International GCSE Physics · Pearson Edexcel 4PH1 · Topic 4
Energy resources and energy transfers
171 free, high-quality past-paper-style questions on Energy resources and energy transfers for Edexcel IGCSE Physics, each with a full mark scheme. Covers: (b) Energy Transfers; (c) Work and Power; (d) Energy Resources and Electricity Generation. For when you have finished the past papers, or want to practise one topic at a time.Notes 4(b) →Notes 4(c) →Notes 4(d) →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.902 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · every answer re-solved before publishing · How questions are checked
1IGCSE-PHY-4-0014(b) Energy TransfersPapers 1 & 2Grade 4–51 mark
An archer pulls back the string of a bow and holds it still. The archer then releases the string and the arrow is fired horizontally. Which of these describes the main energy transfer as the arrow is fired?
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C (elastic energy store of the bow → kinetic energy store of the arrow);
A is incorrect: the chemical store of the archer was used earlier, while pulling the string back, not as the arrow is fired
B is incorrect: this is the reverse of the transfer that happens
D is incorrect: the arrow is fired horizontally so its height (and gravitational store) does not change significantly
3IGCSE-PHY-4-0034(b) Energy TransfersPapers 1 & 2Grade 4–56 marks
A child turns the key of a clockwork toy car. Turning the key winds up a spring inside the car. The child puts the car on a carpet and lets it go. The car moves forwards, then slows down and stops.
(a)
Name the energy store that decreases in the child as the child turns the key.
(1)
(b)
Name the energy store that increases in the spring as it is wound up.
(1)
(c)
State how energy is transferred from the spring to the kinetic energy store of the car. Choose from: electrically, mechanically, by heating, by radiation.
(1)
(d)
Describe what happens to the energy in the kinetic energy store of the car as it slows down on the carpet.
(2)
(e)
State the principle of conservation of energy.
(1)
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(a)
chemical;
Do not accept
kinetic
food (without chemical)
(b)
elastic (potential);
Allow
strain
Do not accept
kinetic
spring energy
(c)
mechanically;
Allow
by a force (doing work)
Do not accept
electrically
(d)
kinetic (energy) store (of car) decreases / energy transferred out of kinetic store;
(energy transferred) to thermal store of surroundings/carpet/car (by friction);
Allow
MP2: allow sound transferred to surroundings
MP2: allow 'heats up the carpet/wheels'
Do not accept
energy is lost / used up / destroyed
energy turned into friction
(e)
energy cannot be created or destroyed (only transferred/stored);
Allow
total energy (of a closed system) stays constant
Do not accept
energy is never lost (alone, without reference to creation/destruction or total)
4IGCSE-PHY-4-0044(c) Work and PowerPapers 1 & 2Grade 4–57 marks
A gardener pushes a wheelbarrow along a level path. The gardener applies a constant horizontal force of 85 N to the wheelbarrow and it moves 24 m in the direction of the force. The journey takes 16 s.
(a)
Calculate the work done by the gardener on the wheelbarrow.
(2)
(b)
State the amount of energy transferred by the gardener to the wheelbarrow.
(1)
(c)
Describe what is meant by power.
(1)
(d)
Calculate the power of the gardener in pushing the wheelbarrow. Give the unit.
(3)
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(a)
substitution;
evaluation;
e.g.
(W =) 85 × 24
(W =) 2040 (J)
Allow
2000 (J)
(b)
2040 J / same as (a);
Allow
ECF from (a)
(c)
(power is) the rate of transfer of energy / rate of doing work;
Allow
energy transferred per second / work done per unit time
5IGCSE-PHY-4-0054(b) Energy TransfersPapers 1 & 2Grade 4–56 marks
The Sankey diagram shows the energy transfers in an electric toothbrush in a short time. The diagram is drawn to scale on a grid. The width of each arrow is proportional to the energy it represents. The total electrical energy input is 50 J.
(a)
Use the diagram to determine the energy transferred to the kinetic energy store of the brush head.
(1)
(b)
Use the diagram to determine the energy transferred as sound.
(1)
(c)
State the relationship between efficiency, useful energy output and total energy output.
(1)
(d)
Calculate the efficiency of the toothbrush.
(2)
(e)
Explain why the toothbrush can never have an efficiency of 100%.
(1)
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(a)
20 J;
(b)
10 J;
(c)
efficiency = useful energy output ÷ total energy output (× 100%);
Allow
total energy input for total energy output
useful power output ÷ total power output/input
standard symbols
useful ÷ total (× 100%)
Do not accept
total ÷ useful
useful ÷ wasted
(d)
substitution;
evaluation;
e.g.
(efficiency =) 20 ÷ 50 (× 100%)
(efficiency =) 40 (%)
Allow
0.4
ECF from (a)
Do not accept
2.5 / 250% (inverted)
(e)
(some) energy is always (wasted/dissipated) to thermal store of surroundings / as sound (because of friction);
Allow
idea that not all input energy is transferred usefully
Do not accept
energy is destroyed / lost (without where it goes)
6IGCSE-PHY-4-0214(b) Energy TransfersPapers 1 & 2Grade 4–51 mark
A mobile phone is plugged into a charger connected to the mains supply. Which statement describes the main energy transfer while the phone's battery is charging?
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B (energy is transferred electrically to the chemical energy store of the battery);
A is incorrect because the useful transfer is electrical; heating the battery is wasted energy
C is incorrect because this describes the battery discharging (the phone being used), not charging
D is incorrect because the charger does not transfer energy by radiation and the battery stores energy chemically
8IGCSE-PHY-4-0234(b) Energy TransfersPapers 1 & 2Grade 4–57 marks
A firework rocket is lit. Burning fuel pushes hot gases out of the bottom of the rocket and the rocket accelerates upwards. At the top of its flight the rocket explodes, giving a bright flash of light and a loud bang.
(a)
Name the energy store of the fuel before the rocket is lit.
(1)
(b)
As the rocket accelerates upwards, two energy stores of the rocket increase. Name these two energy stores.
(2)
(c)
Give two ways in which energy is transferred from the explosion to the surroundings.
(2)
(d)
A student says: 'When the firework has finished, all of its energy has been used up.' Explain why the student is wrong.
(2)
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(a)
chemical (energy store);
Do not accept
heat
fuel energy
electrical
(b)
kinetic (energy store);
gravitational (potential energy store);
Allow
GPE / potential energy
thermal (energy store of the rocket) for either mark — max 1 for thermal
Do not accept
elastic
chemical
(c)
any two from:
(by) radiation / light;
sound;
by heating (of the air) / thermal energy;
Allow
infrared for radiation
Do not accept
electrically
mechanically (alone)
(d)
energy cannot be created or destroyed / total energy is conserved;
(energy has been) transferred/dissipated to the (thermal store of the) surroundings / spread out;
10IGCSE-PHY-4-0254(b) Energy TransfersPapers 1 & 2Grade 4–58 marks
The diagram shows a side view of a room. A hot-water radiator is fixed to the wall below the window. Point X is just above the radiator. Point Y is near the ceiling on the opposite side of the room.
(a)
Name the process by which thermal energy passes through the metal wall of the radiator from the hot water inside it.
(1)
(b)
Describe the movement of the air in the room once the radiator has been on for some time. You may refer to points X and Y.
(2)
(c)
The radiator also transfers energy by radiation. Name the type of electromagnetic radiation mainly emitted by the radiator.
(1)
(d)
The radiator is to be repainted. The owner can choose matt black paint or gloss white paint. Explain which paint would make the radiator emit more radiation at the same temperature.
(2)
(e)
A sheet of shiny aluminium foil is fixed to the wall directly behind the radiator. Explain how this reduces unwanted energy transfer.
(2)
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(a)
conduction;
Do not accept
convection
radiation
(b)
(warm) air rises above the radiator / at X (and moves across the ceiling towards Y);
(cooler) air sinks on the far side / near Y and moves back along the floor to the radiator / circulates;
Allow
'convection current' with correct direction for 1 mark if neither MP scored
Do not accept
heat rises
cold air rises
(c)
infrared / IR;
Do not accept
ultraviolet
microwaves
(d)
matt black;
(matt) black / dull surfaces are better emitters (of infrared) (than white/shiny);
Allow
MP2 can be scored from a comparison: gloss white is a poorer emitter
Do not accept
matt black because it absorbs more heat (absorption, not emission)
(e)
(shiny foil) reflects infrared/radiation back into the room;
(so) less energy absorbed by / conducted through the wall (to the outside);
Allow
foil is a poor absorber (of infrared)
Do not accept
foil insulates by trapping air
foil reflects heat (alone) — needs radiation/infrared or into the room
11IGPE-053-00-014(b) Energy TransfersPaper 1PEasy1 mark
A hand-cranked winch on a sailing boat is supplied with 2400 J of energy by the sailor. The winch transfers 1800 J of useful energy to raising the sail. What is the efficiency of the winch?
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Marking point
Mark
Notes
Efficiency = useful energy output ÷ total energy input = 1800 J ÷ 2400 J = 0.75, which is 75 %. 25 % is the fraction wasted, not the useful fraction. 133 % comes from dividing input by output; no device can be more than 100 % efficient. 600 J is the wasted energy, and efficiency has no unit.
✓ 1
Answer B
Answer: B
Common error: Always put the useful output on top. If your efficiency is greater than 1 (or 100 %), you have inverted the fraction.
12IGPE-053-00-054(c) Work and PowerPaper 1PEasy5 marks
In a theatre, an electric motor lifts a piece of scenery vertically above the stage. The weight of the scenery is 1500 N. The motor lifts it through a height of 8.0 m in 12 s at a steady speed.
(a)
State the equation linking work done, force and distance moved.
(1)
(b)
Calculate the work done on the scenery.
(2)
(c)
Calculate the useful power output of the motor.
(2)
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Marking point
Mark
Notes
Part (a)
Work done = force × distance moved (in the direction of the force)
✓ 1
Part (b)
Work done = 1500 × 8.0
✓ 1
= 12 000 J
✓ 1
Part (c)
Power = work done ÷ time = 12 000 ÷ 12
✓ 1
= 1000 W
✓ 1
Exam tip: Carry your answer from one part into the next. If the first value is wrong, a correct method using it can still earn the later marks.
13IGPE2-057-01-014(c) Work and PowerPaper 1PEasy1 mark
A hospital lift raises a patient, a trolley and a nurse, with a total mass of 450 kg, through a vertical height of 12 m in 20 s at constant speed. g = 10 N/kg What is the useful power output of the lift motor?
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Marking point
Mark
Notes
Work done = weight × height = (450 × 10) × 12 = 54 000 J. Power = work done ÷ time = 54 000 ÷ 20 = 2700 W. 270 W uses the mass (450 kg) instead of the weight (4500 N). 54 000 W is the work done, not the work done per second. 1.08 × 106 W multiplies by the time instead of dividing.
✓ 1
Answer B
Answer: B
Common error: Work done lifting an object uses its weight in newtons (mass × g), not its mass in kilograms.
14IGPE2-057-01-044(c) Work and PowerPaper 1PEasy5 marks
An athlete trains on an indoor rowing machine. On each stroke she pulls the handle with an average force of 320 N, in the direction the handle moves, through a distance of 1.4 m.
(a)
State the equation linking work done, force and distance moved.
(1)
(b)
Calculate the work done by the athlete in one stroke.
(2)
(c)
The athlete completes 30 strokes every minute. Calculate her average power output.
(2)
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Marking point
Mark
Notes
Part (a)
work done = force × distance moved (in the direction of the force)
✓ 1
Part (b)
work done = 320 × 1.4
✓ 1
= 448 J
✓ 1
Part (c)
power = (30 × 448) ÷ 60 = 13 440 ÷ 60 OR time per stroke = 60 ÷ 30 = 2 s, so power = 448 ÷ 2
✓ 1
= 224 W
✓ 1
Exam tip: Always convert minutes to seconds before dividing. Power in watts means joules per second.
15IGPE2-057-01-054(b) Energy TransfersPaper 1PEasy4 marks
A battery-powered drone carries a medical package from a hospital to a remote clinic. While climbing, the drone's motors provide a useful power of 150 W to lift the drone and package.
(a)
State the unit of power.
(1)
(b)
During the climb, the motors do 4500 J of useful work. Calculate the time taken for the climb.
(2)
(c)
State the energy store that decreases as the drone climbs.
(1)
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Marking point
Mark
Notes
Part (a)
watt / W (accept J/s)
✓ 1
Part (b)
time = work done ÷ power = 4500 ÷ 150
✓ 1
= 30 s
✓ 1
Part (c)
chemical (energy) store of the battery
✓ 1
Exam tip: Rearrange power = work done ÷ time before substituting. This shows a clear method even if the arithmetic slips.
16IGPE2-057-01-064(c) Work and PowerPaper 1PEasy3 marks
A wheelchair user pushes herself up a long access ramp. She applies an average forward force of 60 N along the ramp and does 540 J of work.
(a)
Calculate the distance she travels along the ramp.
(2)
(b)
The increase in her gravitational potential energy is less than 540 J. Suggest why.
(1)
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Marking point
Mark
Notes
Part (a)
distance = work done ÷ force = 540 ÷ 60
✓ 1
= 9.0 m
✓ 1
Part (b)
any one of the following :
✓ 1
• some work is done against friction / air resistance / rolling resistance, so some energy is transferred to the thermal store of the surroundings
—
• some energy is transferred to her kinetic store if she speeds up
—
Common error: In work done = force × distance, the distance is measured along the direction of the force. Here that means along the ramp, not the vertical height gained.
17IGPE2-057-03-014(b) Energy TransfersPaper 1PEasy1 mark
A battery-powered delivery drone hovers at a constant height above a garden. Which statement correctly describes the main energy transfer while the drone hovers?
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Marking point
Mark
Notes
The drone's height is constant, so its gravitational potential store does not change. Its speed is zero, so its kinetic store does not change. The battery's chemical store still decreases. The rotors push air downwards, which gives the air kinetic energy. Friction and electrical heating in the motors, and the moving air slowing down, transfer energy to the thermal store of the surroundings. Option C is correct. Option A wrongly links hovering with gaining height. Option B confuses the drone not moving with no energy being transferred. Option D treats the drone as if it were rising and slowing down.
✓ 1
Answer C
Answer: C
Common error: No change in height or speed means the drone's own stores are constant. The energy still goes somewhere, so look for what else is being moved or heated.
18IGPE2-057-03-034(b) Energy TransfersPaper 1PEasy1 mark
An astronaut works outside a space station, where there is a vacuum. The outer layer of the spacesuit is shiny and silvery. By which method can energy be transferred between the spacesuit and the surrounding space?
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Mark
Notes
Conduction and convection both need particles. Conduction passes vibrations between neighbouring particles, and convection needs a fluid that can flow. A vacuum has no particles, so only infrared radiation can transfer energy across it. The shiny outer layer is a poor emitter and a good reflector of infrared, which reduces energy transfer by radiation. Options B, C and D all wrongly assume that there is a medium to carry the energy.
✓ 1
Answer A
Answer: A
Common error: "Heat rises" is not an explanation in a vacuum. With no fluid present, there can be no convection current.
19IGPE2-057-03-054(b) Energy TransfersPaper 1PEasy5 marks
A hospital incubator keeps a premature baby warm. An electric heater in the incubator has a power of 250 W.
(a)
State the equation linking power, energy transferred and time taken.
(1)
(b)
The heater is switched on for 12 minutes. Calculate the energy transferred by the heater.
(2)
(c)
The walls of the incubator are made of two layers of clear plastic with a layer of trapped air between them. Explain how this design reduces energy transfer from the incubator to the room.
(2)
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Marking point
Mark
Notes
Part (a)
power = energy transferred ÷ time taken, or P = E/t
✓ 1
Part (b)
time = 12 × 60 = 720 s, and E = 250 × 720
✓ 1
E = 180 000 J (1.8 × 105 J)
✓ 1
Part (c)
air (and plastic) is a poor conductor, so conduction through the walls is reduced
✓ 1
the air is trapped and cannot circulate, so convection currents between the layers are reduced
✓ 1
Common error: Using 12 instead of 720 s gives an answer that is 60 times too small. Always convert minutes to seconds before using P = E/t.
20IGPE2-057-03-064(c) Work and PowerPaper 1PEasy5 marks
On a building site, an electric winch raises a pallet of bricks vertically at a steady speed. The winch exerts an upward force of 1800 N and does 54 kJ of work on the pallet.
(a)
State the equation linking work done, force and distance moved in the direction of the force.
(1)
(b)
Calculate the height through which the pallet is raised.
(2)
(c)
The lift takes 45 s. Calculate the useful power output of the winch.
(2)
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Marking point
Mark
Notes
Part (a)
work done = force × distance moved (in the direction of the force), or W = F × d
✓ 1
Part (b)
d = W ÷ F = 54 000 ÷ 1800
✓ 1
d = 30 m
✓ 1
Part (c)
P = 54 000 ÷ 45
✓ 1
P = 1200 W
✓ 1
Exam tip: Convert kJ to J (54 kJ = 54 000 J) before dividing. Otherwise your answer will be 1000 times too small.
21IGPE2-057-03-124(b) Energy TransfersPaper 1PEasy4 marks
A smartwatch is powered by a small rechargeable battery. When fully charged, the battery stores 4500 J of energy. The watch uses energy at an average rate of 0.050 W.
(a)
Name the energy store of the charged battery.
(1)
(b)
Calculate how long, in hours, the watch can run on a full charge.
(3)
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Marking point
Mark
Notes
Part (a)
chemical store
✓ 1
Part (b)
t = E ÷ P = 4500 ÷ 0.050
✓ 1
t = 90 000 s
✓ 1
t = 90 000 ÷ 3600 = 25 hours
✓ 1
Exam tip: P = E/t gives the time in seconds. Divide by 3600 only at the end to convert to hours.
22IGPE2-057-02-014(d) Energy Resources and Electricity GenerationPaper 1PEasy1 mark
A geothermal power station in a volcanic region pumps water down to hot rocks deep underground. The steam that returns drives a turbine connected to a generator. Which of these shows the main energy transfers in this power station?
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Notes
The hot rocks hold energy in a thermal store. This energy heats water to make steam, and the moving steam makes the turbine spin, so energy goes into the kinetic store of the turbine. The generator then transfers this energy electrically. Option A treats the rocks as a fuel that burns, which is how fossil fuels work. Option C describes hydroelectricity. Option D puts the electrical transfer before the turbine, which is the wrong order.
✓ 1
Answer B
Answer: B
Common error: Geothermal energy is not burned. The first store is the thermal store of the hot rocks, not a chemical store.
23IGPE2-057-02-074(d) Energy Resources and Electricity GenerationPaper 1PEasy5 marks
A football stadium has solar cells covering its roof. At midday in summer they produce 360 kW. For an evening match, the floodlights use 540 kW for 2.0 hours.
(a)
State the useful energy transfer that takes place in a solar cell.
(1)
(b)
State the equation linking power, work done (energy transferred) and time taken.
(1)
(c)
Calculate the energy transferred by the floodlights during the match.
(2)
(d)
Suggest why the solar cells cannot supply the floodlights directly during an evening match.
(1)
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Marking point
Mark
Notes
Part (a)
Energy carried by light is transferred electrically (accept: light energy to electrical energy) .
✓ 1
Part (b)
power = work done ÷ time taken (P = W ÷ t)
✓ 1
Part (c)
E = P × t = 540 000 W × (2.0 × 3600) s
✓ 1
= 3.9 × 109 J (3.888 × 109 J)
✓ 1
Part (d)
Any one of the following :
✓ 1
• there is little or no sunlight in the evening, so the solar cells produce little or no power
—
• even their maximum output (360 kW at midday) is less than the 540 kW the floodlights need.
—
Exam tip: Convert kilowatts to watts and hours to seconds before using P = E ÷ t, so that the energy comes out in joules.
24IGPE2-057-02-114(d) Energy Resources and Electricity GenerationPaper 1PEasy5 marks
A student lists five energy resources: coal, wind, uranium, wood from managed forests, geothermal.
(a)
State what is meant by a renewable energy resource.
(1)
(b)
State two resources from the list that are renewable.
(2)
(c)
Burning wood releases carbon dioxide. Explain why wood from managed forests can still be described as carbon neutral.
(2)
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Marking point
Mark
Notes
Part (a)
A resource that is replaced naturally as fast as it is used, so it will not run out .
✓ 1
Part (b)
Acceptable answers: wind, wood from managed forests, geothermal.
—
First correct resource .
✓ 1
Second correct resource .
✓ 1
Part (c)
New trees are planted to replace the ones that are cut down .
✓ 1
As the new trees grow, they absorb about the same amount of carbon dioxide as was released when the wood was burned, so there is no net increase in atmospheric carbon dioxide .
✓ 1
Common error: Uranium is not renewable, even though nuclear power releases very little carbon dioxide. 'Low-carbon' and 'renewable' do not mean the same thing.
25IGPE2-057-00-014(c) Work and PowerPaper 1PEasy1 mark
A cyclist travelling along a flat road applies the brakes and comes to a stop. Which energy store gains most of the energy that was in the kinetic store of the bicycle and cyclist?
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Notes
Friction between the brake pads and the rims does work against the motion. This transfers energy mechanically and by heating, so the brakes, rims and surroundings warm up. The thermal store gains the energy. The chemical store of the cyclist is not refilled by braking. The road is flat, so the gravitational store does not change. The cables stretch only very slightly and store almost no energy.
✓ 1
Answer C
Answer: C
Common error: A store that decreases, such as the chemical store of the cyclist while pedalling, cannot also be the store that gains the energy. Name a store that has increased.
27IGPE2-057-00-064(b) Energy TransfersPaper 1PEasy5 marks
A cooling fan inside a laptop is supplied with 5.0 J of electrical energy every second. Every second: • 3.2 J is transferred to the kinetic store of the moving air • 1.5 J is transferred to the thermal store of the motor • the rest is transferred as sound.
(a)
Calculate the energy transferred as sound each second.
(1)
(b)
Calculate the efficiency of the fan as a percentage.
(2)
(c)
State the principle of conservation of energy.
(1)
(d)
Describe what eventually happens to the energy that is wasted by the fan.
(1)
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Mark
Notes
Part (a)
5.0 − 3.2 − 1.5 = 0.3 J
✓ 1
Part (b)
Efficiency = 3.2 ÷ 5.0 (× 100)
✓ 1
64%
✓ 1
Part (c)
Energy cannot be created or destroyed; it is only transferred between stores
✓ 1
Part (d)
It is dissipated (spreads out) into the thermal store of the surroundings, which warm up slightly
✓ 1
Exam tip: The useful output of a fan is the kinetic energy of the air, not the sound or the heating of the motor.
28IGPE2-057-00-124(b) Energy TransfersPaper 1PEasy5 marks
An archer pulls back the string of a bow and then releases it. The arrow is fired forwards.
(a)
State the energy store that increases as the archer pulls back the string.
(1)
(b)
Describe the energy transfer that occurs when the string is released.
(2)
(c)
The arrow has a mass of 0.025 kg and leaves the bow at 60 m/s. Calculate the kinetic energy of the arrow.
(2)
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Marking point
Mark
Notes
Part (a)
Elastic (potential) store of the bow
✓ 1
Part (b)
Energy is transferred from the elastic store of the bow
✓ 1
mechanically (by the force of the string) to the kinetic store of the arrow
✓ 1
Part (c)
½ × 0.025 × 602
✓ 1
45 J
✓ 1
Exam tip: When describing a transfer, name the store it comes from, the store it goes to, and how it is transferred (mechanically, electrically, by heating or by radiation).
29IGPE-053-00-084(b) Energy TransfersPaper 1PEasy5 marks
A beekeeper wants to keep a wooden beehive warm in winter. She wraps the hive in panels made of expanded polystyrene, which contains many tiny pockets of trapped air. The outer surface of each panel is covered with shiny aluminium foil.
(a)
State the main method of thermal energy transfer through the solid wooden walls of the hive.
(1)
(b)
Explain why the polystyrene panels reduce energy loss from the hive.
(2)
(c)
Explain why the outer surface of each panel is covered with shiny foil.
(2)
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Notes
Part (a)
Conduction
✓ 1
Part (b)
Air is a poor conductor (good insulator), so conduction through the panel is reduced
✓ 1
The air is trapped in small pockets, so it cannot move freely and convection is reduced
✓ 1
Part (c)
Shiny surfaces are poor emitters of infrared radiation
✓ 1
So less energy is lost from the panel's surface by radiation
✓ 1
Common error: Do not say insulation 'stops the cold getting in'. Energy is transferred from hot to cold, so insulation reduces energy getting out.
31IGPE3-4-14r7-014(b) Energy TransfersPaper 1PEasy1 mark
A library book scanner takes in 800 J of energy in one cycle and wastes 200 J as thermal energy. What is the efficiency of the scanner?
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Mark
Notes
Useful energy = 800 − 200 = 600 J, so efficiency = 600 ÷ 800 = 75%, which is option C. A (25%) is wasted ÷ total. B (33%) is wasted ÷ useful. D (133%) is total ÷ useful, which is inverted; efficiency cannot exceed 100%.
✓ 1
Answer C
Answer: C
Common error: Efficiency is useful output divided by total input, never wasted energy divided by anything.
32IGPE3-4-14r7-034(b) Energy TransfersPaper 1PEasy1 mark
Which outer surface of a spacesuit is best at limiting the thermal energy absorbed from sunlight?
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Mark
Notes
A shiny white surface reflects most of the incident radiation and so absorbs little, option A. B is wrong because a good emitter is also a good absorber, so the suit would heat up in sunlight. C is wrong because strong absorption increases energy gain. D is wrong because conduction through the suit is not the issue, and good conduction would not reduce absorption.
✓ 1
Answer A
Answer: A
Common error: Surfaces that emit well also absorb well; for keeping cool in sunlight, choose a surface that reflects.
35IGPE3-4-14r10-014(b) Energy TransfersPaper 1PEasy1 mark
The motor of a model aircraft transfers 600 J of electrical energy. 150 J is wasted as thermal energy and sound. What is the efficiency of the motor?
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Mark
Notes
Useful energy = 600 − 150 = 450 J, so efficiency = 450 ÷ 600 = 75%. 25% is the fraction wasted, not the useful fraction. 80% comes from dividing 600 by 750, which treats the wasted energy as extra input. 133% inverts the ratio (600 ÷ 450).
✓ 1
Answer B
Answer: B
Common error: Efficiency is useful output divided by total input, so it can never be greater than 100%.
36IGPE3-4-14r10-044(d) Energy Resources and Electricity GenerationPaper 1PEasy1 mark
Which of these energy resources is non-renewable and does not release carbon dioxide while it is generating electricity?
Show mark scheme
Marking point
Mark
Notes
Nuclear fuel (uranium) is a finite resource and fission releases no carbon dioxide in the reactor. Natural gas is non-renewable but releases carbon dioxide when burned. Biofuel is renewable and releases carbon dioxide when burned. Geothermal energy is renewable.
✓ 1
Answer B
Answer: B
Exam tip: Answer in two steps: first decide whether the resource can be replaced, then whether burning it releases carbon dioxide.
37IGPE3-4-14r10-054(b) Energy TransfersPaper 1PEasy6 marks
The diagram below is described in words. An electric bicycle is charged from the mains and then ridden. In the charging stage, the input is 360 kJ of electrical energy from the mains. 288 kJ is stored in the battery and the rest is wasted as thermal energy in the charger. In the riding stage, the battery transfers its 288 kJ to the motor. The motor transfers 201.6 kJ usefully as kinetic energy and gravitational potential energy of the bicycle and rider, and the rest is wasted.
Energy flow diagram (not to scale)
(a)
Calculate the efficiency of the charging stage. Give your answer as a percentage.
(2)
(b)
Calculate the energy wasted by the motor during the riding stage.
(2)
(c)
Calculate the overall efficiency from the mains to the useful output of the bicycle. Give your answer as a percentage.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 288 ÷ 360
✓ 1
A1 80 (%)
✓ 1
Part (b)
M1 288 − 201.6
✓ 1
A1 86.4 (kJ)
✓ 1
Part (c)
M1 201.6 ÷ 360
✓ 1
A1 56 (%)
✓ 1
Exam tip: In a two-stage energy transfer, the overall efficiency is the useful final output divided by the first input, not the average of the two stage efficiencies.
40IGPE3-4-14r18-024(b) Energy TransfersPaper 1PEasy1 mark
A winch motor on a fishing boat receives 2500 J of electrical energy. 700 J is wasted as thermal energy and sound. What is the efficiency of the motor?
Show mark scheme
Marking point
Mark
Notes
Useful output = 2500 − 700 = 1800 J, efficiency = 1800 ÷ 2500 = 72%, so C is correct. A is wasted ÷ input. B is wasted ÷ useful. D is input ÷ useful.
✓ 1
Answer C
Answer: C
Exam tip: Efficiency is always useful output divided by total input, and cannot exceed 100%.
43IGPE3-4-15r33-014(b) Energy TransfersPaper 1PEasy1 mark
Two identical cans each contain the same mass of hot water at the same starting temperature. One can has a matt black surface and the other a shiny silver surface. They are left to cool in the same room. Which statement is correct?
Show mark scheme
Marking point
Mark
Notes
A matt black surface is a better emitter of infrared radiation than a shiny silver surface, so the black can transfers energy to the surroundings faster and cools faster. The first option reverses the roles of the surfaces. The third ignores that the surface affects the rate of radiation even when the masses are equal. The fourth uses the fact that black surfaces absorb well, but in this experiment the water is hotter than the room and the surface is mainly emitting, so good absorbers cool faster here, not slower.
✓ 1
Answer B
Answer: B
Exam tip: In cooling experiments, good absorbers are also good emitters, and the shiny surface is the poorest at both.
44IGPE3-4-15r33-024(d) Energy Resources and Electricity GenerationPaper 1PEasy1 mark
A pump in a data centre cooling system has an input power of 40 W and is 35% efficient. What is the useful output power of the pump?
Show mark scheme
Marking point
Mark
Notes
Useful output power = efficiency × input power = 0.35 × 40 = 14 W. 26 W is the wasted power if the efficiency were 65%, a misreading of the percentage. 114 W comes from dividing the input by the efficiency, which gives an impossible output larger than the input. 5 W comes from subtracting 35 from 40, treating the percentage as a power in watts.
✓ 1
Answer C
Answer: C
Common error: Output cannot exceed input; if your useful output is larger than the input, you have divided instead of multiplied.
45IGPE3-4-15r33-054(b) Energy TransfersPaper 1PEasy6 marks
A student compares how quickly two identical cans of hot water cool. One can is painted matt black and the other is shiny silver. Both cans are filled with water at 90 °C and left in the same room. After 10 minutes the black can is at 62 °C and the silver can is at 74 °C.
(a)
State two variables that the student should keep the same so that the test is fair.
(2)
(b)
Explain why the black can cools faster than the silver can.
(2)
(c)
Calculate the average rate of cooling of the black can, in °C per minute.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
One valid variable kept the same, e.g. starting temperature of the water; volume or mass of the water; size and shape of the cans; position or room temperature; lid or no lid
✓ 1
A second, different valid variable from the same list
✓ 1
Part (b)
The matt black surface is a better emitter / radiator of infrared radiation than the shiny silver surface
✓ 1
so the black can transfers energy to the surroundings at a faster rate
✓ 1
Part (c)
M1 temperature fall = 90 − 62 = 28 °C and rate = fall ÷ time
✓ 1
A1 28 ÷ 10 = 2.8 °C per minute
✓ 1
Exam tip: Name the quantity being kept the same (for example, 'the volume of water') rather than just saying 'the water'.
46IGPE3-4-15r40-014(b) Energy TransfersPaper 1PEasy1 mark
In a practical, two identical metal plates are placed the same distance from a radiant heater. One plate has a matt black front surface and the other has a shiny silver front surface. A temperature sensor is fixed to the back of each plate and the heater is switched on. Which statement describes the results?
Show mark scheme
Marking point
Mark
Notes
The matt black surface is the better absorber of infrared radiation, so more energy is transferred to it each second and its temperature rises faster. The plates do receive the same radiation, but they do not absorb the same fraction of it, so equal rates are wrong. Reflected radiation leaves the silver plate and does not warm it. Emission is small at the start while the plates are cool and the heater is far hotter, so absorption controls the rate of warming.
✓ 1
Answer B
Answer: B
Common error: Do not write that black surfaces 'attract' heat. Say that a matt black surface absorbs more infrared radiation than a shiny silver one.
47IGPE3-4-15r40-054(b) Energy TransfersPaper 1PEasy6 marks
A student investigates how well different surfaces absorb infrared radiation. She fixes a temperature sensor to the back of two identical metal plates. One plate has a matt black front and the other has a shiny silver front. Each plate is placed 10 cm from an infrared heater, and the heater is switched on for 5.0 minutes. The black plate rises from 20 °C to 38 °C. The silver plate rises from 20 °C to 26 °C.
Diagram not to scale
(a)
State two variables that the student should keep the same.
(2)
(b)
Calculate the mean rate of temperature rise of the black plate, in °C per minute.
(2)
(c)
Explain the difference between the temperature rises of the two plates.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Distance of each plate from the heater
✓ 1
Starting temperature of the plates
✓ 1
Accept any two different valid variables, for example size or thickness of the plates, material of the plates, heating time, power of the heater
—
Part (b)
M1 temperature rise = 38 − 20 = 18 °C
✓ 1
A1 18 ÷ 5.0 = 3.6 °C per minute
✓ 1
Part (c)
The matt black surface absorbs more of the infrared radiation than the silver surface
✓ 1
The shiny silver surface reflects more of the infrared radiation, so less energy is transferred to the plate
✓ 1
Exam tip: In a control-variable question, name the quantity and say it is kept the same, for example the distance from the heater.
50IGPE3-4-16r42-014(d) Energy Resources and Electricity GenerationPaper 1PEasy1 mark
A farmer uses several energy resources. Which one is a renewable energy resource?
Show mark scheme
Marking point
Mark
Notes
Biogas is produced from waste that is continually replaced by the animals on the farm, so it is renewable. Diesel and coal are fossil fuels formed over millions of years and cannot be replaced on a human timescale. Uranium is a finite resource that will run out.
✓ 1
Answer B
Answer: B
Common error: Do not assume that a resource is renewable just because it produces little carbon dioxide: nuclear fuel is low-carbon but not renewable.
51IGPE3-4-16r42-034(b) Energy TransfersPaper 1PEasy1 mark
A loudspeaker at an outdoor concert has an electrical power input of 500 W and a sound power output of 40 W. What is the efficiency of the loudspeaker?
Show mark scheme
Marking point
Mark
Notes
Efficiency = useful power output ÷ total power input = 40 ÷ 500 = 0.080, which is 8.0%. The value 12.5 comes from dividing input by output. The value 92% is the percentage of power wasted, not the useful fraction. The value 0.080% forgets to multiply the ratio by 100.
✓ 1
Answer A
Answer: A
Common error: Efficiency is useful output divided by total input, so it can never be more than 100%.
52IGPE3-4-16r42-054(d) Energy Resources and Electricity GenerationPaper 1PEasy4 marks
A farm uses a wind turbine to generate electricity. In one test the kinetic energy of the wind passing the turbine is transferred at a rate of 80 kW and the electrical power output of the turbine is 28 kW.
(a)
State one advantage of using wind turbines to generate electricity.
(1)
(b)
State one disadvantage of using wind turbines to generate electricity.
(1)
(c)
Calculate the efficiency of the wind turbine. Give your answer as a percentage.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
any one from: no fuel needed / renewable / no greenhouse gases or polluting gases produced when operating / low running costs
✓ 1
Part (b)
any one from: depends on the wind so output is not reliable / noise or visual impact / large number needed for large power output
✓ 1
Part (c)
M1 efficiency = useful power output ÷ total power input = 28 ÷ 80
✓ 1
A1 35%
✓ 1
Exam tip: Make sure both powers are in the same unit before dividing; here both are in kW.
54IGCSE-PHY-4-0064(b) Energy TransfersPapers 1 & 2Grade 6–71 mark
An electric motor in a food mixer is supplied with 250 J of electrical energy. 90 J of this energy is wasted as thermal energy and sound. What is the efficiency of the motor?
Show mark scheme
B (64%);
A is incorrect: 90 ÷ 250 × 100 = 36% is the percentage wasted, not the useful percentage
C is incorrect: 250 ÷ 160 × 100 = 156% inverts the ratio (efficiency cannot exceed 100%)
D is incorrect: 160 ÷ 250 = 0.64 is a ratio; writing it with a % sign forgets to multiply by 100
56IGCSE-PHY-4-0084(b) Energy TransfersPapers 1 & 2Grade 6–77 marks
A hotel replaces its filament lamps with LED lamps. Each LED lamp has an input power of 8.0 W and an efficiency of 35%. Each filament lamp has an input power of 60 W and a useful light output power of 3.0 W.
(a)
Calculate the useful light output power of one LED lamp.
(2)
(b)
Calculate the efficiency of one filament lamp.
(2)
(c)
Calculate the energy wasted by one filament lamp when it is switched on for 2.0 hours. Give your answer in kJ.
57IGCSE-PHY-4-0094(b) Energy TransfersPapers 1 & 2Grade 6–79 marks
A student investigates how well different materials conduct thermal energy. She uses four rods of the same length. One end of each rod is placed in a container of hot water. A pin is stuck to the other end of each rod with a small blob of wax. She measures the time taken for each pin to fall off. Rod / material: copper, aluminium, steel, glass Time for pin to fall / s: 38, 55, 160, over 600
(a)
State the independent variable in this investigation.
(1)
(b)
State two variables that the student should control.
(2)
(c)
Describe what the results show about the four materials.
(2)
(d)
Describe how thermal energy is transferred by conduction along the copper rod.
(2)
(e)
Suggest one reason why the time for the steel rod may not be reliable, and how the student could improve this.
(2)
Show mark scheme
(a)
(type of) material (of rod);
Do not accept
time
temperature
(b)
any two from:
thickness/diameter of rod;
length of rod;
amount/mass of wax (or size of wax blob);
(starting) temperature of water;
length of rod in the water;
mass of pins;
Do not accept
material
time
(c)
copper is the best conductor / shortest time;
glass is the worst conductor / insulator / longest time;
Allow
full ranking copper > aluminium > steel > glass scores 2 marks
rate of conduction greatest in copper
Do not accept
copper heats slowest
(d)
particles at the hot end gain (kinetic) energy / vibrate more;
(vibrations) pass energy to neighbouring particles / free electrons carry energy along rod;
Allow
MP2: allow delocalised electrons collide with particles further along
Do not accept
heat rises along the rod
particles move along the rod
(e)
any one reason:
hot water cools down during the (long) time / wax amount varies / judging when pin falls;
matching improvement: keep water at constant temperature (e.g. top up / use kettle/water bath) / repeat and find mean / weigh the wax;
58IGCSE-PHY-4-0104(b) Energy TransfersPapers 1 & 2Grade 6–78 marks
A teacher places a small crystal of potassium permanganate at the bottom of a beaker of cold water, close to one side. The crystal dissolves slowly and colours the water purple. The teacher gently heats the beaker with a small Bunsen flame directly below the crystal.
(a)
Describe the movement of the purple colour that the students should see.
(2)
(b)
Explain why the water above the flame moves in this way.
(3)
(c)
The heating element of an electric kettle is at the bottom of the kettle. Explain why this allows all the water in the kettle to be heated.
(2)
(d)
Suggest why the teacher uses a small crystal and a small flame.
(1)
Show mark scheme
(a)
(purple colour) rises above the flame/crystal;
moves across (top) and down the other side (forming a loop/circulation);
Allow
MP2: allow description of a circular path
Do not accept
colour spreads evenly in all directions
(b)
water (above flame) is heated and expands / particles move further apart;
(so) density (of this water) decreases;
less dense water rises AND cooler/denser water sinks (to replace it);
Allow
MP3: allow hot water floats on cooler water
Do not accept
heat rises
particles expand
particles become less dense
(c)
(heated) water near element becomes less dense and rises;
cooler (denser) water sinks to the element / convection current mixes all the water;
Do not accept
heat rises
element heats water by radiation only
(d)
any one from:
colour (trail) is easier to see / not too dark;
convection current is slow (enough) to observe / water does not boil;
59IGCSE-PHY-4-0114(b) Energy TransfersPapers 1 & 2Grade 6–75 marks
A picnic cool box keeps food cold on a sunny day. The walls of the box are plastic with a thick layer of polystyrene foam inside. The outside of the box is shiny white. The lid fits tightly.
(a)
Explain how the polystyrene foam reduces the energy transferred into the box.
(2)
(b)
Explain why the outside of the box is shiny white rather than matt black.
(2)
(c)
Explain why the food stays cold for longer if the lid fits tightly.
(1)
Show mark scheme
(a)
(foam/air) is a poor conductor / insulator;
air is trapped (in bubbles) so convection (currents) cannot occur;
Allow
MP1: allow 'good insulator'
Do not accept
foam keeps the cold in
foam reflects heat
(b)
shiny/white surfaces are poor absorbers / good reflectors of (infrared) radiation;
so less energy is absorbed (from the Sun) than with matt black / matt black is a good absorber;
Do not accept
white reflects the cold
shiny surface absorbs more
(c)
prevents (warm) air entering / prevents convection (currents) into the box;
60IGCSE-PHY-4-0124(b) Energy TransfersPapers 1 & 2Grade 6–78 marks
The battery of an electric bicycle is charged from the mains. The battery is then used to drive the motor. The Sankey diagram shows the energy transfers for both stages. It is drawn to scale: each small square on the grid is 40 J.
(a(i))
Use the diagram to determine the total energy taken from the mains.
(1)
(a(ii))
Use the diagram to determine the energy stored in the battery.
(1)
(b)
Calculate the efficiency of the motor.
(2)
(c)
Calculate the overall efficiency of the whole process, from the mains to the kinetic energy store.
(2)
(d)
A student says the overall efficiency is found by adding the efficiencies of the charger and the motor. Explain why the student is wrong.
(2)
Show mark scheme
(a(i))
400 J;
(a(ii))
320 J;
Allow
ECF: (a)(i) − 80 J
(b)
substitution;
evaluation;
e.g.
(efficiency =) 240 ÷ 320 (× 100)
(efficiency =) 75 (%)
Allow
0.75
ECF from (a)(ii)
Do not accept
60% (uses mains input)
(c)
substitution;
evaluation;
e.g.
(efficiency =) 240 ÷ 400 (× 100)
(efficiency =) 60 (%)
Allow
0.80 × 0.75 (× 100) for substitution
ECF from (a)(i)
Do not accept
40% (wasted fraction)
(d)
(adding gives 155%, but) efficiency cannot be more than 100% / output cannot exceed input;
energy is wasted in both stages, so overall efficiency is less than either / is the product of the two efficiencies;
Allow
MP2: allow motor's input is only the energy stored in the battery, not the mains input
61IGCSE-PHY-4-0164(c) Work and PowerPapers 1 & 2Grade 6–79 marks
A building-site crane lifts a load of mass 450 kg vertically. The graph shows how the height of the load changes with time. The electric motor of the crane has an input power of 9.0 kW while the load is moving.
(a)
Use the graph to determine the speed of the load while it is moving upwards.
(2)
(b)
Calculate the gravitational potential energy gained by the load.
(2)
(c)
Calculate the useful power output of the crane while the load is moving.
(2)
(d)
Calculate the efficiency of the crane.
(3)
Show mark scheme
(a)
use of gradient / height ÷ time (for moving section);
62IGCSE-PHY-4-0264(c) Work and PowerPapers 1 & 2Grade 6–71 mark
An empty truck has a mass of 6000 kg and travels at 20 m/s. When it is fully loaded, the truck has a mass of 12 000 kg and travels at 10 m/s. Which statement about the kinetic energy of the loaded truck is correct?
Show mark scheme
A (half the kinetic energy of the empty truck);
B is incorrect because it assumes doubling the mass cancels halving the speed (speed is not squared)
C is incorrect because it applies the effect of halving the speed (× ¼) but ignores doubling the mass
D is incorrect because it squares the mass factor instead of the speed factor (2² × ½ = 2)
63IGCSE-PHY-4-0274(c) Work and PowerPapers 1 & 2Grade 6–71 mark
A mobile phone of mass 0.20 kg falls from a shelf 1.8 m above the floor. Air resistance can be ignored. g = 10 N/kg. What is the speed of the phone just before it hits the floor?
Show mark scheme
C (6.0 m/s);
A is incorrect because 36 is v² (2 × 10 × 1.8); the square root has not been taken
B is incorrect because it uses v² = g × h (= 18), leaving out the factor of 2 from ½mv²
D is incorrect because it uses v = g × h (10 × 1.8) with no square root and no factor of 2
64IGCSE-PHY-4-0284(b) Energy TransfersPapers 1 & 2Grade 6–76 marks
The Sankey diagram shows the energy transfers in a hairdryer in 1.0 s when it is on its hot setting. The diagram is drawn to scale on a grid. The width of each arrow is proportional to the energy it represents. In 1.0 s, 180 J of energy is transferred to the kinetic energy store of the air.
(a)
Determine the energy represented by the width of one small square on the grid.
(1)
(b)
Determine the total electrical energy input to the hairdryer in 1.0 s.
(1)
(c)
Both the thermal energy store of the air and the kinetic energy store of the air are useful outputs. Calculate the efficiency of the hairdryer.
(2)
(d)
Another student calculates the efficiency of the hairdryer as 80%. Suggest the mistake this student made.
(1)
(e)
Determine the power input to the hairdryer.
(1)
Show mark scheme
(a)
60 J;
(b)
1200 J;
Allow
ECF from (a): 20 × (a)
sum of outputs 960 + 180 + 60
(c)
substitution;
evaluation;
e.g.
(efficiency =) (960 + 180) ÷ 1200 (× 100%)
(efficiency =) 95 (%)
Allow
0.95
ECF from (a) and (b)
(1200 − 60) ÷ 1200
Do not accept
80 (%) (thermal output only)
15 (%) (kinetic output only)
105 (%) / 1.05 (inverted)
(d)
only counted the thermal (energy store of the air) as useful / ignored the kinetic energy (of the air);
66IGCSE-PHY-4-0304(c) Work and PowerPapers 1 & 2Grade 6–78 marks
An electric winch pulls a sailing dinghy up a concrete slipway at a steady speed, starting from the water's edge. The slipway is steeper near the water and less steep further up. The graph shows how the force exerted by the winch on the dinghy, along the slipway, changes with the distance moved along the slipway. The mass of the dinghy is 240 kg. Over the whole 14 m, the dinghy rises through a vertical height of 1.5 m.
(a)
Use the graph to calculate the work done by the winch in the first 6.0 m.
(2)
(b)
Calculate the total work done by the winch over the whole 14 m.
(2)
(c)
Calculate the gravitational potential energy gained by the dinghy.
(2)
(d)
The total work done by the winch is greater than the gravitational potential energy gained by the dinghy. Explain why.
(2)
Show mark scheme
(a)
substitution using values from the graph;
evaluation;
e.g.
(W =) 900 × 6.0
(W =) 5400 (J)
Allow
5.4 kJ
Do not accept
150 (N/m) (force ÷ distance)
(b)
work for second section = 650 × 8.0 (= 5200 J);
total evaluated;
e.g.
(W =) 5400 + 5200
(W =) 10 600 (J)
Allow
ECF from (a)
10.6 kJ / 1.06 × 10⁴ J
Do not accept
12 600 (J) (900 × 14)
9100 (J) (650 × 14)
10 850 (J) (775 × 14, average of the two forces)
(c)
substitution;
evaluation;
e.g.
(GPE =) 240 × 10 × 1.5
(GPE =) 3600 (J)
Allow
3530 (J) using g = 9.8 or 9.81
Do not accept
33 600 (J) (uses 14 m)
360 (J) (g omitted)
(d)
work (also) done against friction / resistive forces (between dinghy and slipway);
(this) energy is transferred to the thermal energy store of the dinghy/slipway/surroundings;
Allow
work done against resistive forces / friction (alone) for MP1
dissipated as sound for MP2 only if thermal also mentioned
67IGCSE-PHY-4-0314(b) Energy TransfersPapers 1 & 2Grade 6–79 marks
A Leslie cube is a hollow metal cube. Each of its four vertical faces has a different surface. A student fills the cube with hot water from a kettle. She places an infrared detector 10 cm from one face and records the reading. She then turns the cube so that each face in turn is 10 cm from the detector. Surface / type: matt black, shiny black, dull silver, shiny silver Detector reading / arbitrary units: 9.2, 7.8, 3.0, 1.1
(a)
Explain why all four faces of the cube are at the same temperature.
(1)
(b)
State two other variables that the student should keep the same.
(2)
(c)
Describe what the results show about how the type of surface affects the emission of infrared radiation.
(2)
(d)
The student repeats all four readings 15 minutes later. Every reading is smaller than before. Explain why.
(2)
(e)
The cooling pipes on the back of a refrigerator transfer energy to the air in the kitchen. Use the results to suggest the best surface for these pipes. Give a reason for your answer.
(2)
Show mark scheme
(a)
(all faces are) in contact with the same (hot) water;
Allow
cube contains one body of water at one temperature
(b)
any two from:
distance between detector and face;
angle of detector to the face / detector pointing straight at face;
time between filling and taking readings / readings taken quickly;
same detector;
same room / surroundings / no draughts;
Do not accept
temperature of the water (already stated in a)
type of surface
(c)
black/dark surfaces emit more (infrared) than silver/light surfaces (with the same texture);
matt/dull surfaces emit more than shiny surfaces of the same colour;
Allow
matt black best emitter AND shiny silver worst emitter for 1 mark if neither MP scored
quoting data in support
Do not accept
black absorbs more (absorption not tested)
(d)
the water/cube has cooled / is at a lower temperature;
a cooler surface emits (infrared) radiation at a lower rate / less energy per second;
Do not accept
the cube has run out of heat
the surfaces have changed
(e)
matt black;
best emitter (of infrared) so transfers energy (to the kitchen) fastest;
Allow
MP2 from data: matt black gave the highest reading
68IGCSE-PHY-4-0324(b) Energy TransfersPapers 1 & 2Grade 6–79 marks
A student uses a tall tube of water as a model of a hot-water cylinder in a house. Three thermometers measure the water temperature at the top, middle and bottom of the tube. In experiment A, a small electric heater is placed at the bottom of the tube. In experiment B, the same heater is placed near the top. In each experiment the water starts at 20 °C and the heater is switched on for 10 minutes. The water is not stirred. Thermometer / position: top, middle, bottom Temperature in experiment A / °C: 39, 37, 35 Temperature in experiment B / °C: 68, 22, 20
(a)
State the independent variable in this investigation.
(1)
(b)
Compare the results of experiment A and experiment B.
(2)
(c)
Explain the results of experiment B.
(3)
(d)
In a house, hot water is taken from the top of the cylinder. Suggest one advantage of having the heater near the top of the cylinder when only a small amount of hot water is needed quickly.
(2)
(e)
Explain why the student must not stir the water before taking the readings.
(1)
Show mark scheme
(a)
position of the heater;
Allow
height of the heater
Do not accept
temperature
time
(b)
A: all three temperatures similar / whole tube warmed (35–39 °C);
B: top much hotter (68 °C) but middle and bottom almost unchanged (22 °C and 20 °C);
Allow
comparative statements e.g. top hotter in B than A, bottom hotter in A than B for 1 mark
(c)
heated water (near the top) expands / becomes less dense;
so it stays at the top / does not sink / no convection current reaches the lower water;
water is a poor (thermal) conductor so little energy reaches the bottom by conduction;
Do not accept
heat rises (alone)
cold rises
(d)
only a small volume of water (at the top) is heated;
so it reaches a high temperature quickly / less energy is needed (than heating the whole cylinder);
Do not accept
the heater uses less power
(e)
stirring would mix the water / remove the temperature differences (being measured);
Allow
stirring would make a forced current that is not convection
69IGCSE-PHY-4-0354(b) Energy TransfersPapers 1 & 2Grade 6–79 marks
A student investigates how the type of surface affects the absorption of infrared radiation. She uses two identical aluminium plates. The front of one plate is painted matt black and the front of the other is polished shiny silver. A temperature sensor is attached to the back of each plate. The plates are placed facing an electric radiant heater, one on each side of it, both 15 cm from the heater. She switches on the heater and records the temperature of each plate every minute. The table shows her results. The graph shows the same results plotted. Time / min: 0, 1, 2, 3, 4, 5, 6, 7, 8 Matt black plate temperature / °C: 20, 26, 31, 35, 38, 40, 41.5, 42, 42 Shiny silver plate temperature / °C: 20, 22, 24, 25.5, 27, 28, 28.5, 29, 29
(a)
Explain why both plates are placed the same distance from the heater.
(1)
(b)
Use the graph to determine the increase in temperature of the matt black plate • during the first minute • between 4 minutes and 5 minutes. Use your answers to describe how the rate of temperature rise of the matt black plate changes.
(3)
(c)
The student concludes that matt black surfaces are better absorbers of infrared radiation than shiny silver surfaces. The heater might not emit infrared radiation equally from both of its sides. Describe how the student could check that her conclusion is valid, and state the result that would support it.
(2)
(d)
Explain why the temperature of each plate eventually stops rising, even though the heater is still on.
(3)
Show mark scheme
(a)
so both receive the same amount of (infrared) radiation per second / same intensity;
Allow
fair test because intensity depends on distance
Do not accept
so it is a fair test (alone)
(b)
first minute: 6 (°C);
between 4 and 5 minutes: 2 (°C);
rate of temperature rise decreases (as the plate gets hotter / as time goes on);
e.g.
26 − 20 = 6 (°C) and 40 − 38 = 2 (°C)
Allow
MP3: allow the plate heats up more slowly later / rise in the 5th minute is a third of the first
MP3: ECF from incorrectly read values if the comparison is consistent
unit not required
Do not accept
26 (°C) / 40 (°C) (temperature read instead of the increase)
rate increases
(c)
swap the positions of the two plates (and repeat the experiment);
matt black plate still has the faster temperature rise / reaches the higher temperature;
Allow
MP1: test each plate in turn in the same position (each starting at room temperature)
MP2: matt black still gives the larger temperature rise in the same time
Do not accept
repeat the experiment (alone, with no change of position)
use a more accurate thermometer
(d)
as temperature rises the plate loses/emits energy (to surroundings) at a greater rate;
(eventually) rate of energy loss = rate of energy absorbed (from heater);
so no net energy gain / energy in = energy out, so temperature stays constant;
Allow
energy lost by radiation, conduction or convection to the air
70IGPE-053-00-024(b) Energy TransfersPaper 1PMedium1 mark
In a walk-in cold store at a fish market, the cooling unit that blows out cold air is fitted near the ceiling rather than near the floor. Which statement explains why this cools the whole room most effectively?
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Notes
Cold air is denser than the warmer air around it, so it sinks. Warmer air is pushed upwards towards the cooling unit, where it is cooled and sinks in turn, setting up a convection current that circulates through the whole store. Cold air does not rise. Air is a poor conductor, so conduction is not the main transfer. There is no such thing as 'cold radiation'; objects emit infrared according to their own temperature.
✓ 1
Answer B
Answer: B
Exam tip: In convection answers, link temperature to density and then to rising or sinking. Warmer means less dense, which means it rises.
71IGPE-053-00-034(c) Work and PowerPaper 1PMedium1 mark
A parcel-delivery drone flying horizontally increases its speed from 6 m/s to 12 m/s. Its mass stays the same. By what factor does the energy in its kinetic store increase?
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Notes
Kinetic energy = ½ × m × v2, so kinetic energy is proportional to the square of the speed. Doubling the speed multiplies the kinetic energy by 22 = 4. '× 2' wrongly assumes kinetic energy is proportional to speed. '× 8' wrongly cubes the speed factor. '× 12' is simply the final speed, not a factor.
✓ 1
Answer B
Answer: B
Common error: Remember to square the speed. Doubling v does not double the kinetic energy; it quadruples it.
72IGPE2-057-01-024(c) Work and PowerPaper 1PMedium1 mark
Which of these units is equivalent to the watt?
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Notes
1 W = 1 J/s, and 1 J = 1 N m (work done = force × distance), so 1 W = 1 N m/s. N m is the joule, a unit of work or energy. J s multiplies energy by time instead of dividing. kg m/s is the unit of momentum.
✓ 1
Answer B
Answer: B
Exam tip: Build derived units from the equations: power = work done ÷ time and work done = force × distance.
73IGPE2-057-01-034(c) Work and PowerPaper 1PMedium1 mark
Two weightlifters each lift an identical 120 kg barbell from the floor to the same height above their heads. Weightlifter A takes 1.5 s. Weightlifter B takes 3.0 s. Which statement is correct?
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Notes
Work done = force × distance. The weight and the height are the same for both, so the work done is the same. Power = work done ÷ time. A takes half the time, so A's power is twice B's. Option B confuses work with power. Option C ignores the time taken. Option D wrongly treats a longer time as giving more power.
✓ 1
Answer A
Answer: A
Common error: Doing a job faster does not change the work done. It only changes the power, which is the rate of doing work.
74IGPE2-057-01-074(c) Work and PowerPaper 1PMedium5 marks
A mechanic uses a hydraulic jack to raise one corner of a car. Each time she pushes the jack handle down, she applies a force of 120 N and the handle moves 0.30 m.
(a)
Calculate the work done by the mechanic in one push of the handle.
(2)
(b)
The jack must lift a load of 4000 N through a height of 0.25 m. The jack is 80% efficient. Calculate the number of pushes needed.
(3)
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Notes
Part (a)
work done = 120 × 0.30
✓ 1
= 36 J
✓ 1
Part (b)
useful work = 4000 × 0.25 = 1000 J
✓ 1
input work = 1000 ÷ 0.80 = 1250 J
✓ 1
number of pushes = 1250 ÷ 36 = 34.7, so 35 pushes
✓ 1
Exam tip: When a count must be a whole number, round up. 34 pushes would not quite lift the car to the required height.
75IGPE2-057-01-084(c) Work and PowerPaper 1PMedium6 marks
A patient recovering from knee surgery uses a leg-press machine. Each repetition lifts a weight stack of mass 25 kg through a vertical height of 0.40 m. The patient completes 12 repetitions in 30 s. g = 10 N/kg
(a)
Calculate the weight of the stack.
(1)
(b)
Calculate the work done lifting the stack in one repetition.
(1)
(c)
Calculate the average useful power output of the patient during the 30 s.
(2)
(d)
The physiotherapist wants the patient's average useful power to increase to 60 W. The patient will still complete 12 repetitions through 0.40 m in 30 s. Calculate the new mass of the weight stack.
(2)
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Notes
Part (a)
weight = 25 × 10 = 250 N
✓ 1
Part (b)
work done = 250 × 0.40 = 100 J
✓ 1
Part (c)
total work = 12 × 100 = 1200 J
✓ 1
power = 1200 ÷ 30 = 40 W
✓ 1
Part (d)
Award one correct intermediate step. Any of these earns the mark: total work = 60 × 30 = 1800 J; work per repetition = 150 J; weight = 375 N; or power ratio 60 ÷ 40 = 1.5
✓ 1
mass = 37.5 kg (accept 25 × 1.5 = 37.5 kg)
✓ 1
Exam tip: Work backwards one step at a time: power → total work → work per repetition → force → mass. Write each step down so that every intermediate value can earn credit.
76IGPE2-057-01-104(c) Work and PowerPaper 1PMedium5 marks
A robotic arm on a Mars rover lifts a rock sample of mass 2.4 kg through a vertical height of 0.80 m into a storage container. The gravitational field strength on Mars is 3.7 N/kg.
(a)
Calculate the work done by the arm in lifting the sample.
(2)
(b)
The arm's motor has a useful power output of 0.50 W. Calculate the time taken to lift the sample.
(2)
(c)
State why the arm would need to do more work to lift the same sample through the same height on Earth.
(1)
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Notes
Part (a)
work done = 2.4 × 3.7 × 0.80
✓ 1
= 7.1 J
✓ 1
Part (b)
time = work done ÷ power = 7.1 ÷ 0.50
✓ 1
= 14 s
✓ 1
Part (c)
the gravitational field strength on Earth is greater, so the sample's weight is greater
✓ 1
Common error: Use the value of g given for the planet in the question. Do not automatically use 10 N/kg.
77IGPE2-057-03-024(b) Energy TransfersPaper 1PMedium1 mark
An LED stage lamp has an efficiency of 30%. It gives out 45 W of useful light. What is the total input power to the lamp?
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Notes
efficiency = useful power output ÷ total power input. So total power input = useful power output ÷ efficiency = 45 ÷ 0.30 = 150 W. Option A (13.5 W) comes from multiplying 45 by 0.30 instead of dividing. Option B (1.5 W) comes from dividing by 30 instead of by 0.30. Option C (105 W) is the wasted power, 150 − 45, not the input power.
✓ 1
Answer D
Answer: D
Exam tip: Change a percentage efficiency to a decimal before rearranging. Then check your answer: the input must always be larger than the useful output.
78IGPE2-057-03-044(c) Work and PowerPaper 1PMedium8 marks
A pole vaulter of mass 70 kg sprints along the runway. She reaches a speed of 8.0 m/s just before she plants the pole.
(a)
The pole bends as she takes off. State the energy store that increases as the pole bends.
(1)
(b)
Calculate her kinetic energy just before she plants the pole.
(2)
(c)
Assume that all of this kinetic energy is transferred to her gravitational potential store. Calculate the maximum height that her centre of mass could rise. Use g = 10 N/kg.
(3)
(d)
Suggest two reasons why her centre of mass actually rises less than the height calculated in (c).
(2)
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Notes
Part (a)
elastic potential (strain) store of the pole
✓ 1
Part (b)
KE = ½ × 70 × 8.02
✓ 1
KE = 2240 J (accept 2200 J)
✓ 1
Part (c)
GPE gained = KE from (b), or mgh = ½mv2 is used
✓ 1
h = 2240 ÷ (70 × 10), or h = 8.02 ÷ (2 × 10)
✓ 1
h = 3.2 m (allow 3.1 m with error carried forward from 2200 J)
✓ 1
Part (d)
First valid reason
✓ 1
Second valid reason
✓ 1
Valid reasons include:
—
• she still has some kinetic energy (horizontal speed) at the top of the vault
—
• some energy is transferred to the thermal store of the pole as it bends and straightens
—
• work is done against air resistance, so some energy goes to the thermal store of the surroundings
—
• some energy is transferred as sound, or by heating, where the pole hits the box
—
Exam tip: In (c) the mass cancels, since h = v²/2g. Use it as a quick check: 64 ÷ 20 = 3.2 m.
79IGPE2-057-03-074(b) Energy TransfersPaper 1PMedium7 marks
An electric scooter is ridden along a flat path at a steady speed. Each second: • 480 J of energy is transferred from the chemical store of the battery • 24 J of this is wasted heating the battery and the connecting wires • 60 J of this is wasted heating the motor windings • the rest is useful mechanical work that keeps the scooter moving.
(a)
Describe what eventually happens to the wasted energy.
(2)
(b)
Calculate the useful power output of the motor.
(1)
(c)
Calculate the efficiency of the scooter's electrical system.
(2)
(d)
When fully charged, the battery stores 1.8 MJ of energy. Calculate the maximum time for which the battery can transfer energy at this rate.
(2)
Show mark scheme
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Notes
Part (a)
it is transferred (by heating) to the thermal store of the surroundings
✓ 1
it spreads out (is dissipated), so it becomes less useful
✓ 1
Part (b)
480 − 24 − 60 = 396 W
✓ 1
Part (c)
efficiency = 396 ÷ 480
✓ 1
efficiency = 0.825 or 82.5% (allow 83%)
✓ 1
Part (d)
t = E ÷ P = 1 800 000 ÷ 480
✓ 1
t = 3750 s (about 63 minutes)
✓ 1
Common error: In (d) use the total power of 480 W transferred from the battery, not the useful 396 W. The battery supplies all of the energy, including the part that is wasted.
80IGPE2-057-03-084(b) Energy TransfersPaper 1PMedium6 marks
Vaccines must be kept cold while they are carried to remote clinics. A vaccine cold box has these features: • an outer case of glossy white plastic • walls 5 cm thick, made of expanded polystyrene foam full of tiny trapped air bubbles • a lid with a rubber seal that fits tightly • frozen ice packs clipped to the underside of the lid, above the vaccines.
(a)
Explain how the features of the cold box keep the vaccines cold. Refer to the ways energy is transferred.
(6)
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Notes
Part (a)
• the glossy white surface is a good reflector and a poor absorber of infrared radiation, so less energy is absorbed from the Sun and surroundings by radiation
✓ 1
• polystyrene foam and trapped air are poor conductors (insulators), so energy is conducted through the walls slowly
✓ 1
• the walls are thick (5 cm), which further reduces the rate of energy transfer by conduction
✓ 1
• the air is trapped in tiny bubbles and cannot circulate, so convection currents within the walls are prevented
✓ 1
• the tight rubber seal stops warm air from outside entering and cold air escaping, which reduces energy transfer by convection
✓ 1
• air next to the ice packs at the top is cooled, becomes denser and sinks, so a convection current carries cold air down over the vaccines (the ice also absorbs any energy that does get in)
✓ 1
Exam tip: For each feature, name the energy transfer method it affects and explain why that transfer is reduced. Naming a feature without linking it to conduction, convection or radiation earns little credit.
81IGPE2-057-02-024(d) Energy Resources and Electricity GenerationPaper 1PMedium1 mark
A national grid operator wants a renewable energy resource that can be switched on within minutes to meet a sudden rise in demand at any time of day or night. Which resource is most suitable?
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Notes
Water stored behind a dam can be released through the turbines whenever it is needed, so a hydroelectric dam gives renewable electricity on demand with a very short start-up time. Solar cells produce nothing at night. Wind farms depend on the wind blowing, so their output cannot be guaranteed. Nuclear power is not renewable, and it takes a long time to change its output.
✓ 1
Answer C
Answer: C
Exam tip: Check both conditions in the question. Here the resource must be renewable and it must be available on demand. Rule out each option that fails either one.
82IGPE2-057-02-034(b) Energy TransfersPaper 1PMedium1 mark
A satellite's solar panels have a total area of 12 m2. In orbit, each square metre of panel receives 1400 W of light. The efficiency of the panels is 0.30. What is the electrical power output of the panels?
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Mark
Notes
Total light power received = 1400 W/m2 × 12 m2 = 16 800 W. Useful electrical power = efficiency × input power = 0.30 × 16 800 W = 5040 W ≈ 5.0 kW. 16.8 kW leaves out the efficiency. 56 kW comes from dividing by the efficiency instead of multiplying. 0.035 kW comes from dividing by the area instead of multiplying.
✓ 1
Answer B
Answer: B
Common error: The useful output must always be smaller than the input. If your answer is larger than the input power, you have divided by the efficiency when you should have multiplied.
84IGPE2-057-02-134(d) Energy Resources and Electricity GenerationPaper 1PMedium7 marks
A rover on Mars is powered by a radioisotope generator rather than by solar panels. Energy from a radioactive source heats the generator, which then produces an electrical power output of 110 W. The efficiency of the generator is 5.5%.
(a)
Calculate the rate at which thermal energy is supplied to the generator by the radioactive source.
(2)
(b)
Suggest two reasons why this generator was chosen instead of solar panels for the rover.
(2)
(c)
Suggest how some of the wasted thermal energy can be made useful on the rover.
(1)
(d)
One Martian day lasts 8.9 × 104 s. Calculate the electrical energy the generator supplies in one Martian day.
(2)
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Notes
Part (a)
input power = useful power output ÷ efficiency = 110 ÷ 0.055
✓ 1
= 2000 W (2.0 × 103 W)
✓ 1
Part (b)
Any two different reasons, one mark each, chosen from: it produces power at night and through the long Martian winter / dust storms or dust settling on panels would reduce the output of solar panels / sunlight on Mars is weaker than on Earth because Mars is further from the Sun / it gives a steady output for many years.
—
First valid reason
✓ 1
Second, different valid reason
✓ 1
Part (c)
It can be used to keep the rover's electronics and instruments warm in the very cold Martian conditions
✓ 1
Part (d)
E = P × t = 110 × 8.9 × 104
✓ 1
= 9.8 × 106 J
✓ 1
Common error: Convert a percentage efficiency to a decimal before you use it: 5.5% = 0.055. Dividing by 5.5 instead gives an answer that is 100 times too small.
85IGPE2-057-00-024(b) Energy TransfersPaper 1PMedium1 mark
An LED stage lamp has an efficiency of 0.40. It transfers 18 W of useful power as light. What is the input power to the lamp?
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Notes
Efficiency = useful power output ÷ total power input. Rearranging: input power = useful output ÷ efficiency = 18 ÷ 0.40 = 45 W. A multiplies by the efficiency instead of dividing. B divides by the wasted fraction, 0.60. D adds 40% to the output.
✓ 1
Answer C
Answer: C
Exam tip: The input power must always be larger than the useful output. Use this to check any efficiency rearrangement.
86IGPE2-057-00-034(b) Energy TransfersPaper 1PMedium1 mark
The outside of a spacecraft is covered with shiny, reflective foil to reduce energy transfer between the spacecraft and space. Which statement explains why the foil is shiny?
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Notes
Space is a vacuum. Conduction and convection to the surroundings need particles, so they cannot happen. Infrared radiation is therefore the only way energy is transferred to or from space. A shiny surface reflects most infrared, so it absorbs little, and it also emits little. This reduces transfer in both directions. A and B describe processes that cannot occur in a vacuum. D is wrong because shiny surfaces are poor emitters, not good ones.
✓ 1
Answer C
Answer: C
Common error: Shiny surfaces are poor emitters as well as poor absorbers. Dull black surfaces are good at both.
87IGPE2-057-00-054(b) Energy TransfersPaper 1PMedium6 marks
A portable defibrillator is used in hospitals to restart a patient's heart. A battery charges a capacitor inside the device. The capacitor then delivers a short pulse of energy to the patient's chest.
(a)
Describe the energy transfer that takes place while the battery charges the capacitor.
(2)
(b)
The battery transfers 260 J for every 200 J that the capacitor delivers to the patient. Calculate the efficiency of the device.
(2)
(c)
The 200 J pulse is delivered in 4.0 ms. Calculate the power of the pulse.
(2)
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Notes
Part (a)
Energy is transferred from the chemical store of the battery
✓ 1
electrically to the electrostatic store of the capacitor
✓ 1
Part (b)
Efficiency = useful energy output ÷ total energy input = 200 ÷ 260
✓ 1
0.77 (or 77%)
✓ 1
Part (c)
Power = 200 ÷ 0.0040 (correct conversion of ms to s)
✓ 1
50 000 W (or 5.0 × 104 W)
✓ 1
Common error: Convert milliseconds to seconds first: 4.0 ms = 0.0040 s. Dividing by 4.0 gives an answer 1000 times too small.
88IGPE2-057-00-074(b) Energy TransfersPaper 1PMedium7 marks
A space capsule of mass 2900 kg re-enters the Earth's atmosphere at a speed of 7.6 km/s.
(a)
State the equation linking kinetic energy, mass and speed.
(1)
(b)
Calculate the kinetic energy of the capsule as it enters the atmosphere.
(2)
(c)
The capsule slows down as it falls through the atmosphere. Explain why the outside of the capsule becomes very hot.
(2)
(d)
The capsule has a heat shield on its outer surface. Suggest one property that the heat shield material should have, and explain why.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
KE = ½ × mass × speed2
✓ 1
Part (b)
Substitution with speed in m/s: ½ × 2900 × 76002
✓ 1
8.4 × 1010 J
✓ 1
Part (c)
Work is done against air resistance (friction with the air)
✓ 1
so energy is transferred from the kinetic store of the capsule to the thermal store of the capsule's surface and the air
✓ 1
Part (d)
A suitable property, e.g. low thermal conductivity (good thermal insulator) OR high melting point OR high specific heat capacity
✓ 1
A matching explanation, e.g. little energy is conducted through to the inside, so the crew and equipment stay cool OR the shield does not melt at the very high surface temperature OR it can absorb a lot of energy for only a small rise in temperature
✓ 1
Common error: Convert 7.6 km/s to 7600 m/s before squaring. Remember that only the speed is squared, not the mass.
89IGPE-053-00-064(b) Energy TransfersPaper 1PMedium6 marks
Every second, the motor of an electric scooter is supplied with 900 J of electrical energy. Each second it transfers 630 J usefully to the kinetic store of the scooter and 30 J as sound. The rest is transferred to the thermal store of the motor.
(a)
Calculate the energy transferred to the thermal store of the motor each second.
(1)
(b)
Calculate the efficiency of the motor. Give your answer as a percentage.
(2)
(c)
Describe what eventually happens to the wasted energy.
(2)
(d)
Suggest one way the efficiency of the motor could be improved.
(1)
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Mark
Notes
Part (a)
900 − 630 − 30 = 240 J
✓ 1
Part (b)
Efficiency = 630 ÷ 900 (× 100)
✓ 1
= 70 %
✓ 1
Part (c)
It is transferred to (the thermal store of) the surroundings, which warm up slightly
✓ 1
It becomes spread out / dissipated, so it is less useful
✓ 1
Part (d)
Any one: lubricate the moving parts / bearings to reduce friction, or use wires with lower resistance to reduce heating
✓ 1
Common error: The sound energy is also wasted. Subtract every non-useful output, not just the thermal one, when you find the remaining energy.
90IGPE-053-00-074(c) Work and PowerPaper 1PMedium6 marks
A rider of mass 72 kg starts at rest at the top of a zip line. When the rider reaches the end of the zip line, they have dropped through a vertical height of 35 m and are moving at 18 m/s. Use g = 10 N/kg.
Diagram not to scale
(a)
Calculate the energy lost from the rider's gravitational potential energy store.
(2)
(b)
Calculate the kinetic energy of the rider at the end of the zip line.
(2)
(c)
State where the remaining energy has been transferred to.
(1)
(d)
Calculate the energy dissipated during the ride.
(1)
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Notes
Part (a)
GPE = m × g × h = 72 × 10 × 35
✓ 1
= 25 200 J
✓ 1
Part (b)
KE = ½ × 72 × 182
✓ 1
= 11 664 J ≈ 11 700 J
✓ 1
Part (c)
To the thermal store of the pulley, cable and surrounding air, because of friction and air resistance
✓ 1
Part (d)
25 200 − 11 664 = 13 536 J ≈ 13 500 J
✓ 1
Exam tip: Energy dissipated = energy lost from one store − energy gained by the useful store. Keep unrounded values until the final step.
91IGPE3-4-14r7-024(c) Work and PowerPaper 1PMedium1 mark
A lifeguard drops a phone of mass 0.20 kg from a height of 1.8 m onto the poolside. Ignoring air resistance and using g = 10 N/kg, what is the speed of the phone just before it lands?
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Notes
Energy conservation: mgh = ½mv², so v = √(2gh) = √(2 × 10 × 1.8) = √36 = 6.0 m/s, option B. A (4.2 m/s) uses v² = gh and forgets the factor 2 from ½mv². C (18) stops at gh and does not complete the calculation. D (36) is v², with no square root taken.
✓ 1
Answer B
Answer: B
Exam tip: The mass cancels when GPE becomes kinetic energy, so the speed does not depend on it.
92IGPE3-4-14r7-044(b) Energy TransfersPaper 1PMedium1 mark
An electric bicycle battery is charged by a charger that is 80% efficient. During a ride, the motor system is 70% efficient at transferring the stored energy to useful energy. What is the overall efficiency from mains to useful output?
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Notes
Efficiencies multiply for stages in series: 0.80 × 0.70 = 0.56 = 56%, option A. B (70%) takes only the second stage. C (75%) averages the two efficiencies, which is not valid. D (80%) takes only the first stage.
✓ 1
Answer A
Answer: A
Exam tip: For two stages in a row, the overall efficiency is the product of the stage efficiencies.
93IGPE3-4-14r7-064(b) Energy TransfersPaper 1PMedium7 marks
An electric bicycle battery is charged from the mains. The charger takes in 800 kJ of electrical energy and wastes 160 kJ as thermal energy, and the rest is stored in the battery. During a ride, all the stored energy is transferred by the motor and drive system: 192 kJ is wasted as thermal energy and the rest is useful energy.
(a)
Calculate the energy stored in the battery.
(1)
(b)
Calculate the efficiency of the riding stage.
(2)
(c)
Calculate the overall efficiency from mains to useful output.
(2)
(d)
Explain why the overall efficiency is lower than the efficiency of either stage.
(2)
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Notes
Part (a)
A1 800 − 160 = 640 kJ
✓ 1
Part (b)
M1 useful = 640 − 192 = 448 kJ
✓ 1
A1 448 ÷ 640 = 0.70 or 70% (ECF from (a))
✓ 1
Part (c)
M1 448 ÷ 800
✓ 1
A1 0.56 or 56%
✓ 1
Part (d)
Energy is wasted in both stages
✓ 1
so the useful fraction remaining after the first stage is reduced again by the second stage
✓ 1
Common error: Take the second-stage efficiency from the energy entering that stage (640 kJ), not from the original 800 kJ.
95IGPE3-4-14r7-134(d) Energy Resources and Electricity GenerationPaper 1PMedium7 marks
A coastal town needs about 18 MW of electrical power on average. Two options are considered. Option A: a wind farm of 20 turbines, each with an average output of 0.9 MW. The output varies with the wind and falls to almost zero in calm weather. The turbines make noise and are visible from the shore. Option B: a gas-fired power station with a constant output of 18 MW that can be varied on demand. It burns a fossil fuel and releases carbon dioxide.
(a)
Calculate the average output of the wind farm.
(1)
(b)
Evaluate which option is better for the town.
(6)
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Notes
Part (a)
A1 20 × 0.9 = 18 MW
✓ 1
Part (b)
Wind farm average output matches the town's average demand
✓ 1
but output is not constant and may fall almost to zero in calm weather, so supply is not guaranteed
✓ 1
Wind is renewable and uses no fuel that runs out
✓ 1
No carbon dioxide is released in running the wind farm, whereas gas releases carbon dioxide that adds to global warming
✓ 1
Gas station gives reliable output on demand
✓ 1
Valid conclusion that weighs the points, e.g. wind with storage or back-up, or gas for reliability
✓ 1
Exam tip: In an evaluate question, give points for both options and finish with a conclusion that is backed by those points.
96IGPE3-4-14r10-024(b) Energy TransfersPaper 1PMedium1 mark
Two identical metal cans are filled with the same volume of hot water at the same temperature. One can has a matt black outer surface and the other has a shiny silver outer surface. Both are left to cool in the same room. Which statement is correct?
Show mark scheme
Marking point
Mark
Notes
Matt black surfaces are good emitters (and absorbers) of infrared radiation, so the matt black can transfers energy to the surroundings faster and cools faster. A is wrong because a shiny surface is a poor emitter, so it loses energy more slowly. C ignores the surface effect. D confuses absorbing radiation from outside with emitting radiation when hotter than the surroundings.
✓ 1
Answer B
Answer: B
Exam tip: For cooling by radiation, compare how good each surface is at emitting infrared radiation; a hot object emits more when its surface is dark and matt.
97IGPE3-4-14r10-034(c) Work and PowerPaper 1PMedium1 mark
A farm worker uses a motor to lift a 40 kg sack through a vertical height of 1.5 m in 6.0 s at constant speed. The gravitational field strength is 10 N/kg. What is the useful power output of the motor?
Show mark scheme
Marking point
Mark
Notes
Energy gained = mgh = 40 × 10 × 1.5 = 600 J. Power = 600 ÷ 6.0 = 100 W. 10 W comes from leaving out g. 600 W is the energy in joules, not divided by time. 3600 W multiplies by the time instead of dividing.
✓ 1
Answer B
Answer: B
Common error: Power is energy transferred per second; check that you divide by the time rather than multiply.
98IGPE3-4-14r10-064(d) Energy Resources and Electricity GenerationPaper 1PMedium7 marks
A coal-fired power station burns fuel at a rate that supplies 1000 MW of chemical energy. It produces 360 MW of electrical power. The rest of the energy is wasted as thermal energy, mainly in the cooling towers and flue gases. A combined heat and power (CHP) plant is a power station that also pipes 300 MW of its waste thermal energy to nearby buildings for heating.
(a)
Calculate the efficiency of the power station when it is used only to generate electricity.
(2)
(b)
The same power station is converted to CHP operation, still using 1000 MW of chemical energy. Calculate the overall efficiency of the CHP plant.
(3)
(c)
Explain why a CHP plant makes better use of the fuel than a power station that only generates electricity.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 360 ÷ 1000
✓ 1
A1 36 (%)
✓ 1
Part (b)
M1 useful output = 360 + 300
✓ 1
A1 660 (MW)
✓ 1
A1 66 (%) ECF
✓ 1
Part (c)
B1 some of the thermal energy that would otherwise be wasted is used to heat buildings
✓ 1
B1 so a greater fraction of the input energy is useful (or less fuel is needed to supply the same heating and electricity)
✓ 1
Exam tip: When a Sankey diagram has two useful outputs, add them together before dividing by the total input.
99IGPE3-4-14r10-074(b) Energy TransfersPaper 1PMedium8 marks
A student investigates how the surface of a can affects how fast hot water cools. She has two identical cans, one with a matt black outer surface and one with a shiny silver outer surface. She fills each with the same volume of hot water at 90 °C and records the temperature every 2 minutes.
Time (min)
0
2
4
6
8
10
Matt black can (°C)
90
80
72
66
61
57
Shiny silver can (°C)
90
83
77
72
68
64
(a)
State two variables the student should control.
(2)
(b)
State which can cools faster. Use data from the table to support your answer.
(2)
(c)
Explain, in terms of radiation, the difference in cooling between the two cans.
(2)
(d)
Calculate the mean rate of cooling of the matt black can during the first 4 minutes.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 starting temperature (or volume/mass of water)
✓ 1
B1 any other: room temperature / same type and size of can / same lid / same position in the room
✓ 1
Part (b)
B1 the matt black can
✓ 1
B1 it falls by 33 °C in 10 minutes but the shiny silver can falls by only 26 °C (or valid comparison at one time)
✓ 1
Part (c)
B1 matt black surfaces are better emitters of infrared radiation than shiny silver surfaces
✓ 1
B1 so the matt black can transfers energy to the surroundings faster
✓ 1
Part (d)
M1 (90 − 72) ÷ 4
✓ 1
A1 4.5 (°C/min)
✓ 1
Exam tip: A rate of cooling is a change in temperature divided by the time taken, so state the unit as °C per minute.
100IGPE3-4-14r10-084(b) Energy TransfersPaper 1PMedium7 marks
A farmer wants to compare how well three materials (sheep wool, bubble wrap and foam sheet) insulate a metal can of hot water. She has three identical cans with lids, a thermometer, a kettle, a stopwatch, a measuring cylinder and equal-sized pieces of each material.
(a)
State the independent variable and the dependent variable in this investigation.
(2)
(b)
State two control variables and explain how the farmer would keep each one constant.
(2)
(c)
Describe how the farmer should use her apparatus to decide which material is the best insulator.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 independent variable: the insulating material used
✓ 1
B1 dependent variable: the temperature of the water (or temperature fall in a fixed time, or time to cool by a fixed amount)
✓ 1
Part (b)
B1 starting temperature of the water, controlled by using water from the kettle at the same temperature (measured with the thermometer)
✓ 1
B1 volume of water, controlled by using the measuring cylinder to put the same volume in each can (or same thickness of material)
✓ 1
Part (c)
B1 wrap each can in a different material and fill with the same volume of hot water at the same starting temperature, then fit lids
✓ 1
B1 measure the temperature at regular time intervals using the stopwatch, or after a fixed time such as 10 minutes
✓ 1
B1 the material giving the smallest temperature fall is the best insulator; repeat and take a mean to improve reliability
✓ 1
Exam tip: When asked to plan, always name what you will change, what you will measure and what you will keep the same.
101IGPE3-4-14r18-034(c) Work and PowerPaper 1PMedium1 mark
A winch lifts a net of fish of mass 250 kg through a height of 6.0 m in 20 s at constant speed. The gravitational field strength is 10 N/kg. What is the useful power output of the winch?
Show mark scheme
Marking point
Mark
Notes
Work done = mgh = 250 × 10 × 6.0 = 15 000 J; power = 15 000 ÷ 20 = 750 W, so B is correct. A leaves out g. C is the energy, not divided by time. D multiplies by time instead of dividing.
✓ 1
Answer B
Answer: B
Exam tip: Power is energy transferred divided by time, so check the units give watts.
102IGPE3-4-14r18-074(d) Energy Resources and Electricity GenerationPaper 1PMedium6 marks
In a coal-fired power station, for every 400 MJ of energy transferred from the coal, 140 MJ is transferred as electrical energy, 40 MJ is lost through the chimney and 220 MJ is transferred to the air by cooling towers.
(a)
Calculate the efficiency of the power station as a percentage.
(2)
(b)
In a combined heat and power plant, 180 MJ of the 220 MJ is instead used to heat nearby homes. Calculate the efficiency of the combined heat and power plant as a percentage.
(3)
(c)
Explain why a combined heat and power plant needs less coal than a power station for the same useful energy output.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 140 ÷ 400
✓ 1
A1 35%
✓ 1
Part (b)
M1 useful energy = 140 + 180 = 320 MJ
✓ 1
M1 320 ÷ 400
✓ 1
A1 80%
✓ 1
Part (c)
more of the input energy is transferred usefully, so less fuel is needed for the same useful output
✓ 1
Exam tip: In efficiency questions, list every useful output before dividing by the total input.
103IGPE3-4-14r18-094(c) Work and PowerPaper 1PMedium7 marks
A robot arm in a clothing factory lifts a 3.5 kg component through a height of 0.80 m. The arm's motor transfers 40 J of electrical energy during the lift.
(a)
Calculate the gravitational potential energy gained by the component.
(2)
(b)
Calculate the efficiency of the arm's motor as a percentage.
(2)
(c)
The component is dropped from the raised position. Calculate its speed just before it hits the floor. Ignore air resistance and use 28 J.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 mgh = 3.5 × 10 × 0.80
✓ 1
A1 28 J
✓ 1
Part (b)
M1 28 ÷ 40 (ECF)
✓ 1
A1 70%
✓ 1
Part (c)
M1 kinetic energy gained = potential energy lost = 28 J
✓ 1
M1 ½ × 3.5 × v2 = 28, so v2 = 16
✓ 1
A1 4.0 m/s
✓ 1
Exam tip: Without air resistance, all the lost gravitational potential energy becomes kinetic energy.
105IGPE3-4-15r33-034(c) Work and PowerPaper 1PMedium1 mark
A skateboarder of mass 40 kg starts from rest at the top of a half-pipe, 1.8 m above the lowest point. At the lowest point the skateboarder has a speed of 5.0 m/s. Use g = 10 N/kg. How much energy is transferred to thermal energy stores by friction and air resistance on the way down?
Show mark scheme
Marking point
Mark
Notes
Gravitational potential energy lost = mgh = 40 × 10 × 1.8 = 720 J. Kinetic energy gained = ½mv² = 0.5 × 40 × 5.0² = 500 J. Energy dissipated = 720 − 500 = 220 J. 500 J is only the kinetic energy at the bottom. 720 J is the total energy lost from the gravitational store, ignoring that most became kinetic energy. 1220 J adds the two energies together instead of finding the difference.
✓ 1
Answer D
Answer: D
Common error: Energy wasted is the difference between the energy lost from one store and the energy usefully gained in another.
106IGPE3-4-15r33-044(d) Energy Resources and Electricity GenerationPaper 1PMedium1 mark
A wind turbine is in a steady wind. The air density and the area swept by the blades stay the same. The wind speed doubles. By what factor does the kinetic energy of the air reaching the blades each second change?
Show mark scheme
Marking point
Mark
Notes
When the speed doubles, twice as much air reaches the blades each second, so the mass per second doubles. Each kilogram of air also has four times the kinetic energy because kinetic energy depends on speed squared. The overall factor is 2 × 4 = 8. A factor of 2 only accounts for the extra mass of air. A factor of 4 only accounts for the v² term. A factor of 16 squares both effects.
✓ 1
Answer A
Answer: A
Exam tip: Check whether more than one quantity changes when a speed changes: more air arrives each second and each kilogram carries more energy.
110IGPE3-4-15r40-024(b) Energy TransfersPaper 1PMedium1 mark
Two identical metal cans each contain 250 g of water at 80 °C. One can is painted matt black and the other has a polished silver surface. Both are left in a cool room. Considering energy transfer by infrared radiation only, which statement is correct?
Show mark scheme
Marking point
Mark
Notes
A matt black surface is a better emitter of infrared radiation than a polished silver surface, so at the same temperature the black can transfers energy away faster and cools faster. Silver is a poor emitter, so the first statement is reversed. The surface finish changes the rate of emission, so equal temperatures do not give equal cooling rates. The room is cooler than the can, so even though black also absorbs well, the can emits much more than it absorbs and cooling is still faster.
✓ 1
Answer D
Answer: D
Exam tip: Good absorbers are also good emitters, and poor emitters are also good reflectors. Link the surface to emission when a hot object is cooling.
111IGPE3-4-15r40-034(b) Energy TransfersPaper 1PMedium1 mark
A cooling fan in a data centre is supplied with 240 J of electrical energy. Of this, 180 J is wasted as thermal energy in the motor and bearings. What is the efficiency of the fan?
Show mark scheme
Marking point
Mark
Notes
Useful energy output = 240 − 180 = 60 J, so efficiency = 60 ÷ 240 = 0.25, which is 25%. Dividing the wasted energy by the total input gives 75%, which is the fraction wasted. Dividing useful output by wasted energy gives 33%, which uses the wrong denominator. Dividing total input by useful output gives 400% and inverts the ratio; efficiency can never exceed 100%.
✓ 1
Answer C
Answer: C
Common error: Efficiency is useful output divided by total input. If your answer is above 100% the ratio is upside down.
112IGPE3-4-15r40-064(b) Energy TransfersPaper 1PMedium6 marks
At a desert excavation, a fragile clay vessel is covered with shiny silver foil during the day to keep it cooler. Sunlight contains infrared radiation.
(a)
Explain why the silver foil keeps the vessel cooler than a matt black cover would.
(2)
(b)
In the evening the vessel is at 35 °C and the air is at 15 °C. Explain why there is a net transfer of energy from the vessel to the surroundings by infrared radiation.
(2)
(c)
An archaeologist suggests wrapping the vessel in the silver foil overnight. Explain whether this would help to keep the vessel warm.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Shiny silver is a poor absorber (and good reflector) of infrared radiation
✓ 1
so less energy is transferred to the vessel than under a black cover
✓ 1
Part (b)
Both the vessel and the surroundings emit and absorb infrared radiation
✓ 1
The vessel is hotter, so it emits more infrared radiation than it absorbs
✓ 1
Part (c)
Shiny silver is a poor emitter of infrared radiation
✓ 1
so energy is transferred from the vessel more slowly, so yes it would help
✓ 1
Common error: Hot objects both emit and absorb infrared radiation. Describe the net transfer rather than saying only that the hot object emits.
113IGPE3-4-15r40-084(b) Energy TransfersPaper 1PMedium6 marks
A pump in a data-centre cooling system is supplied with 2500 J of electrical energy in 50 s. During this time 1900 J is wasted as thermal energy and sound.
(a)
Calculate the useful energy output of the pump.
(2)
(b)
Calculate the input power of the pump.
(2)
(c)
Calculate the efficiency of the pump. Give your answer as a percentage.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 useful energy = total energy input − wasted energy
✓ 1
A1 2500 − 1900 = 600 J
✓ 1
Part (b)
M1 power = energy ÷ time
✓ 1
A1 2500 ÷ 50 = 50 W
✓ 1
Part (c)
M1 efficiency = useful energy output ÷ total energy input (× 100%)
✓ 1
A1 600 ÷ 2500 × 100 = 24% (ECF from (a))
✓ 1
Common error: When given the wasted energy, first subtract it from the total input to find the useful output. Do not divide the wasted energy by the total and call it efficiency.
114IGPE3-4-15r40-094(c) Work and PowerPaper 1PMedium6 marks
A cyclist and her bicycle have a total mass of 80 kg. She rides at a steady speed up a hill, gaining 45 m in height in 3.0 minutes. Assume that all of her useful energy output goes to the gravitational potential energy store. g = 10 N/kg.
(a)
Calculate the gain in gravitational potential energy.
(2)
(b)
Calculate her useful power output.
(2)
(c)
Her body is 25% efficient at transferring energy from its chemical store to useful work. Calculate the chemical energy released during the climb, in kJ.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 gain in GPE = mass × g × height
✓ 1
A1 80 × 10 × 45 = 36 000 J
✓ 1
Part (b)
M1 power = energy ÷ time with time = 3.0 × 60 = 180 s
115IGPE3-4-16r42-024(b) Energy TransfersPaper 1PMedium1 mark
Two identical metal cans contain the same mass of water at 90 °C. One can has a matt black surface and the other has a shiny silver surface. They cool in the same room. Which statement is correct?
Show mark scheme
Marking point
Mark
Notes
A matt black surface is a better emitter of infrared radiation than a shiny silver one, so the black can transfers energy to the surroundings faster and cools faster. A shiny surface is a poor emitter, so the silver can cools more slowly, not faster. The cans cool at different rates because the surface affects radiation even when the water temperature is the same. Black surfaces absorb well, but they also emit well, and here the cans are hotter than the room so emission is what matters.
✓ 1
Answer D
Answer: D
Exam tip: Good absorbers are also good emitters; shiny surfaces are poor absorbers and poor emitters.
116IGPE3-4-16r42-084(c) Work and PowerPaper 1PMedium5 marks
The supercapacitor on another tram stores 2.4 MJ of energy. When the tram is stopped at a station, the supercapacitor supplies the lighting and the doors, which use energy at an average rate of 8.0 kW.
(a)
Calculate the time for which the supercapacitor can supply the lighting and doors.
(2)
(b)
Give your answer to part (a) in minutes.
(1)
(c)
Suggest two reasons why a supercapacitor is suitable for a tram that stops and starts frequently.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 time = energy ÷ power = 2.4 × 106 ÷ 8000
✓ 1
A1 300 s
✓ 1
Part (b)
5.0 minutes
✓ 1
Part (c)
can be charged very quickly during braking, or can release energy quickly to help the tram accelerate, or can be charged and discharged many times without wearing out
✓ 1
a second different valid reason from this list
✓ 1
Common error: Convert MJ and kW to J and W before using energy = power × time.
117IGPE3-4-16r42-094(b) Energy TransfersPaper 1PMedium6 marks
A student compares the rate of cooling of a matt black can and a shiny silver can. Each can is filled with water at 90 °C and left to cool in the same room. After 10 minutes the water in the matt black can is at 41 °C and the water in the shiny silver can is at 56 °C.
(a)
State two variables that the student should keep the same so that the test is fair.
(2)
(b)
Calculate how much more the temperature of the water in the black can fell than the temperature of the water in the silver can.
(2)
(c)
Explain why the temperature of the water in the black can fell by more.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
one variable from: volume (or mass) of water, starting temperature of water, size and shape of the cans, thickness of can walls, room temperature or position in the room, use of lids
✓ 1
a second different variable from this list
✓ 1
Part (b)
M1 temperature falls of 49 °C (black) and 34 °C (silver)
✓ 1
A1 15 °C
✓ 1
Part (c)
a matt black surface is a better emitter of (infrared) thermal radiation than a shiny silver surface
✓ 1
so more energy is transferred to the surroundings each second from the black can
✓ 1
Exam tip: In a fair test only one variable is changed (here, the can surface); everything else must be kept the same.
118IGPE3-4-16r42-104(b) Energy TransfersPaper 1PMedium7 marks
An electric bicycle battery is charged from the mains. During charging, 720 kJ of electrical energy is transferred from the mains and 576 kJ is stored in the battery. The cyclist then rides the bicycle, and the 576 kJ stored in the battery is transferred as 432 kJ of useful kinetic energy and the rest as wasted energy.
(a)
Calculate the efficiency of the charging stage. Give your answer as a percentage.
(2)
(b)
Calculate the wasted energy during the riding stage.
(1)
(c)
Calculate the overall efficiency of charging and riding. Give your answer as a percentage.
(2)
(d)
Describe how a two-stage Sankey diagram would show these energy transfers.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 576 ÷ 720
✓ 1
A1 80%
✓ 1
Part (b)
576 − 432 = 144 kJ
✓ 1
Part (c)
M1 overall efficiency = useful energy out ÷ total energy in = 432 ÷ 720
✓ 1
A1 60%
✓ 1
Part (d)
an arrow narrowing from 720 kJ to 576 kJ to 432 kJ, with the width of each arrow drawn proportional to the energy
✓ 1
wasted-energy arrows of 144 kJ at each stage branching off the main arrow
✓ 1
Exam tip: Overall efficiency is not the sum or the average of the two stage efficiencies: use useful energy out of the whole process divided by total energy in.
119IGCSE-PHY-4-0134(c) Work and PowerPapers 1 & 2Grade 8–99 marks
A skateboarder of total mass 55 kg (including the skateboard) starts from rest at the top of a half-pipe. The top is 3.2 m above the lowest point. He rolls down the curved ramp without pushing. The distance he travels along the ramp to the lowest point is 5.0 m. At the lowest point his speed is 7.4 m/s.
(a)
Show that the gravitational potential energy lost by the skateboarder is about 1800 J.
(2)
(b)
Calculate the average resistive force acting on the skateboarder as he moves along the ramp to the lowest point.
(5)
(c)
Explain, without further calculation, why the skateboarder cannot reach a height of 3.2 m on the other side of the half-pipe if he does not push.
(2)
Show mark scheme
(a)
substitution;
evaluation to at least 3 s.f.;
e.g.
(GPE =) 55 × 10 × 3.2
(GPE =) 1760 (J)
Allow
use of g = 9.8 / 9.81 giving 1720 / 1730 (J)
Do not accept
1800 (J) without working
(b)
substitution into KE = ½ × m × v²;
evaluation of KE;
work done against resistive force = GPE lost − KE gained;
substitution into W = F × d (with d = 5.0 m);
evaluation of force;
e.g.
KE = 0.5 × 55 × 7.4² = 1506 (J)
work = 1760 − 1506 = 254 (J)
254 = F × 5.0
(F =) 51 (N)
Allow
ECF from (a) and from calculated KE
50 (N) from rounded values (1760 − 1510) ÷ 5.0
58 (N) from (1800 − 1510) ÷ 5.0 scores 4 marks
use of vertical drop 3.2 m for d giving 79 (N) scores 4 marks
KE not squared (204 J) used correctly thereafter: max 3 marks
Do not accept
352 (N) (GPE ÷ 5.0, ignores KE)
(c)
(at the bottom) KE is less than the GPE needed to rise 3.2 m / some energy already transferred to thermal store;
(and) work is done against resistive forces/friction on the way up too, (so even less energy available for GPE);
120IGCSE-PHY-4-0144(b) Energy TransfersPapers 1 & 2Grade 8–910 marks
A student investigates how the type of surface affects the rate of energy transfer by radiation. She uses two identical aluminium cans. One is painted matt black and the other is polished shiny silver. She pours hot water into each can and measures the temperature every 2 minutes. The graph and table show her results. Time / min: 0, 2, 4, 6, 8, 10 Matt black can temperature / °C: 80, 72, 65, 59, 54, 50 Shiny silver can temperature / °C: 80, 76, 72, 68.5, 65.5, 63
(a)
State two variables the student must keep the same to make this a fair test.
(2)
(b)
Calculate the average rate of decrease in temperature of the water in the matt black can over the 10 minutes. Give the unit.
(3)
(c)
The student claims: 'The matt black can transfers energy to the surroundings twice as fast as the shiny silver can.' Evaluate this claim. Use data from the table in your answer.
(3)
(d)
Explain why the rate of decrease in temperature of each can becomes smaller as time goes on.
(2)
Show mark scheme
(a)
any two from:
volume/mass of water;
starting temperature of water;
size/shape/material of can;
room temperature / position in room;
both with lid or both without lid;
Do not accept
colour of can
time
(b)
(temperature drop =) 30 (°C);
evaluation (30 ÷ 10);
°C/min (or °C per minute);
e.g.
(rate =) (80 − 50) ÷ 10
(rate =) 3.0
Allow
0.05 °C/s with unit scores 3 marks
unit mark independent
Do not accept
5 (final temperature ÷ time)
(c)
rate (or drop) for shiny can calculated: 1.7 °C/min OR 17 °C;
comparison: black/shiny ratio ≈ 1.8 / black drop is less than twice shiny drop (30 < 34);
conclusion: black can cools faster but claim (of twice as fast) is not supported;
Allow
ECF from (b)
MP3: allow comparison at a single time interval if correctly calculated, e.g. first 2 min: 8 vs 4 °C so twice at start but not overall
MP3: allow 'not a valid conclusion because some energy transferred by conduction/convection, not only radiation'
Do not accept
black is better (alone) without data
(d)
(as water cools) the temperature difference between can and surroundings decreases;
(cooler) surface emits (infrared) radiation at a lower rate / transfers energy more slowly;
121IGCSE-PHY-4-0154(b) Energy TransfersPapers 1 & 2Grade 8–96 marks
A student wants to find out which of four materials is the best thermal insulator: cotton wool, bubble wrap, newspaper and felt. She has access to standard laboratory equipment, a kettle and a supply of water.
(a)
Describe an investigation the student could carry out to find out which material is the best thermal insulator. You should include the equipment she should use, the measurements she should take, how she makes it a fair test and how she uses her results to reach a conclusion.
(6)
Show mark scheme
(a)
Indicative content:
MP1. use identical beakers/cans, each wrapped in one of the materials;
MP2. same thickness / same number of layers of each material;
MP3. same volume/mass of hot water (measured with measuring cylinder) in each;
MP4. measure starting temperature with a thermometer / same starting temperature;
MP5. measure temperature after a fixed time (e.g. 10 minutes) using a stopwatch / at regular intervals;
MP6. cover with lid (of same material) / same room temperature / same surface;
MP7. include an unwrapped beaker as a control (for comparison);
MP10. smallest temperature drop / slowest rate of cooling = best insulator;
MP11. safety: care with hot water / stand beakers on heatproof mat;
Allow
award 1 mark per MP to a maximum of 6, but use the levels to judge the overall quality
MP5: allow data logger with temperature probe
Do not accept
best insulator is the one that gets hottest
measure temperature of the material instead of the water
Level 0 (0)
No rewardable content.
Level 1 (1–2)
Basic method with little detail, e.g. wrap beakers and measure temperature. Few or no control variables; no clear way of reaching a conclusion.
Level 2 (3–4)
A workable method with measurements of temperature and time and at least one control variable; conclusion may be stated but not fully linked to the measured data.
Level 3 (5–6)
Detailed, logically sequenced method including measuring instruments, several control variables (or a control beaker/repeats) and a clear way of using temperature drop or rate of cooling to identify the best insulator.
122IGCSE-PHY-4-0334(c) Work and PowerPapers 1 & 2Grade 8–98 marks
A cyclist and her bicycle have a total mass of 80 kg. She cycles up a hill at a steady speed. The road is 1200 m long and rises through a vertical height of 60 m. She takes 4.0 minutes to reach the top. The average resistive force (air resistance and friction) on the cyclist and bicycle is 25 N, acting along the road.Diagram not to scale
(a)
Calculate the average useful power output of the cyclist as she climbs the hill.
(5)
(b)
The efficiency of the cyclist's muscles is 25%. Calculate the rate at which energy is transferred from the chemical energy store of her body while she climbs.
(2)
(c)
Explain why a change in kinetic energy is not included in your answer to (a).
(1)
Show mark scheme
(a)
GPE gained = 80 × 10 × 60 (= 48 000 J);
work done against resistive force = 25 × 1200 (= 30 000 J);
total useful energy = 78 000 (J);
time converted to 240 s;
evaluation of power;
e.g.
(P =) 78 000 ÷ 240
(P =) 325 (W)
Allow
330 (W)
200 (W) (GPE only, 48 000 ÷ 240) scores 3 marks
125 (W) (resistive work only) scores 3 marks
19 500 (W) (time not converted) scores 4 marks
Do not accept
work against resistance calculated with 60 m (25 × 60)
(b)
substitution into efficiency relationship (useful ÷ total);
evaluation;
e.g.
(rate =) 325 ÷ 0.25
(rate =) 1300 (W)
Allow
ECF from (a)
1320 (W) from 330 W
1.3 kW / 1300 J/s
Do not accept
81 (W) (325 × 0.25)
(c)
(steady speed so) kinetic energy (store) does not change / no change in KE;
123IGCSE-PHY-4-0344(c) Work and PowerPapers 1 & 2Grade 8–99 marks
A city tram has a mass of 32 000 kg. It is travelling at 15 m/s when it brakes and comes to rest at a tram stop. The tram uses regenerative braking: its motors act as generators, and 60% of the kinetic energy of the tram is transferred electrically to a bank of supercapacitors on board the tram. The supercapacitors store the energy.
(a)
Calculate the energy transferred to the supercapacitors as the tram comes to rest.
(3)
(b)
When the tram leaves the stop, all of the energy stored in the supercapacitors is transferred to the motors. The motors transfer 75% of this energy to the kinetic energy store of the tram. Calculate the speed of the tram when all of the stored energy has been used. Ignore any other energy transfers.
(4)
(c)
Describe what happens to the other 40% of the kinetic energy of the tram during braking.
124IGCSE-PHY-4-0364(b) Energy TransfersPapers 1 & 2Grade 8–96 marks
An astronaut on a spacewalk wears a spacesuit. Space is almost a vacuum. In direct sunlight, the Sun's radiation can make the outside of the suit very hot. In the Earth's shadow, the suit can lose energy to cold space. The suit has: • a white outer layer • several thin sheets of shiny, aluminium-coated plastic inside the outer layer, with vacuum between the sheets • an inner garment containing thin tubes, through which cool water is pumped next to the astronaut's skin.
(a)
Explain how the design of the spacesuit keeps the astronaut at a safe temperature. Your answer should refer to conduction, convection and radiation.
(6)
Show mark scheme
(a)
Indicative content:
MP1. space is a vacuum / has no particles so no conduction or convection between suit and space;
MP2. so energy is transferred to or from the suit (in space) only by radiation;
MP3. white outer layer is a poor absorber / good reflector of the Sun's radiation, reducing heating in sunlight;
MP4. shiny (aluminium) sheets are poor emitters of infrared, reducing energy loss in shadow;
MP5. shiny sheets reflect infrared back (towards the astronaut or away from the inside);
MP6. vacuum between sheets has no particles so prevents conduction and convection between layers;
MP7. thin sheets with little contact reduce conduction through the suit;
MP8. the astronaut's body continually transfers energy to its thermal store, which cannot escape through the insulated suit;
MP9. cool water absorbs energy from the skin by conduction (heating) and carries it away as it flows;
MP10. overall, the rate of energy gain is balanced by the rate of energy removal so temperature stays constant;
Allow
award 1 mark per MP to a maximum of 6, but use the levels to judge the overall quality
MP3: allow 'white surface reflects light/radiation from the Sun'
MP9: allow 'water carries energy away by convection/flow'
Do not accept
vacuum reflects heat
cold is kept out by the suit
white surface is a good absorber
shiny sheets trap air (there is vacuum, not air, between them)
Level 0 (0)
No rewardable content.
Level 1 (1–2)
Isolated statements about one or two features, e.g. 'white reflects heat' or 'the vacuum stops heat', with little or no correct reference to the type of transfer.
Level 2 (3–4)
Explains how at least two features reduce a named type of energy transfer (e.g. vacuum prevents conduction and convection; white/shiny surfaces reduce absorption or emission of radiation). Some links may be incomplete.
Level 3 (5–6)
A coherent explanation covering conduction, convection and radiation, linking each design feature to the transfer it reduces, recognising that only radiation occurs in space AND explaining why the water cooling is needed to remove energy produced by the body.
125IGPE-053-00-104(c) Work and PowerPaper 1PHard8 marks
An electric winch on a harbour quay lifts a crate of mass 450 kg vertically at a steady speed of 0.40 m/s. The input power to the winch motor is 2.4 kW. Use g = 10 N/kg.
(a)
Calculate the weight of the crate.
(1)
(b)
Calculate the useful power output of the winch.
(2)
(c)
Calculate the efficiency of the winch.
(2)
(d)
The crate is lifted through a height of 6.0 m. Calculate the energy wasted by the winch during this lift.
(3)
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Notes
Part (a)
Weight = 450 × 10 = 4500 N
✓ 1
Part (b)
In 1 s the crate rises 0.40 m, so work done each second = 4500 × 0.40
✓ 1
Useful power = 1800 W
✓ 1
Part (c)
Efficiency = 1800 ÷ 2400
✓ 1
= 0.75 or 75 %
✓ 1
Part (d)
Time = 6.0 ÷ 0.40 = 15 s
✓ 1
Wasted power = 2400 − 1800 = 600 W
✓ 1
Energy wasted = 600 × 15 = 9000 J
✓ 1
Common error: Convert kW to W before comparing with power found in watts: 2.4 kW = 2400 W.
126IGPE2-057-01-114(c) Work and PowerPaper 1PHard7 marks
An office-block lift is driven by an electric motor with an input power of 12 kW. The lift car and its passengers have a total mass of 900 kg. The lift rises at constant speed, and the motor system is 75% efficient. g = 10 N/kg
(a)
Calculate the useful power output of the motor system.
(2)
(b)
Calculate the speed at which the lift rises.
(2)
(c)
Calculate the energy wasted while the lift rises through 36 m.
(3)
Show mark scheme
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Mark
Notes
Part (a)
useful power = 0.75 × 12 000
✓ 1
= 9000 W
✓ 1
Part (b)
in 1 s the useful work is 9000 J; weight = 900 × 10 = 9000 N
✓ 1
height risen per second = 9000 ÷ 9000, so speed = 1.0 m/s
127IGPE2-057-01-124(b) Energy TransfersPaper 1PHard6 marks
An archer draws back the string of a bow through a distance of 0.60 m. The average force she applies is 150 N. She then releases an arrow of mass 0.025 kg.
(a)
Calculate the work done by the archer in drawing the bow.
(2)
(b)
80% of this work is transferred to the kinetic energy store of the arrow. Calculate the speed of the arrow as it leaves the bow.
(3)
(c)
Suggest one reason why not all of the work done is transferred to the arrow.
(1)
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Notes
Part (a)
work done = force × distance = 150 × 0.60
✓ 1
= 90 J
✓ 1
Part (b)
kinetic energy = 0.80 × 90 = 72 J
✓ 1
v2 = (2 × 72) ÷ 0.025 = 5760
✓ 1
v = 76 m/s (allow 75.9 m/s)
✓ 1
Part (c)
any one of: energy is transferred to the kinetic energy store of the moving bow limbs / string, or energy is dissipated to the thermal store of the surroundings / as sound when the string vibrates
✓ 1
Common error: Remember to take the square root at the end. 5760 m²/s² is v², not the speed.
128IGPE2-057-01-134(c) Work and PowerPaper 1PHard6 marks
A delivery van drives first along a flat road and then up a steep hill. On both sections it travels at the same constant speed. Explain, in terms of forces, work done and power, why the van's engine must produce a greater power output on the hill. Explain also why the van may have to slow down on the hill if its engine is already working at maximum power.
(a)
Write an extended answer covering both parts of the question.
(6)
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Mark
Notes
Part (a)
at constant speed the resultant force is zero, so on the flat the driving force equals friction / air resistance
✓ 1
on the hill, part of the van's weight acts down the slope, so a larger driving force is needed to balance resistive forces plus this component of weight
✓ 1
the van gains height, so energy is transferred to its gravitational potential energy store as well as being dissipated against resistive forces
✓ 1
at the same speed the van covers the same distance each second, and work done = force × distance, so more work is done each second
✓ 1
power = work done ÷ time (or power = force × velocity), so a greater power output is needed at the same speed
✓ 1
at maximum power, the driving force available is limited (force = power ÷ speed); if it is less than the force needed on the hill, the resultant force acts backwards and the van decelerates until, at the lower speed, the same power gives a large enough driving force
✓ 1
(accept equivalent wording or reasoning that uses P = Fv throughout)
—
Exam tip: In extended answers, link each step clearly: force → work done per second → power. Use connecting words such as 'so' and 'therefore'. Power = force × speed is a quick way to connect the force and power ideas.
129IGPE2-057-03-094(c) Work and PowerPaper 1PHard9 marks
A cyclist and her bicycle have a total mass of 80 kg. She rides up a long hill at a steady speed and gains 150 m in height in 10 minutes. Use g = 10 N/kg. Assume that all of her useful work output goes into gravitational potential energy.
(a)
Calculate the gain in gravitational potential energy of the cyclist and bicycle.
(2)
(b)
Calculate her useful power output.
(2)
(c)
The efficiency of her muscles is 25%. Calculate the rate at which energy is transferred from the chemical store in her body.
(3)
(d)
Explain why the cyclist sweats heavily during the climb.
(2)
Show mark scheme
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Mark
Notes
Part (a)
GPE = 80 × 10 × 150
✓ 1
GPE = 120 000 J (1.2 × 105 J)
✓ 1
Part (b)
P = 120 000 ÷ 600
✓ 1
P = 200 W
✓ 1
Part (c)
efficiency = useful output ÷ total input, so input = useful output ÷ efficiency
✓ 1
input = 200 ÷ 0.25
✓ 1
input = 800 W
✓ 1
Part (d)
the wasted 600 W (75%) is transferred to the thermal store of her body
✓ 1
evaporation of sweat transfers energy away from her skin, which stops her body temperature rising too much
✓ 1
Common error: Multiplying 200 W by 0.25 gives 50 W. The total input must be bigger than the useful output, so you divide by the efficiency.
130IGPE2-057-02-054(d) Energy Resources and Electricity GenerationPaper 1PHard8 marks
A pumped-storage power station has an upper reservoir 320 m above its turbines. When the upper reservoir is full, it holds 4.0 × 109 kg of water. g = 10 N/kg
Diagram not to scale
(a)
Describe how a pumped-storage power station helps the electricity supply meet changes in demand.
(2)
(b)
Calculate the gravitational potential energy lost when all the water falls from the upper reservoir to the turbines.
(2)
(c)
The station converts 85% of this energy into electrical energy. Calculate the electrical energy output.
(2)
(d)
The station produces a steady electrical power of 1.5 GW. Calculate how long it can do this before the upper reservoir is empty. Give your answer in hours.
(2)
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Notes
Part (a)
When demand is low, spare electricity is used to pump water up to the upper reservoir .
✓ 1
When demand is high, the water is released through the turbines to generate electricity within minutes .
✓ 1
Part (b)
GPE = m × g × h = 4.0 × 109 × 10 × 320
✓ 1
= 1.28 × 1013 J
✓ 1
Part (c)
Useful energy output = efficiency × total energy input
✓ 1
= 0.85 × 1.28 × 1013 = 1.1 × 1013 J, which is 1.088 × 1013 J unrounded
✓ 1
Part (d)
t = E ÷ P = 1.088 × 1013 ÷ (1.5 × 109) = 7250 s (7253 s; using 1.1 × 1013 J gives 7300 s)
✓ 1
7250 ÷ 3600 = 2.0 hours
✓ 1
Exam tip: 1 GW = 1 × 10⁹ W. Write the power in watts before you divide, so that the time comes out in seconds. Then divide by 3600 to convert seconds to hours.
131IGPE2-057-02-064(d) Energy Resources and Electricity GenerationPaper 1PHard6 marks
A remote island of 3000 people gets its electricity from old diesel generators. The island council must choose a replacement. It is considering two options: • a wind farm built on the island's exposed hills • a gas-fired power station supplied with gas delivered by ship.
(a)
Evaluate the use of a wind farm and a gas-fired power station for the island. You should give a conclusion.
(6)
Show mark scheme
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Notes
Part (a)
Wind advantage: wind is renewable and will not run out, or a wind farm releases no carbon dioxide while generating .
✓ 1
Wind advantage linked to the island: it needs no fuel, so there is no reliance on ship deliveries or on changing fuel prices, or the exposed hills are likely to be windy, or running costs are low .
✓ 1
Wind disadvantage: wind is variable, so there is no output on calm days and storage or backup is needed, or it has a high initial cost, or it causes visual impact, noise or harm to birds .
✓ 1
Gas advantage: it gives a reliable supply whose output can be changed to match demand, or it can be started quickly, or it is cheaper and quicker to build .
✓ 1
Gas disadvantage: gas is non-renewable, or burning it releases carbon dioxide that adds to global warming, or supply depends on ships that bad weather may delay, or it is exposed to changing gas prices .
✓ 1
A conclusion for either option, supported by a reason linked to the island. For example: wind with storage or backup, because the hills are windy and it ends reliance on imported fuel; or gas, because it is reliable whatever the weather .
✓ 1
Exam tip: In an 'evaluate' answer, give at least one advantage and one disadvantage of each option. Finish with a clear choice and give a reason that links to the context, such as the remote island and its windy hills.
132IGPE2-057-02-104(d) Energy Resources and Electricity GenerationPaper 1PHard9 marks
A large offshore wind turbine has blades that sweep an area of 5000 m2. The wind blows straight at the turbine with a speed of 12 m/s. The density of air is 1.2 kg/m3. Each second, the air reaching the blades fills a cylinder. The cylinder's cross-sectional area equals the swept area, and its length equals the distance the wind travels in one second.
(a)
Show that about 72 000 kg of air reaches the blades each second.
(2)
(b)
Calculate the kinetic energy carried by this air each second.
(3)
(c)
The turbine produces an electrical output power of 2.1 MW. Calculate its efficiency at this wind speed, as a percentage.
(2)
(d)
Explain why halving the wind speed reduces the kinetic energy reaching the blades each second to one eighth.
(2)
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Notes
Part (a)
Volume of air per second = 5000 × 12 = 60 000 m3
✓ 1
mass = density × volume = 1.2 × 60 000 = 72 000 kg
✓ 1
Part (b)
KE = ½ × m × v2
✓ 1
= 0.5 × 72 000 × 122
✓ 1
= 5.2 × 106 J (accept 5.184 × 106 J)
✓ 1
Part (c)
Input power = 5.184 × 106 W (kinetic energy per second). Efficiency = useful output power ÷ input power = 2.1 × 106 ÷ 5.184 × 106
✓ 1
= 0.41, or 41% (accept 40% from using 5.2 × 106; allow error carried forward from (b))
✓ 1
Part (d)
The cylinder of air is half as long, so half the mass of air reaches the blades each second .
✓ 1
KE depends on v2, so each kilogram has one quarter of the kinetic energy, and ½ × ¼ = ⅛ .
✓ 1
Exam tip: Kinetic energy delivered per second is a power: J/s = W. Compare it directly with the output power when you calculate efficiency. Carry your unrounded value through to later parts.
133IGPE2-057-00-084(c) Work and PowerPaper 1PHard8 marks
An electric motor raises a lift cabin of mass 800 kg through a height of 45 m at a steady speed. The journey takes 30 s. The input power to the motor is 18 kW. Take the gravitational field strength g = 10 N/kg.
(a)
Calculate the gain in gravitational potential energy of the cabin.
(2)
(b)
Calculate the efficiency of the lift system.
(3)
(c)
An engineer adds a counterweight of mass 600 kg. The counterweight descends 45 m while the cabin rises 45 m. The efficiency of the system stays the same. Calculate the energy that must now be supplied to the motor for one journey.
Exam tip: The counterweight loses GPE while the cabin gains it. The motor only has to supply the difference, divided by the efficiency. Keep the unrounded efficiency in your calculator so that rounding does not shift your final answer.
134IGPE2-057-00-104(b) Energy TransfersPaper 1PHard6 marks
At the end of a marathon on a cold, windy day, runners are given survival blankets. A survival blanket is a thin sheet of plastic with a shiny metallic coating. Runners wrap it loosely around their bodies.
(a)
Explain how the survival blanket reduces the rate of energy transfer from a runner's body to the surroundings. Refer to conduction, convection and radiation in your answer.
(6)
Show mark scheme
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Notes
Part (a)
Radiation: the shiny surface is a poor emitter of infrared radiation
✓ 1
and it reflects infrared emitted by the body back towards the runner
✓ 1
Conduction: a layer of air is trapped between the blanket and the body
✓ 1
air is a poor conductor (a good insulator), so little energy is conducted away
✓ 1
Convection: the blanket stops the warmed air rising away and being replaced by cold air
✓ 1
it also shields the body from the wind, which would carry warm air away
✓ 1
Exam tip: For a six-mark answer, name each method of energy transfer and link a feature of the blanket to how it reduces that method.
135IGPE2-057-00-134(c) Work and PowerPaper 1PHard8 marks
A pile driver is used to push steel posts into the ground on a building site. A hammer of mass 500 kg is raised and then released. It falls freely through 3.0 m onto the top of a post. Take the gravitational field strength g = 10 N/kg.
Diagram not to scale
(a)
Calculate the gravitational potential energy lost by the hammer as it falls 3.0 m.
(2)
(b)
Assume that air resistance is negligible. Calculate the speed of the hammer just before it hits the post.
(3)
(c)
After impact, the hammer and post move 0.12 m further into the ground. 80% of the hammer's kinetic energy is used doing work against the ground. Ignore any further change in gravitational potential energy. Calculate the average resistive force exerted by the ground on the post.
(3)
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Notes
Part (a)
GPE = m × g × h = 500 × 10 × 3.0
✓ 1
= 15 000 J
✓ 1
Part (b)
KE gained = GPE lost, so 15 000 = ½ × 500 × v2 (allow answer to (a) carried forward)
✓ 1
v2 = 60 (m2/s2)
✓ 1
v = 7.7 m/s (accept 7.75 m/s)
✓ 1
Part (c)
Work done against ground = 0.80 × 15 000 = 12 000 J (allow answer carried forward from earlier parts)
✓ 1
Force = work done ÷ distance = 12 000 ÷ 0.12
✓ 1
= 1.0 × 105 N (100 000 N)
✓ 1
Common error: In (c), use only the energy that does work on the ground (80%). Using the full 15 000 J gives a force that is too large: 1.25 × 10⁵ N.
136IGPE3-4-14r7-114(b) Energy TransfersPaper 1PHard7 marks
A swimmer swims one 50 m length of a pool at a steady speed of 1.4 m/s. The drag force from the water on the swimmer is 45 N. The swimmer's body is 25% efficient at transferring chemical energy to the work done against drag.
(a)
Calculate the work done against drag in one length.
(2)
(b)
Calculate the useful power output of the swimmer.
(2)
(c)
Calculate the chemical energy the swimmer transfers in one length.
(2)
(d)
Suggest what happens to the energy that is wasted by the swimmer.
(1)
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Notes
Part (a)
M1 W = F × d = 45 × 50
✓ 1
A1 2250 J
✓ 1
Part (b)
M1 P = F × v = 45 × 1.4
✓ 1
A1 63 W
✓ 1
Part (c)
M1 input = useful ÷ efficiency = 2250 ÷ 0.25
✓ 1
A1 9000 J (ECF from (a))
✓ 1
Part (d)
The wasted 6750 J is transferred as thermal energy to the swimmer's body and the surrounding water
✓ 1
Exam tip: Check the units: force in N times distance in m gives energy in J, while force in N times speed in m/s gives power in W.
137IGPE3-4-14r10-134(c) Work and PowerPaper 1PHard9 marks
A small aircraft of mass 1200 kg climbs at constant speed from a height of 500 m to a height of 1700 m in 6.0 minutes. Assume that all of the useful energy output of the engine becomes gravitational potential energy of the aircraft. The engine converts 24% of the chemical energy in its fuel into useful energy. The fuel releases 40 MJ per kg when burned. The gravitational field strength is 10 N/kg.
(a)
Calculate the gain in gravitational potential energy of the aircraft.
(2)
(b)
Calculate the useful power output of the engine during the climb.
(2)
(c)
Calculate the chemical energy that must be released by the fuel during the climb.
(2)
(d)
Calculate the mass of fuel burned during the climb.
(2)
(e)
State why the kinetic energy of the aircraft does not need to be included in the calculation.
(1)
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Notes
Part (a)
M1 1200 × 10 × (1700 − 500)
✓ 1
A1 14 400 000 (J)
✓ 1
Part (b)
M1 14 400 000 ÷ (6.0 × 60)
✓ 1
A1 40 000 (W) ECF from (a)
✓ 1
Part (c)
M1 14 400 000 ÷ 0.24
✓ 1
A1 60 000 000 (J) ECF from (a)
✓ 1
Part (d)
M1 60 000 000 ÷ 40 000 000
✓ 1
A1 1.5 (kg) ECF from (c)
✓ 1
Part (e)
B1 the speed is constant, so the kinetic energy does not change
✓ 1
Exam tip: Use the height gained (1700 − 500), not the final height, when calculating the gain in potential energy.
138IGPE3-4-14r18-064(c) Work and PowerPaper 1PHard7 marks
A winch pulls a boat up a slipway along a straight path. The force from the cable is a constant 600 N for the first 4.0 m. Over the next 4.0 m the force increases steadily from 600 N to 1000 N.
(a)
Calculate the work done by the cable over the first 4.0 m.
(2)
(b)
Calculate the work done over the next 4.0 m.
(2)
(c)
Determine the total work done over the whole 8.0 m.
(1)
(d)
The boat takes 28 s to move the 8.0 m. Calculate the average power of the winch.
(2)
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Notes
Part (a)
M1 work done = force × distance = 600 × 4.0
✓ 1
A1 2400 J
✓ 1
Part (b)
M1 average force = (600 + 1000) ÷ 2 = 800 N, multiplied by 4.0 m
✓ 1
A1 3200 J
✓ 1
Part (c)
A1 5600 J (ECF from (a) and (b))
✓ 1
Part (d)
M1 5600 ÷ 28 (ECF)
✓ 1
A1 200 W
✓ 1
Common error: When the force changes steadily, use the average force; do not use the final force for the whole distance.
140IGPE3-4-15r33-114(b) Energy TransfersPaper 1PHard8 marks
A skateboarder and board have a total mass of 62 kg. The skateboarder is released from rest at a point 1.50 m above the lowest point of a half-pipe. The skateboarder rolls down and up the other side, stopping momentarily at a height of 1.20 m above the lowest point. Use g = 10 N/kg.
Diagram not to scale
(a)
Calculate the energy transferred to thermal energy stores between the release point and the highest point on the other side.
(2)
(b)
Calculate the percentage of the initial gravitational potential energy that was transferred to thermal energy stores.
(2)
(c)
The skateboarder travels 4.65 m along the curved surface. Calculate the average resistive force.
(2)
(d)
Explain why the skateboarder reaches a lower height on each successive rise if no pushing is done.
(2)
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Mark
Notes
Part (a)
M1 energy lost = mgΔh with Δh = 1.50 − 1.20 = 0.30 m
141IGPE3-4-15r40-044(c) Work and PowerPaper 1PHard1 mark
An electric motor has an input power of 400 W and an efficiency of 60%. It lifts a load of mass 12 kg at a constant speed. Take g = 10 N/kg. What is the speed of the load?
Show mark scheme
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Notes
Useful power = 0.60 × 400 = 240 W. The weight of the load is 12 × 10 = 120 N. For constant speed, useful power = force × speed, so speed = 240 ÷ 120 = 2.0 m/s. Using the full 400 W ignores the wasted energy and gives 3.3 m/s. Using the wasted 40% (160 W) gives 1.3 m/s. Dividing the weight by the useful power inverts the relationship and gives 0.50 m/s.
✓ 1
Answer A
Answer: A
Exam tip: For a motor lifting a load, work out the useful power first, then use power = force × speed with the weight as the force.
142IGPE3-4-15r40-104(c) Work and PowerPaper 1PHard7 marks
A builder pushes a wheelbarrow of total mass 55 kg at constant speed up a straight ramp 6.0 m long. The top of the ramp is 1.2 m above the bottom. The builder pushes parallel to the ramp. The resistive force from friction and air is constant at 25 N. The builder takes 9.0 s. g = 10 N/kg.
Diagram not to scale
(a)
Calculate the gain in gravitational potential energy of the wheelbarrow.
(2)
(b)
Calculate the force with which the builder pushes the wheelbarrow.
(3)
(c)
Calculate the power developed by the builder.
(2)
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Notes
Part (a)
M1 gain in GPE = mass × g × height
✓ 1
A1 55 × 10 × 1.2 = 660 J
✓ 1
Part (b)
M1 work done against the resistive force = 25 × 6.0 = 150 J
✓ 1
M1 total work done by the builder = 660 + 150 = 810 J
✓ 1
A1 force = 810 ÷ 6.0 = 135 N (ECF from (a))
✓ 1
Part (c)
M1 power = work done ÷ time
✓ 1
A1 810 ÷ 9.0 = 90 W (ECF from (b))
✓ 1
Exam tip: At constant speed the kinetic energy does not change, so the work done by the pusher equals the energy gained by the gravitational store plus the energy wasted against resistive forces.
143IGPE3-4-16r42-134(b) Energy TransfersPaper 1PHard8 marks
An electric pump lifts water from a borehole into a storage tank on a farm. In 15 minutes the pump lifts 12 000 kg of water through a height of 18 m. The electrical power supplied to the pump is 4.0 kW. The gravitational field strength is 10 N/kg.
(a)
Calculate the gain in gravitational potential energy of the water.
(2)
(b)
Calculate the electrical energy supplied to the pump in 15 minutes.
(2)
(c)
Calculate the efficiency of the pump system. Give your answer as a percentage.
(2)
(d)
State where the wasted energy goes and suggest one way of improving the efficiency of the system.
(2)
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Notes
Part (a)
M1 gain in GPE = m × g × h = 12 000 × 10 × 18
✓ 1
A1 2 160 000 J (accept 2.16 MJ)
✓ 1
Part (b)
M1 energy = power × time = 4000 × (15 × 60)
✓ 1
A1 3 600 000 J (accept 3.6 MJ)
✓ 1
Part (c)
M1 efficiency = useful energy out ÷ total energy in = 2 160 000 ÷ 3 600 000
✓ 1
A1 60% (ECF from (a) and (b))
✓ 1
Part (d)
wasted energy is transferred as thermal energy (and sound) to the surroundings (due to friction in the pump and resistance in the motor)
✓ 1
any one from: lubricate moving parts / use a wider pipe to reduce friction / use a more efficient motor
✓ 1
Exam tip: Check units before dividing: energy in joules and time in seconds, so 15 minutes is 900 s.
145IGCSE-PHY-4-0374(d) Energy Resources and Electricity GenerationPaper 2 onlyGrade 4–57 marks
The block diagram shows the main parts of a nuclear power station. In the reactor, the fission of uranium fuel heats water in the boiler to make high-pressure steam. The steam drives a turbine, which turns a generator.
(a)
Name the energy store of the uranium fuel, and state how energy is transferred from the reactor to the water in the boiler.
(2)
(b)
A student writes: 'In the turbine, the thermal energy of the steam is changed into electrical energy, which is stored in the generator.' Identify two mistakes in the student's statement.
(2)
(c)
Name one energy resource that can generate electricity without heating water to make steam.
(1)
(d)
Give one advantage and one disadvantage of using nuclear power stations to generate electricity on a large scale.
(2)
Show mark scheme
(a)
nuclear (energy store);
(transferred) by heating;
Allow
thermally for by heating
Do not accept
chemical (store)
radioactive energy
by radiation (alone)
mechanically
(b)
any two from:
(steam/turbine) energy goes to the kinetic (energy store) of the turbine / is transferred mechanically, not into electrical energy;
the generator does not store (electrical) energy / electrical energy is not a store;
(it is) the generator (not the turbine) that transfers energy electrically (to the grid);
Allow
a correct rewritten statement that removes both mistakes scores 2 marks
Do not accept
the steam is not thermal energy (alone)
the turbine stores kinetic energy (alone) with no reference to the mistake
(c)
wind / hydroelectric / water / tidal / wave / solar cells;
Allow
photovoltaic for solar cells
Do not accept
geothermal
solar heating
coal / oil / gas / biofuel / nuclear
solar (unqualified)
(d)
advantage: no CO₂/greenhouse gases (when generating) / reliable or constant output / large energy from small mass of fuel;
disadvantage: radioactive waste (long-lived, needs storage) / high building or decommissioning cost / risk of accident or leak / non-renewable fuel;
Allow
'does not contribute to global warming' for advantage
146IGCSE-PHY-4-0184(d) Energy Resources and Electricity GenerationPaper 2 onlyGrade 6–78 marks
The diagram shows a geothermal power station. Cold water is pumped down to hot rocks deep underground. Hot water and steam return to the surface.
(a)
Describe the energy transfers involved in generating electricity in a geothermal power station.
(3)
(b)
Give one disadvantage of geothermal power stations.
(1)
(c)
A house has a solar heating system and a solar cell panel on its roof. Describe how the energy transfer in the solar heating system is different from the energy transfer in the solar cells.
(2)
(d)
On a sunny day, the power of the sunlight reaching the solar cell panel is 1.6 kW. The electrical power output of the panel is 0.30 kW. Calculate the efficiency of the solar cell panel.
(2)
Show mark scheme
(a)
energy (from thermal store of) hot rocks transferred by heating to water / turns water to steam;
steam turns turbine / kinetic store of turbine;
generator transfers energy electrically (to the grid);
Allow
MP3: allow 'generator produces electricity'
Do not accept
rocks burn
steam turns generator directly with no turbine (MP2)
(b)
any one from:
only available in a few places / volcanic regions / where hot rocks near surface;
high cost of drilling deep wells;
can release harmful gases (e.g. hydrogen sulfide/CO₂) from underground;
Do not accept
it is non-renewable
does not work at night
(c)
solar heating system: (infrared) radiation absorbed to heat water / increase thermal store of water;
solar cells: (light) radiation transferred (directly) electrically / produces electricity;
147IGCSE-PHY-4-0384(d) Energy Resources and Electricity GenerationPaper 2 onlyGrade 6–79 marks
The diagram shows a pumped-storage hydroelectric power station. When electricity is needed, water flows from the upper reservoir down a pipe to a turbine and generator, then into the lower reservoir. At times of low demand, electric pumps lift water back up to the upper reservoir. g = 10 N/kg
(a)
Describe the energy transfers that take place as water flows down from the upper reservoir and electricity is generated.
(3)
(b)
When the station is generating, 90 000 kg of water passes through the turbines every second. The water falls through a vertical height of 300 m. Calculate the gravitational potential energy lost by the water each second.
(2)
(c)
The electrical power output of the station is 230 MW. Calculate the efficiency of the station when generating.
(2)
(d)
Give one advantage and one disadvantage of a pumped-storage scheme for electricity supply.
(2)
Show mark scheme
(a)
gravitational (potential energy store of water) decreases and kinetic (store of water) increases;
(moving water) transfers energy mechanically to the kinetic (energy store) of the turbine;
generator transfers energy electrically (to the grid);
advantage: stores energy at times of low demand (e.g. surplus from wind/nuclear at night) and can supply it very quickly at peak demand / no fuel or CO₂ at the station;
disadvantage: more energy needed to pump water up than is generated / few suitable sites / flooding of land or harm to habitats / high building cost;
148IGPE2-057-03-114(d) Energy Resources and Electricity GenerationPaper 2PMedium7 marks
A small island community plans to generate electricity using a tidal barrage across a river estuary. At high tide, water is trapped behind the barrage. It is released through turbines when the tide outside has gone out.
(a)
Describe the energy transfers that take place as the barrage generates electricity.
(3)
(b)
The island could use solar panels instead. Give one advantage and one disadvantage of the tidal barrage compared with solar panels for this island.
(2)
(c)
During one tide, 4.0 × 109 kg of water falls through an average height of 2.5 m. Calculate the gravitational potential energy transferred by this water. Use g = 10 N/kg.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
the trapped water has energy in its gravitational potential store
✓ 1
as the water falls through the barrage, this energy is transferred to the kinetic store of the water
✓ 1
the moving water turns the turbines and generators, which transfer the energy electrically to the island's homes
✓ 1
Part (b)
advantage: e.g. tides are predictable, unlike sunshine, or it can generate at night and in cloudy weather
✓ 1
disadvantage: e.g. much higher building cost, or it damages estuary habitats, or it blocks boats and shipping, or it needs a suitable estuary whereas panels can go on roofs
✓ 1
Part (c)
GPE = 4.0 × 109 × 10 × 2.5
✓ 1
GPE = 1.0 × 1011 J
✓ 1
Exam tip: In comparison questions, make each point relative. For example, write "tides are predictable, unlike sunshine" rather than just "tides are renewable", because solar is renewable too.
149IGPE2-057-00-094(d) Energy Resources and Electricity GenerationPaper 2PMedium7 marks
In a hydroelectric power station, water flows from a high reservoir down through turbines. Each second, 2.0 × 104 kg of water falls through a vertical height of 120 m. Take the gravitational field strength g = 10 N/kg.
(a)
Describe the energy transfers that take place as electricity is generated in the power station.
(3)
(b)
Calculate the gravitational potential energy lost by the water each second.
(2)
(c)
The power station has an efficiency of 85%. Calculate the electrical power output.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Energy in the gravitational store of the water decreases as it falls and is transferred to the kinetic store of the water
✓ 1
Energy is transferred mechanically (the water does work) to the kinetic store of the turbines and generator
✓ 1
The generator transfers energy electrically to consumers / the National Grid, with some wasted as thermal energy to the surroundings
✓ 1
Part (b)
GPE = mgh = 2.0 × 104 × 10 × 120
✓ 1
= 2.4 × 107 J
✓ 1
Part (c)
Useful power output = 0.85 × 2.4 × 107 (allow their answer to (b))
✓ 1
= 2.0 × 107 W (accept 2.04 × 107 W)
✓ 1
Exam tip: The energy transferred each second is numerically equal to the power in watts, because 1 W = 1 J/s.
150IGPE3-4-14r7-094(b) Energy TransfersPaper 2PMedium8 marks
A spacesuit has a liquid cooling garment. Water, with specific heat capacity 4200 J/kg °C, circulates through tubes in the garment and absorbs thermal energy from the astronaut.
(a)
Calculate the thermal energy absorbed when 3.0 kg of water warms from 14 °C to 17 °C.
(3)
(b)
The astronaut's body transfers thermal energy to the water at a rate of 140 W. Calculate the time for the water to absorb 37 800 J.
(2)
(c)
The outer layer of the suit is shiny white. Explain how this helps the astronaut in sunlight.
(2)
(d)
State why the suit cannot lose thermal energy by convection to its surroundings in space.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 ΔQ = mcΔθ
✓ 1
M1 3.0 × 4200 × 3
✓ 1
A1 37 800 J
✓ 1
Part (b)
M1 time = energy ÷ power = 37 800 ÷ 140
✓ 1
A1 270 s (ECF from (a))
✓ 1
Part (c)
A shiny white surface is a good reflector and a poor absorber of radiation
✓ 1
so less thermal energy is absorbed from the Sun and the suit stays cooler
✓ 1
Part (d)
Convection needs a fluid (particles that can flow) and space is a vacuum
✓ 1
Exam tip: Temperature change is final minus initial temperature; 17 − 14 = 3 °C, not the sum of the two.
151IGPE3-4-14r10-104(d) Energy Resources and Electricity GenerationPaper 2PMedium5 marks
A greenhouse has a water tank containing 150 kg of water at 12 °C. An electric immersion heater of power 2000 W heats the water to 30 °C. The specific heat capacity of water is 4200 J/kg °C. The heater transfers 90% of its electrical energy to the water.
(a)
Calculate the energy needed to raise the temperature of the water from 12 °C to 30 °C.
(2)
(b)
Calculate the time the heater takes to heat the water. Give your answer in seconds.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 150 × 4200 × (30 − 12)
✓ 1
A1 11 340 000 (J)
✓ 1
Part (b)
M1 useful power = 0.90 × 2000 = 1800 W
✓ 1
M1 time = energy ÷ power = 11 340 000 ÷ 1800
✓ 1
A1 6300 (s)
✓ 1
Common error: Dividing by the full 2000 W instead of the 90% that reaches the water gives a time that is too short.
152IGPE3-4-14r10-114(b) Energy TransfersPaper 2PMedium6 marks
In a workshop that makes drums, an aluminium plate of mass 0.80 kg is heated from 20 °C by an electric heater of power 60 W for 5.0 minutes. The specific heat capacity of aluminium is 900 J/kg °C. Only 80% of the electrical energy supplied is transferred to the plate.
(a)
Calculate the electrical energy supplied by the heater in 5.0 minutes.
(2)
(b)
Calculate the energy transferred to the plate.
(1)
(c)
Calculate the final temperature of the plate.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 60 × (5.0 × 60) (time converted to 300 s)
✓ 1
A1 18 000 (J)
✓ 1
Part (b)
B1 0.80 × 18 000 = 14 400 (J) ECF from (a)
✓ 1
Part (c)
M1 14 400 = 0.80 × 900 × Δθ
✓ 1
A1 Δθ = 20 (°C)
✓ 1
A1 final temperature 40 (°C)
✓ 1
Exam tip: Convert minutes to seconds before using power × time, and add the temperature rise to the starting temperature at the end.
154IGPE3-4-15r33-064(b) Energy TransfersPaper 2PMedium9 marks
A student measures the temperature of the water in a matt black can and in a shiny silver can as they cool. Each can contains 0.25 kg of water. The specific heat capacity of water is 4200 J/kg °C. Ignore the energy stored in the cans themselves.
Time / min
0
5
10
15
20
Black can, temperature / °C
85
70
59
51
45
Silver can, temperature / °C
85
76
69
63
58
(a)
Calculate the average rate of cooling of the silver can in the first 10 minutes, in °C per minute.
(2)
(b)
Explain why the rate of cooling of the black can becomes smaller as time passes.
(2)
(c)
Calculate the energy transferred from the water in the black can during the 20 minutes.
(3)
(d)
Calculate the average power with which the black can transfers energy during the 20 minutes. Use the unrounded answer from (c).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 temperature fall = 85 − 69 = 16 °C divided by 10 min
✓ 1
A1 1.6 °C per minute
✓ 1
Part (b)
The temperature difference between the can and the surroundings gets smaller as the can cools
✓ 1
so the energy transferred to the surroundings each second decreases
155IGPE3-4-15r33-124(d) Energy Resources and Electricity GenerationPaper 2PMedium8 marks
A wind turbine supplies a remote archaeological camp. The blades sweep out an area of 50 m2. The wind speed is 8.0 m/s and the density of air is 1.2 kg/m3. Assume that the air is brought completely to rest by the blades.
(a)
Calculate the volume of air that reaches the blades each second.
(2)
(b)
Calculate the mass of air that reaches the blades each second.
(1)
(c)
Calculate the kinetic energy of this mass of air. Give your answer to 3 significant figures.
(3)
(d)
The turbine transfers 40% of this kinetic energy each second into electrical energy. Calculate the electrical power output. Use the unrounded value from (c).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 volume per second = area × wind speed
✓ 1
A1 50 × 8.0 = 400 m3
✓ 1
Part (b)
mass = density × volume = 1.2 × 400 = 480 kg (ECF from (a))
✓ 1
Part (c)
M1 use of KE = ½mv²
✓ 1
M1 ½ × 480 × 8.02 (ECF from (b))
✓ 1
A1 15 400 J (unrounded value 15 360 J)
✓ 1
Part (d)
M1 power = 0.40 × 15 360 J each second
✓ 1
A1 6144 W (6.1 kW)
✓ 1
Exam tip: Kinetic energy depends on the square of speed, so square only the speed, not the mass or the product.
156IGPE3-4-15r40-114(b) Energy TransfersPaper 2PMedium6 marks
A student fills two identical metal cans with 0.200 kg of water at 80 °C. One can has a matt black surface and the other has a shiny silver surface. After 10 minutes in a cool room, the water in the black can is at 52 °C and the water in the silver can is at 66 °C. The specific heat capacity of water is 4200 J/kg °C.
(a)
State which can is the better emitter of infrared radiation and give the evidence.
(2)
(b)
Calculate the energy transferred away from the water in the black can.
(2)
(c)
Calculate the mean rate at which energy is transferred away from the water in the black can.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
The matt black can
✓ 1
Its water cooled by more (28 °C compared with 14 °C) in the same time
✓ 1
Part (b)
M1 energy = mass × specific heat capacity × temperature change, with change = 80 − 52 = 28 °C
✓ 1
A1 0.200 × 4200 × 28 = 23 520 J
✓ 1
Part (c)
M1 power = energy ÷ time with time = 600 s
✓ 1
A1 23 520 ÷ 600 = 39.2 W (ECF from (b))
✓ 1
Common error: Use the temperature change, not the final temperature, in the specific heat capacity equation.
157IGPE3-4-15r40-124(d) Energy Resources and Electricity GenerationPaper 2PMedium7 marks
A data centre has a solar farm with 250 m2 of solar panels. Sunlight delivers 800 W of power to each square metre of the panels. The panels are 18% efficient.
(a)
Calculate the power of the sunlight arriving at the panels.
(2)
(b)
Calculate the electrical power output of the panels.
(2)
(c)
Calculate the electrical energy produced in 5.0 hours. Give your answer in kWh.
(2)
(d)
Suggest one reason why the solar farm alone cannot supply a data centre that runs day and night.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 power = 800 × 250
✓ 1
A1 200 000 W (200 kW)
✓ 1
Part (b)
M1 output power = efficiency × input power
✓ 1
A1 0.18 × 200 000 = 36 000 W (36 kW) (ECF from (a))
✓ 1
Part (c)
M1 energy = power in kW × time in hours
✓ 1
A1 36 × 5.0 = 180 kWh (ECF from (b))
✓ 1
Part (d)
There is no sunlight at night, or cloud reduces the output, so the supply is not continuous
✓ 1
Exam tip: When energy is asked in kWh, use power in kW and time in hours. Do not convert to joules and back.
158IGPE3-4-16r42-114(b) Energy TransfersPaper 2PMedium7 marks
The outer layers of an astronaut's spacesuit are designed to control thermal energy transfers. The outside of the suit is shiny white, and there is a vacuum gap between layers of the suit. The astronaut's body transfers energy at a rate of 420 W to a liquid cooling garment. The water in the garment has mass 2.0 kg and specific heat capacity 4200 J/kg°C.
(a)
Explain why the outside of the suit is shiny white rather than matt black.
(2)
(b)
Explain why a vacuum gap between layers of the suit reduces the thermal energy transferred.
(2)
(c)
The astronaut's body transfers energy to the water for 60 s. Calculate the increase in temperature of the water, assuming that all of the energy goes into the water and that none is lost.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
a shiny white surface is a poor absorber of radiation from the Sun
✓ 1
so the suit absorbs less thermal energy / reflects most radiation, keeping the astronaut cooler
✓ 1
Part (b)
a vacuum contains no particles, so there can be no conduction
✓ 1
and no convection
✓ 1
Part (c)
M1 energy = power × time = 420 × 60 = 25 200 J
✓ 1
M1 change in temperature = energy ÷ (mass × specific heat capacity) = 25 200 ÷ (2.0 × 4200)
✓ 1
A1 3.0 °C
✓ 1
Exam tip: Use the equation ΔQ = mcΔθ and make sure the energy is in joules before substituting.
159IGCSE-PHY-4-0194(d) Energy Resources and Electricity GenerationPaper 2 onlyGrade 8–911 marks
A gas-fired power station supplies a steady electrical power of 600 MW. Engineers propose replacing it with offshore wind turbines. The table shows the mean power output of one turbine in each of 8 weeks. Week / number: 1, 2, 3, 4, 5, 6, 7, 8 Mean power of one turbine / MW: 5.6, 1.2, 3.8, 0.6, 2.9, 4.4, 5.1, 2.0
(a)
Describe the energy transfers in a gas-fired power station, from the fuel to the electricity supplied to the grid.
(3)
(b)
Calculate the mean power output of one turbine over the 8 weeks.
(2)
(c)
Calculate the number of turbines needed so that their mean power output over the 8 weeks equals 600 MW.
(2)
(d)
A council member says: 'If we build the number of turbines calculated in (c), they will always supply as much electricity as the gas-fired power station.' Evaluate this statement. Use a calculation in your answer.
(4)
Show mark scheme
(a)
chemical (energy) store of gas/fuel transferred by burning/combustion (to thermal store);
water heated / turned into steam (thermal store of steam), which turns the turbine (kinetic store);
generator transfers energy electrically;
Do not accept
gas turns the turbine directly (MP2)
(b)
sum ÷ 8 / substitution;
evaluation;
e.g.
(mean =) 25.6 ÷ 8
(mean =) 3.2 (MW)
(c)
substitution;
evaluation rounded up to a whole number;
e.g.
(N =) 600 ÷ 3.2 = 187.5
(N =) 188
Allow
ECF from (b) (rounded up)
187.5 scores 1 mark
187 scores 1 mark
(d)
calculation of output in a low week, e.g. week 4: 188 × 0.6 = 113 (MW) / week 2: 188 × 1.2 = 226 (MW);
comparison: much less than 600 MW (e.g. about 19%) / output sometimes exceeds 600 MW (e.g. week 1: 1053 MW);
statement is incorrect because wind (output) varies / depends on the weather (only the mean matches);
need backup / energy storage / keep (part of) gas station for calm periods;
Allow
ECF from (c)
MP4: allow balanced point that wind is renewable/no CO₂ emissions/no fuel cost but this does not make the statement correct
160IGCSE-PHY-4-0204(d) Energy Resources and Electricity GenerationPaper 2 onlyGrade 8–96 marks
A small island gets all of its electricity from diesel generators. Diesel is a fossil fuel brought to the island by ship. The island council is considering three alternatives: • a large array of solar cells (a solar farm) • a tidal barrage across a bay • a small nuclear power station
(a)
Discuss the advantages and disadvantages of each of the three alternatives compared with the diesel generators.
(6)
Show mark scheme
(a)
Indicative content:
MP1. diesel: available on demand / reliable BUT non-renewable and releases CO₂ (greenhouse gas) / pollution;
MP2. diesel: fuel must be shipped to island (cost/supply can be interrupted / spill risk);
MP3. solar cells: renewable / no fuel cost / no CO₂ released when generating;
MP4. solar cells: no output at night and less on cloudy days, so storage (e.g. batteries) or backup needed;
MP5. solar cells: need a large area of land on a small island / high initial cost for the output obtained;
MP6. tidal: renewable and predictable / tides occur twice a day and times known in advance;
MP7. tidal: high building cost / damages habitats in the bay / affects shipping; output varies through the tidal cycle;
MP8. nuclear: reliable (large constant output) and no CO₂ released when generating / small mass of fuel;
MP9. nuclear: radioactive waste must be stored safely for a long time / risk of accident / non-renewable;
MP10. nuclear: high cost of building and decommissioning / may be too large for small island demand;
MP11. sensible conclusion supported by points, e.g. solar + tidal with some backup or storage;
Allow
award 1 mark per MP to a maximum of 6, but use the levels to judge the overall quality
MP4: allow 'depends on the weather / time of day'
Do not accept
nuclear power releases CO₂ / causes global warming when generating
tides are unpredictable
solar cells generate electricity at night
Level 0 (0)
No rewardable content.
Level 1 (1–2)
Isolated advantages or disadvantages, mostly for one alternative, with little comparison to diesel.
Level 2 (3–4)
Advantages and disadvantages given for at least two alternatives with some comparison to diesel; reasoning partly developed.
Level 3 (5–6)
Balanced discussion of all three alternatives, with both advantages and disadvantages linked to the island's situation and compared with diesel; may reach a supported conclusion.
161IGCSE-PHY-4-0394(d) Energy Resources and Electricity GenerationPaper 2 onlyGrade 8–910 marks
On a windy day, air approaches a wind turbine at a speed of 12 m/s. Every second, 8000 kg of air passes through the area swept by the blades. Behind the turbine, the speed of this air has fallen to 9.0 m/s. Assume that all of this air slows from 12 m/s to 9.0 m/s. The electrical power output of the turbine's generator is 200 kW.
(a)
Describe the energy transfers involved in generating electricity using a wind turbine.
(3)
(b)
Calculate the efficiency with which the turbine transfers the kinetic energy lost by the air into electrical energy.
(5)
(c)
An engineer suggests redesigning the blades so that the air leaves the turbine at 0 m/s, so that all of the kinetic energy of the wind is transferred. Explain why this is not possible.
(2)
Show mark scheme
(a)
kinetic (energy store) of the air/wind decreases;
energy transferred mechanically to the kinetic (energy store) of the blades/turbine;
generator transfers energy electrically (to the grid);
Allow
some energy dissipated to thermal store / as sound
Do not accept
wind energy store
chemical
(b)
KE of air arriving per second = 0.5 × 8000 × 12² (= 576 000 J);
KE of air leaving per second = 0.5 × 8000 × 9.0² (= 324 000 J);
KE lost per second = 252 000 (J);
substitution into efficiency relationship (200 000 ÷ 252 000);
evaluation;
e.g.
(efficiency =) 200 000 ÷ 252 000 (× 100%)
(efficiency =) 79 (%)
Allow
0.79
79.4 (%)
ECF for an arithmetic slip in KE lost
35 (%) (200 000 ÷ 576 000, leaving air ignored) scores 2 marks
use of (12 − 9)² giving 36 000 J and an efficiency > 100%: max 1 mark (efficiency substitution)
Do not accept
126 (%) / 1.26 (inverted)
0.079 (%) (kW and W mixed)
(c)
the air would stop / pile up at (or behind) the blades;
so no more air could flow through the turbine / the air must keep moving (with some KE) to leave the turbine;
162IGCSE-PHY-4-0404(d) Energy Resources and Electricity GenerationPaper 2 onlyGrade 8–911 marks
The Sankey diagram shows the energy transfers each second in a coal-fired power station. It is drawn to scale on a grid. The width of each arrow is proportional to the power it represents. The chemical energy from the coal is transferred at a rate of 2000 MW.
(a)
The coal is burned to heat water in a boiler. Name the energy store that decreases as the coal burns and the energy store of the cooling water that increases.
(2)
(b)
Use the diagram to determine the electrical power output, and calculate the efficiency of the power station.
(3)
(c)
The power station is converted to 'combined heat and power'. Hot water from the cooling system is piped to heat nearby homes. After the conversion, the coal input is still 2000 MW, the electrical output falls to 600 MW and 800 MW is supplied to the homes for heating. Before the conversion, these homes were heated by gas boilers with an efficiency of 80%. The electrical output that is lost is now generated by another power station with an efficiency of 40%. A local newspaper claims: 'The conversion reduces the total rate at which energy from fuels is used.' Evaluate this claim. Use calculations in your answer.
(5)
(d)
Give one disadvantage of generating electricity using coal compared with using renewable resources.
(1)
Show mark scheme
(a)
chemical (energy store of the coal) decreases;
thermal (energy store of the cooling water) increases;
Allow
internal energy for thermal
Do not accept
heat energy store
nuclear
(b)
electrical output = 700 MW (from the diagram);
substitution into efficiency relationship;
evaluation;
e.g.
(efficiency =) 700 ÷ 2000 (× 100%)
(efficiency =) 35 (%)
Allow
0.35
ECF for a misread output value
Do not accept
2.9 / 286 (%) (inverted)
50 (%) (cooling water used as useful)
(c)
fuel (gas) energy no longer needed for boilers = 800 ÷ 0.80 = 1000 (MW);
extra fuel energy needed to replace lost electricity = (700 − 600) ÷ 0.40 = 250 (MW);
net reduction = 1000 − 250 = 750 (MW);
conclusion: claim supported because fuel use falls (by 750 MW) while coal input is unchanged;
limitation: heating only needed in cold weather / homes must be near the station / energy lost from the hot-water pipes, so actual saving smaller;
Allow
ECF throughout, including from (b) for the lost electrical output
MP4 for a correct conclusion based on their values
overall efficiency (600 + 800) ÷ 2000 = 70% compared with 35% before scores MP4 only if MP1–MP3 not scored
Do not accept
640 (MW) (800 × 0.80, multiplying by efficiency)
40 (MW) (100 × 0.40)
claim supported because electrical output increases
(d)
any one from:
releases carbon dioxide / greenhouse gas (contributes to global warming / climate change);
163IGPE-053-00-124(d) Energy Resources and Electricity GenerationPaper 2PHard6 marks
A small island community currently gets all its electricity from a diesel generator. The diesel fuel is delivered by boat every two weeks. The council is considering replacing the generator with a large array of solar cells and a bank of rechargeable batteries.
(a)
Evaluate the proposal to replace the diesel generator with solar cells and batteries.
(6)
Show mark scheme
Marking point
Mark
Notes
Part (a)
One advantage stated, e.g. solar energy is renewable, so no fuel needs to be bought or shipped in
✓ 1
A second, different advantage, e.g. no carbon dioxide or other polluting gases are released while generating, unlike burning diesel; or low running costs
✓ 1
One disadvantage stated, e.g. no output at night and less output on cloudy days or in winter, so the supply is not constant
✓ 1
A second, different disadvantage, e.g. high initial cost of the panels and batteries; or the array needs a large area of land; or the batteries wear out, need replacing and must be disposed of safely
✓ 1
A further relevant point (for or against) linked to the island, e.g. supply is no longer disrupted when bad weather delays the fuel boat; or the batteries can only store a limited amount of energy for long cloudy spells
✓ 1
A conclusion consistent with the points made, e.g. a good choice if the island gets enough sunshine and the battery storage is large enough, perhaps keeping the diesel generator as a backup
✓ 1
Exam tip: 'Evaluate' needs points for and against, then a conclusion that is supported by your points. Link your points to the situation given, such as the fuel deliveries by boat. A list of facts with no judgement cannot gain full marks.
164IGPE3-4-14r7-124(b) Energy TransfersPaper 2PHard7 marks
A motor in a library lift raises a trolley of books with total mass 90 kg. The motor takes in electrical power of 600 W and is 70% efficient. It runs for 15 s and all its useful energy output raises the trolley. Use g = 10 N/kg.
(a)
Calculate the electrical energy supplied to the motor in 15 s.
(2)
(b)
Calculate the useful energy output of the motor.
(2)
(c)
State the main form of the wasted energy.
(1)
(d)
Calculate the height through which the trolley is raised.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 E = P × t = 600 × 15
✓ 1
A1 9000 J
✓ 1
Part (b)
M1 0.70 × 9000
✓ 1
A1 6300 J (ECF from (a))
✓ 1
Part (c)
Thermal energy (in the motor and bearings)
✓ 1
Part (d)
M1 ΔGPE = mgh so h = 6300 ÷ (90 × 10)
✓ 1
A1 7.0 m (ECF from (b))
✓ 1
Common error: Only the useful output (6300 J), not the energy supplied (9000 J), raises the trolley.
165IGPE3-4-14r7-144(c) Work and PowerPaper 2PHard7 marks
An astronaut is in the shadow of a spacecraft. The astronaut and spacesuit together have a mass of 90 kg and an average specific heat capacity of 3500 J/kg °C. The astronaut's body transfers thermal energy to the suit system at 85 W and an electric heater in the suit adds 40 W. The suit loses thermal energy to space by radiation at 110 W.
(a)
Calculate the net power gained by the astronaut and suit system.
(2)
(b)
Calculate the energy gained in 45 minutes.
(2)
(c)
Calculate the rise in temperature of the astronaut and suit in this time.
(2)
(d)
Calculate the heater power that would keep the temperature constant.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 85 + 40 − 110
✓ 1
A1 15 W
✓ 1
Part (b)
M1 time = 45 × 60 = 2700 s
✓ 1
A1 15 × 2700 = 40 500 J (ECF from (a))
✓ 1
Part (c)
M1 Δθ = 40 500 ÷ (90 × 3500)
✓ 1
A1 0.13 °C (ECF from (b))
✓ 1
Part (d)
A1 110 − 85 = 25 W
✓ 1
Common error: Convert minutes to seconds before using power × time; using 45 instead of 2700 gives a result 60 times too small.
166IGPE3-4-14r10-144(b) Energy TransfersPaper 2PHard10 marks
A dairy farm has an electrically heated hot-water tank containing 200 kg of water at 60 °C. The specific heat capacity of water is 4200 J/kg °C. When the tank is fitted with a lagging jacket, the temperature of the water falls by 3.0 °C in 10 hours. Without the jacket, the temperature falls by 9.0 °C in 10 hours. Assume the rate of energy loss is constant in each case.
(a)
Calculate the energy lost from the water in 10 hours when the tank has the jacket.
(2)
(b)
Calculate the mean power lost from the tank when it has the jacket.
(2)
(c)
The power lost from the tank without the jacket is 210 W. Calculate the energy saved in 24 hours by using the jacket, in kWh.
(2)
(d)
Electricity costs €0.21 per kWh. Calculate the saving per day, using the unrounded energy saved. Give your answer in euros to the nearest cent.
(2)
(e)
Explain how a lagging jacket reduces the rate of energy loss from the tank.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 200 × 4200 × 3.0
✓ 1
A1 2 520 000 (J)
✓ 1
Part (b)
M1 2 520 000 ÷ (10 × 3600)
✓ 1
A1 70 (W) ECF from (a)
✓ 1
Part (c)
M1 (210 − 70) × 24 = 3360 Wh
✓ 1
A1 3.36 (kWh)
✓ 1
Part (d)
M1 3.36 × 0.21
✓ 1
A1 €0.71 (0.7056)
✓ 1
Part (e)
B1 the jacket traps air (or is made of a poor conductor of thermal energy)
✓ 1
B1 so less energy is transferred by conduction (and convection) from the tank to the surroundings
✓ 1
Exam tip: Keep power in watts and time in seconds for joules, or power in kW and time in hours for kWh, but do not mix the two sets.
167IGPE3-4-14r18-144(d) Energy Resources and Electricity GenerationPaper 2PHard8 marks
A pumped-storage scheme lifts 3.0 × 105 kg of water through a height of 80 m using electrical energy. The pump system is 80% efficient. Later the water falls back through turbines and the generator is 90% efficient.
(a)
Calculate the gravitational potential energy gained by the water.
(2)
(b)
Calculate the electrical energy supplied to the pump system.
(2)
(c)
Calculate the electrical energy produced by the generator when all the water falls back.
(2)
(d)
Calculate the overall efficiency of storing and returning the energy, as a percentage.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 mgh = 3.0 × 105 × 10 × 80
✓ 1
A1 2.4 × 108 J
✓ 1
Part (b)
M1 2.4 × 108 ÷ 0.80 (ECF)
✓ 1
A1 3.0 × 108 J
✓ 1
Part (c)
M1 0.90 × 2.4 × 108 (ECF from (a))
✓ 1
A1 2.16 × 108 J
✓ 1
Part (d)
M1 2.16 × 108 ÷ 3.0 × 108 (ECF)
✓ 1
A1 72%
✓ 1
Common error: Overall efficiency compares final useful output with the original input, not with the stored energy.
168IGPE3-4-15r33-134(d) Energy Resources and Electricity GenerationPaper 2PHard7 marks
A wind turbine supplies a data centre. The wind speed increases from 6.0 m/s to 9.0 m/s. The air density and the swept area of the blades are unchanged.
(a)
Determine the ratio of the mass of air reaching the blades each second at 9.0 m/s to that at 6.0 m/s.
(2)
(b)
Determine the ratio of the kinetic energy of one kilogram of air at 9.0 m/s to that at 6.0 m/s.
(2)
(c)
Hence determine the factor by which the kinetic energy of the air arriving each second increases.
(2)
(d)
State one reason why the electrical power output is smaller than the kinetic energy of the air arriving each second.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 mass per second is proportional to wind speed (volume per second = area × speed)
✓ 1
A1 9.0 ÷ 6.0 = 1.5
✓ 1
Part (b)
M1 kinetic energy is proportional to speed squared
✓ 1
A1 (9.0 ÷ 6.0)2 = 2.25
✓ 1
Part (c)
M1 multiply the two ratios: 1.5 × 2.25 (ECF from (a) and (b))
✓ 1
A1 3.375 (accept 3.4)
✓ 1
Part (d)
Any one from: air must keep moving after the blades; friction in the turbine and generator; thermal energy and sound are produced
✓ 1
Common error: Do not forget that doubling the wind speed also changes the amount of air arriving each second, not just the energy of each kilogram.
169IGPE3-4-15r40-134(c) Work and PowerPaper 2PHard8 marks
An archaeology team uses an electric winch to lift a stone block of mass 360 kg through 2.5 m at a steady speed. The motor in the winch is 80% efficient and the gearbox is 90% efficient. The winch is supplied with electrical energy at a rate of 2.5 kW. g = 10 N/kg.
(a)
Calculate the gain in gravitational potential energy of the block.
(2)
(b)
Show that the overall efficiency of the winch is 72%.
(2)
(c)
Calculate the electrical energy supplied to the winch to lift the block.
(2)
(d)
Calculate the time taken to lift the block.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 gain in GPE = mass × g × height
✓ 1
A1 360 × 10 × 2.5 = 9000 J
✓ 1
Part (b)
M1 overall efficiency = 0.80 × 0.90
✓ 1
A1 = 0.72, which is 72%
✓ 1
Part (c)
M1 energy supplied = useful energy output ÷ efficiency
✓ 1
A1 9000 ÷ 0.72 = 12 500 J (ECF from (a))
✓ 1
Part (d)
M1 time = energy ÷ power, with power = 2500 W
✓ 1
A1 12 500 ÷ 2500 = 5.0 s (ECF from (c))
✓ 1
Common error: Efficiencies of machines in sequence are multiplied, not added. Two stages of 80% and 90% do not give 85%.
170IGPE3-4-16r42-124(d) Energy Resources and Electricity GenerationPaper 2PHard8 marks
A light aircraft makes a flight lasting 2.0 hours. The engine delivers a useful power of 90 kW to the propeller throughout the flight. The engine has an efficiency of 25%. The aircraft fuel releases 4.5 × 107 J of energy per kilogram when burned, and costs £2.50 per kilogram.
(a)
Calculate the useful energy delivered by the engine during the flight.
(2)
(b)
Calculate the total energy that must be released from the fuel during the flight.
(2)
(c)
Calculate the mass of fuel burned during the flight.
(2)
(d)
Calculate the cost of the fuel for the flight.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 time = 2.0 × 3600 = 7200 s
✓ 1
A1 6.48 × 108 J
✓ 1
Part (b)
M1 total energy = useful energy ÷ efficiency = 6.48 × 108 ÷ 0.25
✓ 1
A1 2.592 × 109 J
✓ 1
Part (c)
M1 mass = 2.592 × 109 ÷ 4.5 × 107
✓ 1
A1 57.6 kg
✓ 1
Part (d)
M1 57.6 × 2.50
✓ 1
A1 £144
✓ 1
Common error: Convert hours to seconds before using energy = power × time, and divide by the efficiency (as a decimal) to find the input energy.
171IGPE3-4-16r42-144(b) Energy TransfersPaper 2PHard8 marks
An audience of 400 people sits in a concert hall for a 90 minute performance. Each person transfers thermal energy to the surroundings at an average rate of 100 W. The air in the hall has a total mass of 21 600 kg and a specific heat capacity of 1000 J/kg°C.
(a)
Calculate the total rate at which thermal energy is transferred to the hall by the audience.
(1)
(b)
Calculate the energy transferred to the hall by the audience during the performance.
(2)
(c)
Assume that all of this energy is transferred to the air in the hall. Calculate the increase in temperature of the air.
(3)
(d)
In practice the increase in temperature of the air is smaller than your answer to part (c). Suggest two reasons for this.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
400 × 100 = 40 000 W
✓ 1
Part (b)
M1 90 minutes = 5400 s and energy = power × time
✓ 1
A1 2.16 × 108 J
✓ 1
Part (c)
M1 ΔQ = mcΔθ, so Δθ = ΔQ ÷ (m × c)
✓ 1
M1 2.16 × 108 ÷ (21 600 × 1000)
✓ 1
A1 10 °C (ECF from (b))
✓ 1
Part (d)
one reason from: thermal energy is also transferred to the walls, floor, seats and roof, or is lost through the walls and roof to the outside by conduction, or ventilation or air conditioning removes warm air
✓ 1
a second different valid reason from this list
✓ 1
Common error: Time must be in seconds in energy = power × time: 90 minutes is 5400 s, not 90 s.
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