa)
the graph is a curve (not a straight line) so the fall is not constant, for MP3 only if linked to data
25IGCSE-PHY-5-032Pressure under a marble columnPapers 1 & 2Grade 8–910 marks
A solid marble column stands upright on flat ground, as shown in the diagram. The column is a cylinder of height 8.0 m and diameter 0.60 m. density of marble = 2700 kg/m³ g = 10 N/kg area of a circle = π × radius² volume of a cylinder = area of circular end × height
(a)
Calculate the mass of the column.
(3)
(b)
Calculate the pressure exerted on the ground by the column.
(4)
(c)
An architect says: 'A column of the same height but twice the diameter would exert only half the pressure on the ground, because the area in contact with the ground would be bigger.' Evaluate the architect's statement.
(3)
Show mark scheme
(a)
volume = π × 0.30² × 8.0 (= 2.26 m³);
substitution into m = ρ × V;
e.g. m = 2700 × 2.26
evaluation;
(m =) 6100 (kg)
Allow
6107 kg
6110 kg
Do not accept
24 400 (kg) (diameter used as radius)
(b)
weight = mass × 10 (= 61 000 N);
area = π × 0.30² (= 0.283 m²);
substitution into p = F ÷ A;
e.g. p = 61 000 ÷ 0.283
evaluation;
(p =) 216 000 (Pa)
Allow
ecf from (a)
2.2 × 10⁵ Pa
215 000 to 217 000 Pa
correct answer with no working scores 4
p = F/A = ρ × A × h × g ÷ A = h × ρ × g = 8.0 × 2700 × 10 (= 216 000 Pa) with this justification scores 4
Do not accept
21 600 (Pa) (mass used instead of weight)
(c)
area (in contact with the ground) increases by a factor of 4 (not 2);
volume / mass / weight increases by a factor of 4;
so pressure (= weight ÷ area) is unchanged, so the architect is wrong;
Allow
calculation for the wider column: mass 24 400 kg, area 1.13 m², pressure 216 000 Pa
pressure = height × density × g, so independent of diameter
26IGCSE-PHY-5-033Sealed bottle squashed after a trip from mountain to beachPapers 1 & 2Grade 8–96 marks
A hiker drinks some water from a plastic bottle at the top of a mountain, then screws the cap on tightly. The bottle now contains only air. At the top of the mountain, the atmospheric pressure is 70 kPa and the temperature is 2 °C. The hiker travels down to a beach, where the atmospheric pressure is 100 kPa and the temperature is 32 °C. When the hiker arrives at the beach, the bottle has been squashed inwards.
(a)
Explain why the bottle is squashed, even though the air inside it is now warmer than it was at the top of the mountain. Your answer should refer to the air molecules inside the bottle and to the air outside the bottle. You may include a calculation.
(6)
Show mark scheme
(a)
Indicative content:
MP1. air molecules inside move randomly and collide with the walls, exerting a force / pressure;
MP2. when sealed at the summit, pressure inside = atmospheric pressure = 70 kPa;
MP3. temperature rise increases the average kinetic energy / speed of the molecules inside;
MP4. so molecules hit the walls more often / harder, increasing the pressure inside (at constant volume);
MP5. pressure ∝ Kelvin temperature: 275 K and 305 K used;
MP6. inside pressure rises only to about 78 kPa (70 × 305 ÷ 275);
MP7. outside / atmospheric pressure rises to 100 kPa, which is a bigger increase / greater than the pressure inside;
MP8. atmospheric pressure acts on the bottle equally in all directions / over all its surface;
MP9. resultant / unbalanced inward force on the walls, so the bottle is squashed;
MP10. as the volume decreases the inside pressure increases (molecules hit walls more often) until it (nearly) equals the outside pressure;
Allow
award 1 mark per MP to a maximum of 6, subject to the level descriptors
77.6 kPa for MP6
MP6 implies MP5
Do not accept
air shrinks / molecules get smaller when cold
uses Celsius ratio, e.g. 70 × 32 ÷ 2 = 1120 kPa
the heat makes the plastic soft so it collapses (on its own)
Level 0 (0)
No rewardable content.
Level 1 (1–2)
Isolated statements about molecules hitting the walls, or about pressure inside or outside the bottle; little attempt to link temperature and pressure changes.
Level 2 (3–4)
Explains how either the temperature rise affects the pressure inside (via molecular motion) or how the outside pressure change acts on the bottle; some linking of ideas but the comparison of inside and outside pressure is incomplete.
Level 3 (5–6)
Coherent, linked explanation of both effects: the warming raises the inside pressure only a little (molecular reasoning, Kelvin temperatures or calculation) while the outside pressure rises more, so there is a resultant inward force that squashes the bottle.
29IGCSE-PHY-5-034Temperature rise of a copper blockPaper 2 onlyGrade 4–51 mark
A 0.50 kg copper block is heated. 7700 J of energy is transferred to the block. specific heat capacity of copper = 385 J/kg °C What is the increase in temperature of the block?
Show mark scheme
D (40 °C);
A: 10 °C — multiplies by the mass instead of dividing (7700 × 0.50 ÷ 385)
B: 20 °C — ignores the mass (7700 ÷ 385)
C: 1.5 × 10⁶ °C — multiplies all three quantities (0.50 × 385 × 7700)
30IGCSE-PHY-5-035Gallium: particles in solids, liquids and gasesPaper 2 onlyGrade 4–57 marks
Gallium is a metal with a melting point of 30 °C. A gallium spoon is solid at a room temperature of 20 °C but melts when it is held in a warm hand. The diagrams X, Y and Z show the arrangement of particles in a solid, a liquid and a gas, but not in that order.
(a)
State which diagram, X, Y or Z, shows the arrangement of the particles in gallium at 20 °C.
(1)
(b)
Describe the arrangement and motion of the particles in diagram Z.
(2)
(c)
Explain why the gallium melts when it is held in a warm hand.
(3)
(d)
Convert 30 °C to kelvin.
(1)
Show mark scheme
(a)
Y;
Do not accept
Z
(b)
close together / touching but randomly / irregularly arranged;
move around / slide past each other;
Do not accept
vibrate about fixed positions
far apart
move very fast in straight lines
(c)
energy (thermally) transferred from the hand to the gallium / energy stored in gallium increases;
temperature of gallium rises to its melting point (hand is above 30 °C);
particles vibrate more and break free from fixed positions / overcome the forces between them;
Allow
particles gain enough energy to move past each other for MP3
31IGCSE-PHY-5-016Practical: cooling curve for salolPaper 2 onlyGrade 6–78 marks
A student heats a test tube of solid salol (phenyl salicylate) in a hot water bath until it has all melted. The student then removes the test tube from the water bath and records the temperature of the salol every 30 s as it cools. The graph shows the student's results.
(a)
Use the graph to determine the melting point of salol.
(1)
(b)
The student says: 'Between 180 s and 480 s the salol is not losing any energy because its temperature is not changing.' Explain why the student is wrong.
(3)
(c)
Describe how the arrangement and motion of the salol particles change when the salol changes from a liquid to a solid.
(2)
(d)
Give two things the student should do to make sure the temperature readings are accurate.
(2)
Show mark scheme
(a)
43 (°C);
Allow
42 to 44 °C
Do not accept
70 °C
(b)
salol is (still) at a higher temperature than the surroundings so energy is transferred to the surroundings;
salol is changing state / freezing / solidifying;
energy released as (forces/bonds between) particles form a fixed arrangement balances the energy lost, so temperature stays constant (while energy stored decreases);
Allow
energy stored in the system decreases for MP3
freezing releases energy for MP3
Do not accept
salol is gaining energy from the surroundings
heat is trapped
(c)
arrangement changes from irregular/random to a regular pattern (particles stay close together);
motion changes from moving past each other/sliding to vibrating about fixed positions;
Allow
particles stay close together but become fixed in place for MP1
Do not accept
particles stop moving
particles move closer together a lot / from far apart
(d)
any two from:
keep the thermometer bulb fully in the salol / not touching the glass;
stir the salol while it is liquid;
read the thermometer at eye level / perpendicular to the scale;
take readings at exact 30 s intervals using a stopwatch;
use a thermometer with a higher resolution / temperature probe and data logger;
Allow
any other valid procedure that improves accuracy of temperature readings
32IGCSE-PHY-5-017Heating water in an aluminium saucepanPaper 2 onlyGrade 6–78 marks
An aluminium saucepan of mass 0.80 kg contains 1.5 kg of water at 18 °C. The saucepan is heated on a 2.0 kW electric hob until the water and the saucepan reach 100 °C. specific heat capacity of water = 4200 J/kg °C specific heat capacity of aluminium = 900 J/kg °C
(a)
Calculate the energy needed to raise the temperature of the water to 100 °C.
(2)
(b)
Calculate the energy needed to raise the temperature of the saucepan to 100 °C.
(2)
(c)
Calculate the minimum time needed to heat the water and saucepan to 100 °C.
(3)
(d)
The actual time taken is longer than your answer to (c). Give one reason for this.
(1)
Show mark scheme
(a)
substitution;
e.g. ΔQ = 1.5 × 4200 × 82
evaluation;
(ΔQ =) 516 600 (J)
Allow
520 000 J
517 kJ
Do not accept
630 000 J (ΔT = 100 °C)
(b)
substitution;
e.g. ΔQ = 0.80 × 900 × 82
evaluation;
(ΔQ =) 59 040 (J)
Allow
59 000 J
59 kJ
Do not accept
72 000 J (ΔT = 100 °C)
(c)
total energy = 575 640 J;
substitution into power = work done ÷ time;
e.g. t = 575 640 ÷ 2000
evaluation;
(t =) 288 (s)
Allow
ecf from (a) and (b)
290 s / 4.8 minutes
258 s if only water heated (ecf from (a) only scores MP2 and MP3)
Do not accept
288 000 s (kW not converted)
(d)
energy transferred to the surroundings / air / hob / some water evaporates;
33IGCSE-PHY-5-018Practical: specific heat capacity of an aluminium blockPaper 2 onlyGrade 6–79 marks
A student uses the apparatus shown to determine the specific heat capacity of aluminium. The joulemeter measures the energy transferred to the heater. These are the student's results. mass of block = 1.0 kg initial temperature = 21.0 °C final temperature = 36.0 °C joulemeter reading at start = 0.0 kJ joulemeter reading at end = 15.6 kJ
(a)
Calculate the specific heat capacity of aluminium from the student's results.
(3)
(b)
The accepted value for the specific heat capacity of aluminium is 900 J/kg °C. Explain why the student's value is higher than the accepted value.
(2)
(c)
Suggest two changes to the method that would give a more accurate value.
(2)
(d)
The student continues heating the block for a much longer time. Explain why the temperature of the block rises more slowly as the block gets hotter.
(2)
Show mark scheme
(a)
substitution into ΔQ = m × c × ΔT;
e.g. 15 600 = 1.0 × c × 15.0
rearrangement;
e.g. c = 15 600 ÷ (1.0 × 15.0)
evaluation;
(c =) 1040 (J/kg °C)
Allow
1000 J/kg °C
substitution and rearrangement in either order
Do not accept
1.04 (kJ not converted) unless unit kJ/kg °C given
433 J/kg °C (final temperature used as ΔT)
(b)
energy (from the heater) is transferred to the surroundings / not all energy heats the block;
so the temperature rise is smaller (than if no energy were lost) so calculated c is larger;
Allow
thermometer/heater/air also heated for MP1
Do not accept
the block is not pure aluminium
human error
joulemeter is broken
(c)
any two from:
insulate the block (with lagging / foam);
put oil / water in the thermometer hole (for good thermal contact);
put oil / water in the heater hole;
record the highest temperature reached after the heater is switched off;
place the block on an insulating mat / heatproof mat;
Allow
start below room temperature and finish the same amount above it
Do not accept
repeat the experiment (unqualified)
use a bigger block (unqualified)
(d)
the temperature difference between the block and the surroundings increases so the rate of energy transfer to the surroundings increases;
so less of the energy supplied each second raises the temperature of the block;
34IGCSE-PHY-5-020Evaporative cooling in a clay water jugPaper 2 onlyGrade 6–711 marks
In hot countries, drinking water is often kept in unglazed clay jugs. A little water seeps through the clay and evaporates from the outside surface of the jug. A clay jug contains 5.0 kg of water. Over 3 hours the temperature of the water falls from 30 °C to 22 °C. specific heat capacity of water = 4200 J/kg °C
(a)
Describe two differences between evaporation and boiling.
(2)
(b)
Explain, in terms of particles, why evaporation causes the water left in the jug to cool.
(3)
(c)
Calculate the energy transferred from the water in the jug as it cools.
(2)
(d)
Calculate the average rate at which energy is transferred from the water during the 3 hours. Give the unit.
(2)
(e)
Explain why the water in the jug cools faster on a dry, windy day.
(2)
Show mark scheme
(a)
evaporation (only) from the surface but boiling throughout the liquid;
evaporation at any temperature but boiling (only) at the boiling point;
Allow
boiling produces bubbles (of vapour) in the liquid for MP1
evaporation happens below the boiling point for MP2
Do not accept
evaporation needs no energy
evaporation is slower (unqualified)
(b)
(some of the) faster / higher-energy particles (near the surface) escape;
(so) the remaining particles have a lower average kinetic energy / speed;
(lower average kinetic energy means) a lower temperature;
Allow
particles with enough energy overcome the attractive forces for MP1
Do not accept
the slowest particles escape
cold particles are left
particles lose heat to the clay (unqualified)
(c)
substitution;
e.g. ΔQ = 5.0 × 4200 × 8
evaluation;
(ΔQ =) 168 000 (J)
Allow
1.7 × 10⁵ J
168 kJ
Do not accept
630 000 J (ΔT = 30 °C)
462 000 J (ΔT = 22 °C)
(d)
substitution with time in seconds;
e.g. P = 168 000 ÷ (3 × 60 × 60)
evaluation with unit;
(P =) 15.6 W
Allow
ecf from (c)
16 W
15.6 J/s
Do not accept
56 000 (time in hours)
933 (time in minutes)
(e)
wind removes water vapour near the surface / dry air contains little water vapour;
so fewer escaped particles return to the liquid / rate of evaporation increases (so energy is removed faster);
35IGCSE-PHY-5-036Electric storage heater: bricks compared with waterPaper 2 onlyGrade 6–77 marks
An electric storage heater contains 90 kg of clay bricks. Overnight, the bricks are heated from 15 °C to 650 °C. During the day, the bricks release this energy to warm a room. specific heat capacity of clay brick = 840 J/kg °C specific heat capacity of water = 4200 J/kg °C
(a)
Calculate the energy stored in the bricks as they are heated.
(2)
(b)
A designer suggests storing the same amount of energy in a tank of water instead. The water can only be heated from 15 °C to 95 °C. Calculate the mass of water needed.
(3)
(c)
density of clay brick = 1900 kg/m³ density of water = 1000 kg/m³ Use calculations to explain one advantage of using bricks rather than water in the storage heater.
(2)
Show mark scheme
(a)
substitution with ΔT = 635 (°C);
e.g. ΔQ = 90 × 840 × 635
evaluation;
(ΔQ =) 4.8 × 10⁷ (J)
Allow
48 006 000 J
48 MJ
Do not accept
4.9 × 10⁷ (J) (uses 650 °C as ΔT)
(b)
ΔT = 80 (°C) used;
rearrangement / substitution;
e.g. m = 4.8 × 10⁷ ÷ (4200 × 80)
evaluation;
(m =) 143 (kg)
Allow
ecf from (a)
140 to 143 kg
142.9 kg
Do not accept
120 (kg) (uses 95 °C as ΔT)
762 (kg) (uses 15 °C as ΔT)
(c)
volumes calculated: bricks 0.047 m³ and water 0.14 m³ (ecf);
conclusion: bricks take up much less space (about one third) so heater can be smaller;
Allow
alternative MP2: bricks can be heated far above 100 °C without boiling, so a small mass stores a lot of energy (if linked to a calculation)
Do not accept
bricks have a higher specific heat capacity than water
36IGCSE-PHY-5-037Practical: cooling data for molten tin (thermocouple)Paper 2 onlyGrade 6–77 marks
A technician melts some pure tin in a small steel crucible, then lets it cool in the air of the workshop. She measures the temperature of the tin every 30 s using a thermocouple connected to a data logger. Time / s: 0, 30, 60, 90, 120, 150, 180, 210, 240, 270, 300, 330, 360 Temperature / °C: 300, 281, 264, 247, 232, 232, 232, 232, 232, 215, 199, 184, 171
(a)
Use the data to determine the melting point of tin.
(1)
(b)
State the physical state, or states, of the tin at 180 s.
(1)
(c)
The technician repeats the experiment with twice the mass of tin in a similar crucible. Describe and explain how the part of the results where the temperature stays constant would be different.
(3)
(d)
The laboratory's liquid-in-glass thermometers have a range of −10 °C to 110 °C. Give two reasons why the technician uses a thermocouple and data logger instead of one of these thermometers.
(2)
Show mark scheme
(a)
232 (°C);
Do not accept
300 (°C)
171 (°C)
(b)
solid and liquid (both present);
Allow
solidifying / freezing / changing from liquid to solid
Do not accept
solid only
liquid only
gas
(c)
(constant) temperature is the same / 232 °C;
constant temperature lasts longer / about twice as long (about 240 s);
twice as much energy must be transferred (to the surroundings) for all of the tin to change state / solidify;
Allow
more particles must form a fixed arrangement / bonds, so more energy is released, for MP3
Do not accept
melting point is higher because there is more tin
lasts the same time
(d)
any two from:
temperatures (up to 300 °C) are above the range of the thermometer / thermometer would break;
readings taken automatically at exactly 30 s intervals / more often;
higher resolution / reads to 0.1 °C;
no human reading / parallax errors;
technician does not need to stay close to the very hot crucible;
37IGCSE-PHY-5-019Heating curve of ice: power and specific heat capacity from gradientsPaper 2 onlyGrade 8–911 marks
0.25 kg of crushed ice at −20 °C is put into an insulated container. An electric heater supplies energy to the ice at a constant rate. The graph shows how the temperature changes with time. specific heat capacity of water = 4200 J/kg °C Assume all the energy from the heater is transferred to the ice or water.
(a)
Explain why the temperature stays constant between 50 s and 450 s.
(2)
(b)
Use the graph to show that the heater supplies energy at a rate of about 200 W.
(4)
(c)
Determine the energy needed to melt all the ice.
(2)
(d)
Use the graph to calculate the specific heat capacity of ice.
(3)
Show mark scheme
(a)
the ice is melting / changing state (from solid to liquid);
energy supplied is used to break bonds / overcome forces between particles (not to increase their kinetic energy);
Allow
energy increases the potential energy of the particles for MP2
Do not accept
the heater is switched off
no energy is being supplied
the ice has reached its boiling point
(b)
values from graph for the water section;
e.g. 40 °C rise in 200 s
substitution into ΔQ = m × c × ΔT;
e.g. 0.25 × 4200 × 40 (= 42 000 J)
division by time;
e.g. 42 000 ÷ 200
evaluation to more than 1 s.f.;
(P =) 210 (W)
Allow
use of gradient 0.20 °C/s: 0.25 × 4200 × 0.20 = 210 W
any correct pair of values from the water line
Do not accept
200 W with no working
answer of 200 W without an evaluated value to more s.f.
38IGCSE-PHY-5-038Practical: specific heat capacity of water using a kettle and an energy–mass graphPaper 2 onlyGrade 8–910 marks
A student uses an electric kettle to determine the specific heat capacity of water. The power of the kettle is 2.2 kW. She puts a measured mass of water at 20 °C into the kettle, switches it on and measures the time taken for the temperature of the water to reach 60 °C. She empties the kettle and lets it cool to 20 °C between readings. Mass of water / kg: 0.40, 0.60, 0.80, 1.00, 1.20 Time taken / s: 45, 61, 76, 92, 107 She calculates the energy supplied by the kettle for each mass. The graph shows energy supplied against mass of water, with a line of best fit.
(a)
Show that the energy supplied when the mass of water is 1.00 kg is about 200 kJ.
(2)
(b)
Determine the gradient of the line of best fit.
(2)
(c)
Use your answer to (b) to determine the specific heat capacity of water.
(3)
(d)
The line of best fit does not pass through the origin. Explain what the intercept on the energy axis represents, and why the gradient gives a better value for the specific heat capacity of water than a calculation using just one of her results.
(3)
Show mark scheme
(a)
substitution into E = P × t;
e.g. E = 2200 × 92
evaluation to at least 3 s.f.;
(E =) 202 400 (J) / 202 (kJ)
Allow
2.2 × 92 = 202.4 (kJ)
Do not accept
200 kJ with no calculation
202 (J) (kW not converted, unless unit kJ given)
(b)
values from a large triangle on the line;
e.g. (236 − 31) ÷ 1.20
evaluation in range 165 to 176 (kJ/kg);
Allow
165 000 to 176 000 J/kg
unit not required
Do not accept
196 (kJ/kg) (235.4 ÷ 1.20, line taken through origin)
(c)
idea that gradient = energy per kg = c × ΔT / ΔQ ÷ m;
ΔT = 40 (°C) and gradient converted to J/kg (× 1000);
e.g. c = 170 500 ÷ 40
evaluation;
(c =) 4300 (J/kg °C)
Allow
ecf from (b)
4100 to 4400 J/kg °C
4260 J/kg °C
Do not accept
4.3 (J/kg °C) (kJ not converted)
2840 (J/kg °C) (uses 60 °C as ΔT)
(d)
intercept (≈ 31 kJ) is energy supplied that does not heat the water, e.g. energy needed to heat the kettle (element and body);
this energy is (about) the same for every mass of water;
so it does not change the gradient, but a single result includes it, giving a value of c that is too large (e.g. 0.60 kg gives about 5600 J/kg °C);
Allow
energy transferred to the surroundings for MP1
the intercept is the energy supplied when there is no water (for MP1)
Do not accept
the intercept is the energy in the water at the start
the line should go through the origin so the results are wrong
39IGCSE-PHY-5-039Mixing hot and cold water in a bathPaper 2 onlyGrade 8–99 marks
A bath contains 60 kg of cold water at 15 °C. Hot water at 55 °C is added until the temperature of all the water in the bath is 38 °C. specific heat capacity of water = 4200 J/kg °C
(a)
Assume that no energy is transferred to the bath or to the surroundings. Calculate the mass of hot water that must be added.
(4)
(b)
The hot water was heated from 15 °C to 55 °C by a gas boiler with an efficiency of 90%. Calculate the energy supplied by the gas to heat this hot water.
(3)
(c)
In practice, more hot water must be added than your answer to (a). Explain why.
(2)
Show mark scheme
(a)
idea that energy lost by hot water = energy gained by cold water;
temperature changes 17 (°C) and 23 (°C);
substitution;
e.g. m × 4200 × 17 = 60 × 4200 × 23
evaluation;
(m =) 81 (kg)
Allow
81.2 kg
80 kg
Do not accept
44 (kg) (temperature changes swapped)
(b)
energy gained by hot water = 81.2 × 4200 × 40 (= 1.36 × 10⁷ J);
use of efficiency: ÷ 0.90;
(E =) 1.5 × 10⁷ (J);
Allow
ecf from (a)
1.52 × 10⁷ J
15 MJ
Do not accept
1.2 × 10⁷ (J) (multiplies by 0.90)
(c)
energy (from the hot water) is transferred to the bath / air / surroundings;
so less energy goes to the cold water / final temperature would be lower than 38 °C (so more hot water needed);
Do not accept
the hot water has less specific heat capacity
hot water evaporates (without link to energy transfer)
40IGCSE-PHY-5-040Why sand heats up faster than the seaPaper 2 onlyGrade 8–99 marks
On a sunny day, each square metre of a beach absorbs energy from the Sun at a rate of 400 W. The energy is absorbed by the top 0.10 m of sand. density of dry sand = 1600 kg/m³ specific heat capacity of dry sand = 830 J/kg °C
(a)
Calculate the energy absorbed by 1.0 m² of the beach in 1 hour.
(2)
(b)
Calculate the increase in temperature of the top 0.10 m of sand under 1.0 m² of the beach during this hour. Assume that no energy is transferred away from the sand.
(4)
(c)
Each square metre of the sea next to the beach absorbs energy at the same rate. The energy is spread through the top 2.0 m of the sea by waves and currents. density of sea water = 1030 kg/m³ specific heat capacity of sea water = 3900 J/kg °C Explain why the temperature of the sea rises much less than the temperature of the sand. You should include a calculation in your answer.
(3)
Show mark scheme
(a)
substitution with t = 3600 s;
e.g. E = 400 × 3600
evaluation;
(E =) 1.44 × 10⁶ (J)
Allow
1 440 000 J
1.4 MJ
Do not accept
24 000 (J) (t = 60 s)
400 (J) (t = 1 s)
(b)
volume = 1.0 × 0.10 = 0.10 m³;
mass = 1600 × 0.10 = 160 (kg);
substitution into ΔQ = m × c × ΔT;
e.g. 1.44 × 10⁶ = 160 × 830 × ΔT
evaluation;
(ΔT =) 11 (°C)
Allow
ecf from (a)
10.8 °C
volume used as mass with no other error scores 2 (MP1 and MP3)
Do not accept
17349 (°C) (volume used as mass)
(c)
sea water has a (much) greater specific heat capacity (so more energy needed per kg per °C);
energy spread through a (much) greater mass of water (2060 kg compared with 160 kg);
calculated temperature rise of the sea ≈ 0.18 °C (so much smaller than for the sand);
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