ENERGY

AQA GCSE Physics · 8463 & 8464 · Lessons 1–12
⬇ Student Booklet ⬇ Teacher Booklet

Key Terms — Energy

All Terms

01
Energy
DefinitionThe ability to do work. Cannot be created or destroyed, only transferred between stores.
02
Energy store
DefinitionA way in which an object can hold energy (e.g. kinetic, gravitational).
03
Law of Conservation of Energy
DefinitionEnergy cannot be created or destroyed, only transferred between stores.
04
Joule (J)
DefinitionThe SI unit of energy.
05
Gravitational potential energy
DefinitionEnergy stored due to an object’s position in a gravitational field.
06
Gravitational field strength (g)
DefinitionForce per unit mass due to gravity; 9.8 N/kg on Earth.
07
Kinetic energy
DefinitionEnergy stored in a moving object.
08
Speed
DefinitionHow fast an object is moving, regardless of direction (a scalar quantity) — used in the kinetic energy equation.
09
Closed system
DefinitionA system in which no energy enters or leaves.
10
Dissipation
DefinitionEnergy transferred to the surroundings in a less useful form, usually thermal or sound.
11
Elastic potential energy
DefinitionEnergy stored in a stretched or compressed elastic object.
12
Extension
DefinitionThe extra length a spring has stretched beyond its natural length.
13
Spring constant (k)
DefinitionA measure of the stiffness of a spring, in N/m.
14
Temperature
DefinitionThe average kinetic energy store of the particles in a system.
15
Specific heat capacity
DefinitionThe energy needed to raise the temperature of 1 kg of a substance by 1°C.
16
Power
DefinitionThe rate at which energy is transferred.
17
Watt (W)
DefinitionThe SI unit of power; 1 W = 1 J/s.
18
Sankey diagram
DefinitionA diagram showing energy transfers, with arrow width proportional to energy value.
19
Thermal conductivity
DefinitionHow quickly a material allows thermal energy to pass through it.
20
Thermal insulation
DefinitionReduces the rate of energy transfer to the surroundings.
21
Efficiency
DefinitionThe fraction of input energy transferred to a useful output.
22
Non-renewable resource
DefinitionAn energy resource that cannot be replenished at the rate it is used, e.g. fossil fuels.
23
Renewable resource
DefinitionAn energy resource that is naturally replenished and will not run out.
24
Fossil fuels
DefinitionCoal, oil and natural gas — non-renewable resources formed from ancient organisms.
25
Nuclear fission
DefinitionThe splitting of a heavy atomic nucleus, releasing large amounts of energy.

Unit Prefixes

PrefixSymbolMultiplierDecimal
teraT10121,000,000,000,000
gigaG1091,000,000,000
megaM1061,000,000
kilok1031,000
centic10−20.01
millim10−30.001
microµ10−60.000001
nanon10−90.000000001

Lesson 1 · Energy Stores & Gravitational Potential Energy

Do Now

Q1
Calculate: 6 × 9.8 × 3
Model Answer176.4
Q2
What is the SI unit of mass, and what instrument is used to measure it?
Model AnswerThe kilogram (kg); measured using a mass balance or digital scales.
Q3
What is the SI unit of length or height called?
Model AnswerThe metre (m).
Q4
Give an example of an object that stores energy because it has been lifted up high.
Model AnswerAny sensible example, e.g. a book on a high shelf, water in a raised tank, a person at the top of a climbing wall.

Part 1 of 3 · Energy Stores

Energy is the ability to do work. Whenever something happens anywhere in the universe, energy is transferred.

Energy is stored in objects. The main energy stores are: gravitational, kinetic, elastic, thermal, chemical, nuclear, electrostatic and magnetic.

Energy cannot be created or destroyed — it is only transferred between stores. This is the Law of Conservation of Energy.

The unit of energy is the Joule (J).

Worked Example — converting 4500 J using the prefixes table
kilo
45001000 = 4.5 kJ
mega
45001 000 000 = 0.0045 MJ
micro
4500 × 1 000 000 = 4 500 000 000 µJ

Questions

Q1
Define energy. (1 mark)
Model AnswerThe ability to do work.
Q2
Name four energy stores. (2 marks)
Model AnswerAny four from: gravitational, kinetic, elastic, thermal, chemical, nuclear, electrostatic, magnetic.
Q3
State the Law of Conservation of Energy. (2 marks)
Model AnswerEnergy cannot be created or destroyed, only transferred between stores.
Q4
A ball is held above the ground. Which energy store does it have? (1 mark)
Model AnswerGravitational potential energy store.
Q5
The prefix reference table in the Key Terms tab can help with questions 5–9 and 11.
Convert 4500 J into kJ. (1 mark)
Model Answer4.5 kJ.
Q6
Convert 3.2 kJ into J. (1 mark)
Model Answer3200 J.
Q7
Convert 6 800 000 J into MJ. (1 mark)
Model Answer6.8 MJ.
Q8
Convert 0.045 MJ into J. (1 mark)
Model Answer45 000 J.
Q9
Convert 2 500 000 µJ into J. (1 mark)
Model Answer2.5 J.
Q10
A stretched spring and a moving car both store energy, but in different stores. Name the energy store for each. (2 marks)
Model AnswerSpring: elastic potential energy store. Car: kinetic energy store.
Q11
Convert 3 200 000 000 J into GJ. (1 mark)
Model Answer3.2 GJ.

Part 2 of 3 · Gravitational Potential Energy

Gravitational potential energy (GPE) is stored when an object with mass is raised in a gravitational field.

Ep = m × g × h

Ep = gravitational potential energy store (J); m = mass (kg); g = gravitational field strength (N/kg); h = height above the ground (m).

On Earth, g = 9.8 N/kg. GPE increases if mass, height, or g increases.

Diagram of a 75 kg rock held 4 m above the ground
Fig 1.1 — A rock raised 4 m above the ground storing gravitational potential energy.

A rock of mass 75 kg is lifted 4 m. g = 9.8 N/kg. Calculate its gravitational potential energy.

Worked Example
V
Ep = ?   m = 75 kg   g = 9.8 N/kg   h = 4 m
E
Ep = m × g × h
S
Ep = 75 × 9.8 × 4
S
Ep = 2940
U
Joules (J)

Questions

Q12
Define gravitational potential energy store. (1 mark)
Model AnswerEnergy stored due to an object’s position in a gravitational field.
Q13
What does g measure? Give its value on Earth. (2 marks)
Model AnswerGravitational field strength; 9.8 N/kg on Earth.
Q14
A 5 kg book is placed on a shelf 1.5 m high. g = 9.8 N/kg. Calculate its GPE. (3 marks)
Model AnswerE = 5 × 9.8 × 1.5 = 73.5 J.
Q15
If the height of an object doubles, what happens to its GPE? Explain. (2 marks)
Model AnswerGPE doubles, because GPE ∝ h.
Q16
A 60 kg person stands on a 3 m diving board. g = 9.8 N/kg. Calculate their GPE. (3 marks)
Model AnswerE = 60 × 9.8 × 3 = 1764 J.

Part 3 of 3 · Rearranging Ep = mgh

Ep = mgh can also be used to find m or h, if the other values are known.

Always start from the original equation. Substitute the known values in, then multiply any known numbers together on the same line before rearranging — this avoids fractions with more than one term on the bottom.

Worked Example — Finding m
V
Ep = 5403 J   g = 9.8 N/kg   h = 98 m   m = ?
E
Ep = m × g × h
S
5403 = m × 9.8 × 98
S
5403 = m × 960.4
S
m = 5403960.4 = 5.63
U
kg

A buzzard stores 5403 J of GPE at a height of 98 m.

Worked Example — Finding h
V
Ep = 196 J   m = 2 kg   g = 9.8 N/kg   h = ?
E
Ep = m × g × h
S
196 = 2 × 9.8 × h
S
196 = 19.6 × h
S
h = 19619.6 = 10
U
m

A ball has a GPE of 196 J and a mass of 2 kg.

Questions

Q17
A ball has a GPE of 196 J and a mass of 2 kg. g = 9.8 N/kg. Find the height. (3 marks)
Model Answerh = 1962 × 9.8 = 10 m.
Q18
A buzzard stores 5403 J of GPE at a height of 98 m. g = 9.8 N/kg. Find its mass. (3 marks)
Model Answerm = 54039.8 × 98 = 5.63 kg.
Q19
A 4 kg object stores 78.4 J of GPE. g = 9.8 N/kg. Find its height. (3 marks)
Model Answerh = 78.44 × 9.8 = 2 m.
Q20
A crate stores 1176 J of GPE at a height of 6 m. g = 9.8 N/kg. Find its mass. (3 marks)
Model Answerm = 11769.8 × 6 = 20 kg.
Q21
An object of mass 15 kg stores 441 J of GPE. g = 9.8 N/kg. Find its height. (3 marks)
Model Answerh = 44115 × 9.8 = 3 m.
Q22
A book stores 29.4 J of GPE at a height of 0.5 m. g = 9.8 N/kg. Find its mass. (3 marks)
Model Answerm = 29.49.8 × 0.5 = 6 kg.
Q23
A 3 kg ball is dropped from 10 m. g = 9.8 N/kg. Calculate the GPE lost. (3 marks)
Model AnswerE = 3 × 9.8 × 10 = 294 J.
Q24
A rock on the Moon (g = 1.6 N/kg) has a mass of 20 kg and is 8 m high. Find its GPE. (3 marks)
Model AnswerE = 20 × 1.6 × 8 = 256 J.

Exam Question — Zip Wire

Figure 1.2 shows a person sliding down a zip wire from height h.
Total: 7 marks · mark allocations shown as (n) following AQA convention.
Zip wire from a tower showing change in vertical height
Fig 1.2 — Zip wire exam figure showing change in vertical height.
(a)
The change in GPE is 1.47 kJ. The person’s mass is 60 kg, g = 9.8 N/kg. Calculate the change in vertical height. (3)
Mark Scheme1.47 kJ = 1470 J. h = 147060 × 9.8 = 2.5 m.
(b)
As the person moves down the zip wire, the increase in KE is less than the decrease in GPE. Explain why. (2)
Mark SchemeSome GPE is transferred to thermal energy due to friction/air resistance.
(c)
Different people reach different speeds at the bottom. Explain why. (2)
Mark SchemePeople have different masses, so the same GPE change gives different velocities.

Lesson 2 · Kinetic Energy

Do Now

Gravitational Potential Energy: Ep = m × g × h
Q1
State the Law of Conservation of Energy.
Model AnswerEnergy cannot be created or destroyed, only transferred between stores.
Q2
Name four of the eight energy stores.
Model AnswerAny four from: gravitational, kinetic, elastic, thermal, chemical, nuclear, electrostatic, magnetic.
Q3
A 10 kg object is 5 m high. g = 9.8 N/kg. Calculate its GPE.
Model AnswerEp = 10 × 9.8 × 5 = 490 J.
Q4
A 45 kg object is at a height of 6 m. g = 9.8 N/kg. Calculate its GPE.
Model AnswerEp = 45 × 9.8 × 6 = 2646 J.

Part 1 of 3 · Kinetic Energy

Kinetic energy (KE) is the energy stored in a moving object.

Ek = ½ × m × v²

Ek = kinetic energy store (J); m = mass (kg); v = speed (m/s).

Speed (not velocity) is used, because kinetic energy is a scalar quantity — it does not depend on direction.

Block of mass m moving right at speed v
Fig 2.1 — An object of mass 4 kg moving at speed 3 m/s.

An object of mass 4 kg moves at a speed of 3 m/s. Calculate its kinetic energy.

Worked Example
V
Ek = ?   m = 4 kg   v = 3 m/s
E
Ek = ½ × m × v²
S
Ek = 0.5 × 4 × 3²
S
Ek = 0.5 × 4 × 9 = 2 × 9 = 18
U
Joules (J)

Questions

Q1
Define kinetic energy store. (1 mark)
Model AnswerEnergy stored in a moving object.
Q2
Write the formula for kinetic energy. (1 mark)
Model AnswerEk = ½mv².
Q3
Why does a train waiting at a station have zero kinetic energy store? (1 mark)
Model AnswerBecause it is stationary; v = 0.
Q4
A 4 kg object moves at 3 m/s. Calculate its KE. (3 marks)
Model AnswerE = 0.5 × 4 × 9 = 18 J.
Q5
A 0.2 kg ball moves at 5 m/s. Calculate its KE. (3 marks)
Model AnswerE = 0.5 × 0.2 × 25 = 2.5 J.
Q6
A 6 kg object moves at 4 m/s. Calculate its KE. (3 marks)
Model AnswerE = 0.5 × 6 × 16 = 48 J.
Q7
A 1500 kg car travels at 10 m/s. Calculate its KE. (3 marks)
Model AnswerE = 0.5 × 1500 × 100 = 75 000 J.

Part 2 of 3 · Rearranging Ek = ½mv²

Ek = ½mv² can also be used to find m or v, if the other values are known.

Always start from the original equation. Substitute the known values in, then multiply any known numbers together on the same line before rearranging — this avoids fractions with more than one term on the bottom.

A lorry moving at 30 mph has more KE than a car at the same speed because its mass is larger — KE ∝ m.

KE increases with the square of speed — doubling speed quadruples KE.

Worked Example — Finding m
V
Ek = 450 J   v = 6 m/s   m = ?
E
Ek = ½ × m × v²
S
450 = 0.5 × m × 6²
S
450 = m × 18
S
m = 45018 = 25
U
kg
Worked Example — Finding v
V
Ek = 200 J   m = 25 kg   v = ?
E
Ek = ½ × m × v²
S
200 = 0.5 × 25 × v²
S
200 = 12.5 × v²
S
v² = 20012.5 = 16  →  v = √16 = 4
U
m/s

Questions

Q8
Why does a lorry at 30 mph have more KE than a car at the same speed? (1 mark)
Model AnswerThe lorry has a much greater mass, so even at the same speed KE = ½mv² is much larger.
Q9
A car of mass 1200 kg travels at 20 m/s. Calculate its KE. (3 marks)
Model AnswerE = 0.5 × 1200 × 400 = 240 000 J.
Q10
An object has KE = 200 J and mass 25 kg. Find its speed. (3 marks)
Model Answerv = √(2 × 20025) = √16 = 4 m/s.
Q11
An object has KE = 450 J and speed = 6 m/s. Find its mass. (3 marks)
Model Answerm = 2 × 45036 = 25 kg.
Q12
How does doubling the speed of an object affect its KE? Explain. (2 marks)
Model AnswerKE is quadrupled because KE ∝ v².
Q13
An object has KE = 96 J and mass 12 kg. Find its speed. (3 marks)
Model Answerv = √(2 × 9612) = √16 = 4 m/s.
Q14
An object has KE = 250 J and speed = 10 m/s. Find its mass. (3 marks)
Model Answerm = 2 × 250100 = 5 kg.

Part 3 of 3 · Applying Kinetic Energy

KE and GPE are linked by conservation of energy. When an object falls, GPE converts to KE (ignoring air resistance).

If a 3 kg ball falls from 5 m: GPE lost = mgh = 3 × 9.8 × 5 = 147 J = KE gained.

The maximum KE at the bottom equals the GPE at the top (in a closed system).

Questions

Q15
A trampolinist has KE = 8500 J and mass 65 kg. Find their speed. (3 marks)
Model Answerv = √(2 × 850065) = √261.5 ≈ 16.2 m/s.
Q16
A 0.05 kg pancake is tossed with initial KE = 0.6 J. Calculate its initial speed. (3 marks)
Model Answerv = √(2 × 0.60.05) = √24 ≈ 4.9 m/s.
Q17
A 2 kg ball is dropped from 10 m. g = 9.8 N/kg. What is its KE just before hitting the ground? (3 marks)
Model AnswerGPE = 2 × 9.8 × 10 = 196 J = KE.
Q18
A cyclist of mass 70 kg (including bike) reaches 15 m/s. Calculate their KE. (3 marks)
Model AnswerE = 0.5 × 70 × 225 = 7875 J.
Q19
A skateboarder has KE = 720 J and mass 60 kg. Find their speed. (3 marks)
Model Answerv = √(2 × 72060) = √24 ≈ 4.9 m/s.
Q20
A 4 kg object falls from 6 m. g = 9.8 N/kg. Find its KE just before hitting the ground (ignore air resistance). (3 marks)
Model AnswerGPE = 4 × 9.8 × 6 = 235.2 J = KE.

Exam Question — Falling Ball

A ball of mass 0.5 kg is dropped from a height of 8 m. g = 9.8 N/kg. Ignore air resistance.
Total: 6 marks · mark allocations shown as (n) following AQA convention.
(a)
Calculate the gravitational potential energy of the ball before it is dropped. (2)
Mark SchemeE = 0.5 × 9.8 × 8 = 39.2 J.
(b)
State the kinetic energy of the ball just before it hits the ground. (1)
Mark Scheme39.2 J (GPE fully converted to KE).
(c)
Calculate the speed of the ball just before it hits the ground. (3)
Mark Schemev = √(2 × 39.20.5) = √156.8 ≈ 12.5 m/s.

Lesson 3 · Conservation of Energy & Dissipation

Do Now

Kinetic Energy: Ek = ½ × m × v²  ·  Gravitational PE: Ep = m × g × h
Q1
What does “v” represent in the kinetic energy equation, and what is its unit?
Model AnswerSpeed — how fast something is moving, regardless of direction — measured in m/s.
Q2
A 2 kg ball moves at 4 m/s. Calculate its KE.
Model AnswerEk = 0.5 × 2 × 16 = 16 J.
Q3
A 3 kg ball is dropped from 5 m. g = 9.8 N/kg. What is its KE just before hitting the ground?
Model AnswerGPE = 3 × 9.8 × 5 = 147 J = KE (energy is conserved as it falls).
Q4
A ball has KE = 200 J and mass 25 kg. Find its speed.
Model Answerv = √(2 × 20025) = √16 = 4 m/s.

Part 1 of 3 · Conservation of Energy

In a closed system, the total energy remains constant. Energy is transferred between stores but the total never changes.

Example: A ball thrown upward — KE converts to GPE as it rises; GPE converts back to KE as it falls.

In reality, no system is perfectly closed. Some energy is always transferred to the surroundings as thermal energy.

When a ball hits the ground, KE is transferred to sound and thermal energy stores — both are dissipated.

Ball thrown vertically with KE and GPE bar charts at five points in time
Fig 3.1 — A ball thrown straight up, plotted against time: kinetic energy store converts to gravitational potential energy store as it rises (A→C), then GPE converts back to KE as it falls (C→E). The ball’s motion is vertical only — its position along the time axis shows when each snapshot occurs, not how far it has travelled sideways.
Dropped ball showing GPE at top converting to KE at the bottom
Fig 3.2 — A dropped ball: all of its GPE at the top converts to KE just before impact.

A 2 kg ball is dropped from a height of 10 m. g = 9.8 N/kg. Find its speed just before it hits the ground.

Step 1 — Find the GPE lost
V
Ep = ?   m = 2 kg   g = 9.8 N/kg   h = 10 m
E
Ep = m × g × h
S
Ep = 2 × 9.8 × 10
S
Ep = 196
U
Joules (J)
Step 2 — Find the speed (Ek = GPE lost)
V
Ek = 196 J   m = 2 kg   v = ?
E
Ek = ½ × m × v²
S
196 = 0.5 × 2 × v²
S
196 = 1 × v²  →  v = √196 = 14
U
m/s
All of the GPE lost converts to KE as the ball falls (conservation of energy).

Questions

Q1
What is a closed system? (1 mark)
Model AnswerA system in which no energy enters or leaves.
Q2
State the law of conservation of energy. (2 marks)
Model AnswerEnergy cannot be created or destroyed; it can only be transferred between stores.
Q3
A 1 kg ball is thrown upward with KE = 20 J. What is its maximum GPE? Explain. (2 marks)
Model Answer20 J — all KE converts to GPE at the highest point.
Q4
Why is no real system truly closed? (2 marks)
Model AnswerSome energy is always lost to the surroundings (e.g. as heat/sound).
Q5
A 25 kg cannonball is fired upward at 8 m/s. g = 9.8 N/kg. Find its maximum height. (3 marks)
Model AnswerKE = ½ × 25 × 64 = 800 J. h = 80025 × 9.8 = 3.27 m.

Part 2 of 3 · Dissipation

Dissipation is when energy is transferred to the surroundings in a less useful form, usually thermal or sound.

Dissipated energy is “wasted” — it spreads into the surroundings and cannot be recovered easily.

Examples: friction in a car engine (thermal), air resistance on a cyclist (thermal), sound from brakes.

Reducing dissipation: lubrication reduces friction; streamlining reduces air resistance; insulation reduces thermal loss.

Questions

Q6
What does “dissipation” mean in physics? (2 marks)
Model AnswerEnergy transferred to surroundings in a less useful (often thermal) form.
Q7
Give two examples of energy dissipation in everyday life. (2 marks)
Model AnswerAny two: friction in engines, air resistance on vehicles, sound in brakes, heat from light bulbs.
Q8
A bullet is shot upward with KE = 32 J. At its highest point its GPE = 31.8 J. Explain why the values differ. (2 marks)
Model Answer0.2 J was transferred to thermal energy/sound due to air resistance.
Q9
Name two ways to reduce energy dissipation in machines. (2 marks)
Model AnswerLubrication (to reduce friction) and streamlining (to reduce air resistance).
Q10
Why is dissipated energy considered “wasted”? (2 marks)
Model AnswerBecause it cannot easily be recovered and put back to useful work.

Part 3 of 3 · Energy Transfers in Scenarios

When describing energy transfers, state the initial store → mechanism of transfer → final store(s).

Zip wire: GPE → (mechanically) → KE + thermal (friction).

Bouncing ball: GPE → KE (falling) → elastic PE (squash) → KE + thermal + sound (bounce).

The total energy at the end equals the total at the start; only the distribution changes.

Questions

Q11
Describe the energy transfer as a ball falls from a height (ignore air resistance). (2 marks)
Model AnswerGPE → KE.
Q12
Describe the energy transfers for a person braking on a bicycle. (3 marks)
Model AnswerKE in cyclist → thermal energy in brakes (friction) + sound.
Q13
A 60 kg person slides down a zip wire. GPE decreases by 1470 J but KE increases by only 1100 J. How much energy was dissipated? (2 marks)
Model Answer1470 − 1100 = 370 J dissipated as thermal energy/sound.
Q14
Explain why a bouncing ball never bounces back to its original height. (2 marks)
Model AnswerEach bounce transfers some KE to thermal/sound, so less energy remains as KE/GPE.

Exam Question — Zip Wire Dissipation

A person of mass 60 kg slides down a zip wire. The change in vertical height is 2.5 m. g = 9.8 N/kg.
Total: 6 marks · mark allocations shown as (n) following AQA convention.
(a)
Calculate the decrease in GPE. (2)
Mark SchemeE = 60 × 9.8 × 2.5 = 1470 J.
(b)
The increase in KE is 1100 J. Calculate the energy transferred to thermal energy. (2)
Mark Scheme1470 − 1100 = 370 J.
(c)
Explain why different people reach different speeds at the bottom. (2)
Mark SchemeDifferent masses → same GPE loss gives different KE → different velocities.

Lesson 4 · Elastic Potential Energy

Do Now

Q1
What is meant by “dissipation”?
Model AnswerEnergy transferred to the surroundings in a less useful form, usually thermal or sound.
Q2
Give two ways of reducing energy dissipation.
Model AnswerAny two: lubrication (reduces friction), streamlining (reduces air resistance), insulation (reduces thermal loss).
Q3
Describe the energy transfer as a ball falls from a height (ignoring air resistance).
Model AnswerGPE → KE.
Q4
A 60 kg person slides down a zip wire. GPE decreases by 1470 J but KE increases by only 1100 J. How much energy was dissipated?
Model Answer1470 − 1100 = 370 J.

Part 1 of 3 · Elastic Potential Energy

Elastic potential energy (EPE) is stored when an elastic object (e.g. a spring) is stretched or compressed.

Ee = ½ × k × e²

Ee = elastic potential energy (J); k = spring constant (N/m); e = extension (m).

The extension is the extra length stretched — not the total length.

The spring constant k measures the stiffness of the spring. A higher k means a stiffer spring.

This formula is on the AQA formula sheet — you do not need to memorise it.

Spring at natural length beside the same spring stretched by extension e
Fig 4.1 — A spring showing natural length and extension e.

A spring (k = 8 N/m) is stretched 3 m. Calculate its elastic potential energy.

Worked Example
V
Ee = ?   k = 8 N/m   e = 3 m
E
Ee = ½ × k × e²
S
Ee = 0.5 × 8 × 3²
S
Ee = 0.5 × 8 × 9 = 36
U
Joules (J)

Questions

Q1
Define elastic potential energy store. (1 mark)
Model AnswerEnergy stored in a stretched or compressed elastic object.
Q2
What two factors affect the elastic potential energy stored in a spring? (2 marks)
Model AnswerSpring constant (k) and extension (e).
Q3
Do you need the total length of a spring to calculate its EPE? Explain. (2 marks)
Model AnswerNo — only the extension (extra length) is needed, not the total length.
Q4
A spring (k = 2 N/m) is stretched 3 m. Calculate its EPE. (3 marks)
Model AnswerE = 0.5 × 2 × 9 = 9 J.
Q5
A spring (k = 6 N/m) is stretched 2 m. Calculate its EPE. (3 marks)
Model AnswerE = 0.5 × 6 × 4 = 12 J.
Q6
A spring (k = 150 N/m) is extended 20 cm. Calculate its EPE. (3 marks)
Model AnswerE = 0.5 × 150 × 0.04 = 3 J. (0.2 m extension)
Q7
Which spring is easier to stretch: k = 0.5 N/m or k = 40 N/m? Why? (2 marks)
Model Answerk = 0.5 N/m is easier — it has a lower spring constant so less force is needed.

Part 2 of 3 · Rearranging Ee = ½ke²

Ee = ½ke² can also be used to find k or e, if the other values are known.

Always start from the original equation. Substitute the known values in, then multiply any known numbers together on the same line before rearranging — this avoids fractions with more than one term on the bottom.

Worked Example — Finding k
V
Ee = 32 J   e = 4 m   k = ?
E
Ee = ½ × k × e²
S
32 = 0.5 × k × 4²
S
32 = k × 8
S
k = 328 = 4
U
N/m
Worked Example — Finding e
V
Ee = 45 J   k = 10 N/m   e = ?
E
Ee = ½ × k × e²
S
45 = 0.5 × 10 × e²
S
45 = 5 × e²  →  e² = 455 = 9
S
e = √9 = 3
U
m

Questions

Q8
A spring (k = 75 N/m) is extended 40 cm. Calculate its EPE. (3 marks)
Model AnswerE = 0.5 × 75 × 0.16 = 6 J. (0.4 m extension)
Q9
A spring stores 400 J with k = 12 N/m. Find the extension. (3 marks)
Model Answere = √(2 × 40012) = √66.7 ≈ 8.16 m.
Q10
A spring stores 50 J when extended 5 m. Calculate the spring constant. (3 marks)
Model Answerk = 2 × 5025 = 4 N/m.
Q11
A spring (k = 200 N/m) has 180 J of EPE. Find the extension. (3 marks)
Model Answere = √(2 × 180200) = √1.8 ≈ 1.34 m.
Q12
A spring stores 18 J when extended 3 m. Calculate the spring constant. (3 marks)
Model Answerk = 2 × 189 = 4 N/m.
Q13
A spring stores 64 J when extended 4 m. Calculate the spring constant. (3 marks)
Model Answerk = 2 × 6416 = 8 N/m.
Q14
A spring stores 90 J when extended 6 m. Calculate the spring constant. (3 marks)
Model Answerk = 2 × 9036 = 5 N/m.
Q15
A spring (k = 8 N/m) stores 100 J of EPE. Find the extension. (3 marks)
Model Answere = √(2 × 1008) = √25 = 5 m.
Q16
A spring (k = 18 N/m) stores 81 J of EPE. Find the extension. (3 marks)
Model Answere = √(2 × 8118) = √9 = 3 m.
Q17
A spring (k = 20 N/m) stores 250 J of EPE. Find the extension. (3 marks)
Model Answere = √(2 × 25020) = √25 = 5 m.

Part 3 of 3 · Elastic Energy in Context

When a spring is released, elastic PE converts to KE (and some thermal).

A bungee cord stores elastic PE when stretched; when it pulls the jumper back, EPE converts to KE then GPE.

Elastic PE ↔ KE conversions occur in trampolines, bows, catapults and musical instruments.

Kinetic Energy: Ek = ½ × m × v²  ·  Gravitational PE: Ep = m × g × h

A spring (k = 50 N/m) is stretched 0.4 m and used to fire a 0.5 kg ball. Find the energy stored, then the speed.

Step 1 — Find the EPE
V
Ee = ?   k = 50 N/m   e = 0.4 m
E
Ee = ½ × k × e²
S
Ee = 0.5 × 50 × 0.4²
S
Ee = 0.5 × 50 × 0.16 = 4
U
Joules (J)
Step 2 — Find the speed (Ek = EPE)
V
Ek = 4 J   m = 0.5 kg   v = ?
E
Ek = ½ × m × v²
S
4 = 0.5 × 0.5 × v²
S
4 = 0.25 × v²  →  v² = 16
S
v = √16 = 4
U
m/s
All of the EPE converts to KE as the ball is fired.

Questions

Q18
Describe the energy transfers in a bow-and-arrow as the arrow is fired. (3 marks)
Model AnswerChemical energy in archer → EPE in string → KE in arrow (+thermal from friction).
Q19
Give two examples of objects that store elastic potential energy. (2 marks)
Model AnswerAny two: spring, rubber band, trampoline, bungee cord.
Q20
A 0.1 kg ball is fired by a spring (k = 50 N/m, e = 0.2 m). Assuming all EPE converts to KE, find the ball’s speed. (4 marks)
Model AnswerEPE = 0.5 × 50 × 0.04 = 1 J. v = √(2 × 10.1) = √20 ≈ 4.47 m/s.
Q21
A spring (k = 40 N/m, e = 0.3 m) fires a 0.05 kg ball vertically upward. Assuming all EPE converts to GPE, find the maximum height reached. g = 9.8 N/kg. (4 marks)
Model AnswerEPE = 0.5 × 40 × 0.09 = 1.8 J. h = 1.80.05 × 9.83.67 m.
Q22
A catapult (k = 30 N/m, e = 0.2 m) fires a 0.06 kg stone. Assuming all EPE converts to KE, find its speed. (4 marks)
Model AnswerEPE = 0.5 × 30 × 0.04 = 0.6 J. v = √(2 × 0.60.06) = √20 ≈ 4.47 m/s.
Q23
A spring (k = 200 N/m, e = 0.5 m) fires a 2 kg ball vertically upward. Assuming all EPE converts to GPE, find the maximum height reached. g = 9.8 N/kg. (4 marks)
Model AnswerEPE = 0.5 × 200 × 0.25 = 25 J. h = 252 × 9.81.28 m.

Exam Question — Bungee Jump

A bungee cord has spring constant k = 15 N/m. A student of mass 70 kg jumps from a bridge. The unstretched cord length is 20 m.
Total: 6 marks · mark allocations shown as (n) following AQA convention.
Bungee jumper at bridge level and at the lowest point above a river
Fig 4.2 — Bungee jumper before and after jump: student at bridge level (Before) and at lowest point (After).
(a)
Give two reasons why the cord must be appropriate for the student’s weight. (2)
Mark SchemeIf too short/stiff the student won’t reach maximum stretch; if too long/weak they may hit the ground.
(b)
The cord stretches 18 m. Calculate the elastic PE stored. (3)
Mark SchemeE = 0.5 × 15 × 18² = 0.5 × 15 × 324 = 2430 J.
(c)
The student’s KE at the lowest point is zero. Explain why. (1)
Mark SchemeAt the lowest point, speed = 0 so KE = 0.

Lesson 5 · Specific Heat Capacity

Do Now

Elastic Potential Energy: Ee = ½ × k × e²
Q1
What is meant by the “extension” of a spring?
Model AnswerThe extra length a spring has stretched beyond its natural length (not the total length).
Q2
Describe the energy transfers in a bow-and-arrow as the arrow is fired.
Model AnswerChemical energy in archer → EPE in string → KE in arrow (+ some thermal from friction).
Q3
A spring (k = 2 N/m) is stretched 3 m. Calculate its EPE.
Model AnswerEe = 0.5 × 2 × 9 = 9 J.
Q4
A spring (k = 75 N/m) is extended 40 cm. Calculate its EPE.
Model AnswerEe = 0.5 × 75 × 0.16 = 6 J (0.4 m extension).

Part 1 of 4 · What is Specific Heat Capacity?

Putting the same energy into different materials gives different temperature rises. This is described by specific heat capacity.

The specific heat capacity (c) is the energy needed to raise the temperature of 1 kg of a substance by 1°C.

A material with a high specific heat capacity needs more energy to heat up (e.g. water, c = 4200 J/kg°C).

Copper and gold blocks of equal mass each receiving 500 J
Fig 5.1 — Copper (SHC = 386 J/kg°C) and gold (SHC = 126 J/kg°C) blocks of equal mass. Adding 500 J raises gold’s temperature more.

Questions

Q1
Define specific heat capacity. (2 marks)
Model AnswerThe energy needed to raise the temperature of 1 kg of a substance by 1°C.
Q2
State the unit of specific heat capacity. (1 mark)
Model AnswerJ/kg°C.
Q3
Water has a higher specific heat capacity than sand. Explain why a beach gets hot quickly while the sea stays cool. (3 marks)
Model AnswerWater needs more energy to raise its temperature by 1°C, so for the same energy from the Sun the sand heats up more than the sea.
Q4
Two 1 kg blocks, one of copper (c = 385) and one of water (c = 4200), are given the same energy. State which gets hotter and explain. (2 marks)
Model AnswerThe copper, because it has a lower specific heat capacity so the same energy gives a bigger temperature rise.
Q5
Explain why water is used as the coolant in car engines and central heating. (2 marks)
Model AnswerIts high specific heat capacity lets it absorb/carry a large amount of energy with only a small temperature change.

Part 2 of 4 · The Specific Heat Capacity Equation

The energy needed to change the temperature of a substance is:

change in thermal energy = mass × specific heat capacity × temperature change

ΔE = m c ΔT

ΔE = change in thermal energy (J); m = mass (kg); c = specific heat capacity (J/kg°C); ΔT = temperature change (°C).

This equation is on the AQA formula sheet.

Heat 2 kg of water (c = 4200 J/kg°C) by 30°C. Find the energy transferred.

Worked Example
V
m = 2 kg   c = 4200   ΔT = 30°C   ΔE = ?
E
ΔE = m × c × ΔT
S
ΔE = 2 × 4200 × 30
S
ΔE = 252 000
U
Joules (J)

Questions

Q6
Write down the equation for the energy needed to change the temperature of a substance. (1 mark)
Model AnswerΔE = mcΔT.
Q7
Calculate the energy in each case:(a) m = 10 kg, c = 4200, ΔT = 4°C(b) m = 0.5 kg, c = 385, ΔT = 20°C(c) m = 2 kg, c = 540, ΔT = 60°C (3 marks)
Model Answer(a) 168 000 J(b) 3850 J(c) 64 800 J
Q8
0.3 kg of water (c = 4200) is heated by 50°C. Calculate the energy transferred. (3 marks)
Model AnswerΔE = 0.3 × 4200 × 50 = 63 000 J.
Q9
A 1.5 kg block of aluminium (c = 900 J/kg°C) was heated from 20°C to 70°C. Calculate the energy used. (3 marks)
Model AnswerΔE = 1.5 × 900 × 50 = 67 500 J.
Q10
A 0.2 kg block of lead (c = 128 J/kg°C) was heated from 15°C to 85°C. Calculate the energy used. (3 marks)
Model AnswerΔE = 0.2 × 128 × 70 = 1792 J.
Q11
A 5 kg block of concrete (c = 880 J/kg°C) was heated by 12°C. Calculate the energy used. (3 marks)
Model AnswerΔE = 5 × 880 × 12 = 52 800 J.

Part 3 of 4 · Finding Mass or Specific Heat Capacity

ΔE = mcΔT can also be used to find m or c, if the other values are known.

Always start from the original equation. Substitute the known values in, then multiply any known numbers together on the same line before rearranging — this avoids fractions with more than one term on the bottom.

Worked Example — Finding m
V
ΔE = 84 000 J   c = 4200   ΔT = 20°C   m = ?
E
ΔE = m × c × ΔT
S
84 000 = m × 4200 × 20
S
84 000 = m × 84 000
S
m = 84 00084 000 = 1
U
kg
Worked Example — Finding c
V
ΔE = 5000 J   m = 2 kg   ΔT = 5°C   c = ?
E
ΔE = m × c × ΔT
S
5000 = 2 × c × 5
S
5000 = c × 10
S
c = 500010 = 500
U
J/kg°C

Questions

Q12
Rearrange ΔE = mcΔT to make (a) m the subject (b) c the subject. (2 marks)
Model Answer(a) m = ΔEc × ΔT    (b) c = ΔEm × ΔT
Q13
Calculate the mass in each case:(a) ΔE = 16 800 J, c = 4200, ΔT = 4°C(b) ΔE = 3850 J, c = 385, ΔT = 10°C (2 marks)
Model Answer(a) 1 kg    (b) 1 kg.
Q14
Calculate the specific heat capacity in each case:(a) ΔE = 9000 J, m = 2 kg, ΔT = 5°C(b) ΔE = 4000 J, m = 0.5 kg, ΔT = 10°C (2 marks)
Model Answer(a) 900 J/kg°C    (b) 800 J/kg°C.
Q15
18 000 J of energy raises the temperature of a 1 kg block by 20°C. Calculate its specific heat capacity. (3 marks)
Model Answerc = 18 0001 × 20 = 900 J/kg°C.
Q16
How much mass of oil (c = 2000 J/kg°C) can be heated by 15°C using 60 000 J? (3 marks)
Model Answerm = 60 0002000 × 15 = 2 kg.

Part 4 of 4 · Finding the Temperature Change

126 000 J heats 1 kg of water (c = 4200 J/kg°C). Find the temperature change.

Worked Example
V
ΔE = 126 000 J   m = 1 kg   c = 4200   ΔT = ?
E
ΔE = m × c × ΔT
S
126 000 = 1 × 4200 × ΔT
S
126 000 = 4200 × ΔT
S
ΔT = 126 0004200 = 30
U
°C

Questions

Q17
Rearrange ΔE = mcΔT to make ΔT the subject. (1 mark)
Model AnswerΔT = ΔEm × c
Q18
Calculate the temperature change in each case:(a) ΔE = 8400 J, m = 2 kg, c = 4200(b) ΔE = 1900 J, m = 1 kg, c = 380 (2 marks)
Model Answer(a) 1°C    (b) 5°C.
Q19
A 0.5 kg block of iron (c = 450 J/kg°C) is given 9000 J of energy. Calculate its temperature rise. (3 marks)
Model AnswerΔT = 90000.5 × 450 = 40°C.
Q20
2 kg of water (c = 4200 J/kg°C) starts at 15°C and is given 84 000 J of energy. Calculate its final temperature. (3 marks)
Model AnswerΔT = 84 0002 × 4200 = 10°C. Final temperature = 15 + 10 = 25°C.
Q21
A 0.8 kg block of glass (c = 670 J/kg°C) is given 26 800 J of energy. Calculate its temperature rise. (3 marks)
Model AnswerΔT = 26 8000.8 × 670 = 50°C.

Exam Question — Kettle

An electric kettle is used to boil water. After boiling, the temperature of the water decreases by 22°C. The mass of water is 0.50 kg and the specific heat capacity of water is 4200 J/kg°C.
Total: 12 marks · mark allocations shown as (n) following AQA convention.
(a)
Calculate the energy transferred to the surroundings from the water. (3)
Mark SchemeΔE = m × c × ΔT = 0.50 × 4200 × 22 = 46 200 J.
(b)
Explain why the total energy supplied to the kettle is greater than the energy used to heat the water. (2)
Mark SchemeSome energy is transferred to the surroundings (e.g. heating the kettle itself and the air) rather than the water, so more energy must be supplied than is used to heat the water.
(c)
A new design of kettle is made from two layers of plastic separated by a vacuum. After the water in this kettle has boiled, the water stays hot for at least 2 hours. The energy transferred from the water to the surroundings in 2 hours is 46 200 J, and its initial temperature is 100°C. Calculate the temperature of the water after 2 hours. (4)
Mark SchemeΔT = 46 2000.50 × 4200 = 22°C. Temperature after 2 hours = 100 − 22 = 78°C.
(d)
Calculate the average power output from the water to the surroundings over the 2 hours. (3)
Mark SchemeP = Et = 46 2002 × 36006.4 W (to 2 s.f.).

Lesson 6 · Investigating Specific Heat Capacity

Do Now

Specific Heat Capacity: ΔE = m × c × ΔT
Q1
Write down the equation for specific heat capacity.
Model AnswerΔE = mcΔT.
Q2
A 2 kg block absorbs 5000 J of energy and its temperature rises by 5°C. Find its specific heat capacity.
Model Answerc = 50002 × 5 = 500 J/kg°C.
Q3
A 0.5 kg block of iron (c = 450 J/kg°C) is given 9000 J of energy. Calculate its temperature rise.
Model AnswerΔT = 90000.5 × 450 = 40°C.
Q4
Why might a measurement be repeated during an experiment?
Model AnswerTo check the result is reliable/repeatable, and to average out random error.

Part 1 of 3 · Aim and Equipment

Required practical: determine the specific heat capacity of a material by transferring a known amount of energy to it and measuring its temperature change.

A block with a mass of exactly 1 kg is normally used — this simplifies the calculation, since m = 1 kg makes c easy to find directly from the results.

Insulated metal block with immersion heater, thermometer, joulemeter and power supply
Fig 6.1 — The equipment used to measure the specific heat capacity of a metal block.

Equipment

A 1 kg block of the material, with two holes — one for a heater, one for a thermometer.

Electric heater, power supply and joulemeter (a voltmeter and ammeter are sometimes used instead, but require the total energy to be calculated separately).

Thermometer, pipette, insulation, stopwatch, balance, heatproof mat.

Questions

Q1
State the aim of this required practical. (2 marks)
Model AnswerTo determine the specific heat capacity of a material by measuring its temperature change when a known amount of energy is transferred to it.
Q2
List the equipment required for this experiment. (5 marks)
Model AnswerThe block, heater, power supply (or joulemeter), thermometer, pipette, insulation, stopwatch, balance and heatproof mat (accept ammeter and voltmeter in place of a joulemeter).
Q3
Explain why a drop of water is placed in the hole with the thermometer. (2 marks)
Model AnswerIt improves the thermal contact between the block and the thermometer, so the temperature reading is more accurate.
Q4
Explain why the block is placed on a heatproof mat. (1 mark)
Model AnswerTo avoid damaging the surface underneath and to reduce the risk of burns/fire, since the block gets hot.
Q5
State the equation for specific heat capacity. (1 mark)
Model AnswerΔE = mcΔT.
Q6
Suggest why a block with a mass of exactly 1 kg is a convenient choice for this experiment. (2 marks)
Model AnswerWith m = 1 kg, the equation ΔE = mcΔT simplifies so that c can be found directly from the energy and temperature change, without extra division by mass.

Part 2 of 3 · Method

Set up the equipment as shown in Fig 6.1: wrap the block in insulation and place it on a heatproof mat. Fit the heater snugly into one hole, and the thermometer (with a drop of water) into the other, then connect the heater to a joulemeter and power supply.

Measure and record the mass of the block on the balance.

Record the starting temperature of the block.

Switch on the power supply and start the stopwatch at the same time.

Record the temperature of the block and the energy used at regular time intervals (e.g. every 30 s). Switch off the heater once you have enough readings (after 10 minutes, or once the block reaches 50°C).

Safety: the block and heater get hot — do not touch them during or straight after heating, and clean up any water spills near the power supply immediately.

Questions

Q7
State one safety precaution that should be taken during this experiment. (1 mark)
Model AnswerAny sensible precaution, e.g. do not touch the block or heater while hot, or clean up water spills near the power supply immediately.

Part 3 of 3 · Analysing the Results

Normally we plot the independent variable (the thing we change) on the x-axis and the dependent variable (the thing we measure) on the y-axis.

However, this time is an exception, to make our analysis easier — we are going to plot ΔE on the y-axis and ΔT on the x-axis.

This is because the equation of a straight line is:

y = mx + c

and our equation has the same form:

ΔE = m c ΔT

Because our block has a mass of m = 1 kg, making y = ΔE and x = ΔT leaves the gradient equal to c:

y = mx   →   ΔE = c ΔT

The graph often curves at the start — some energy heats the heater itself before the block responds. Use only the straight-line part to find the gradient.

Energy is still lost to the surroundings even with insulation, and there is a time lag between energy being supplied and the temperature rising — both make the measured c less accurate.

Blank graph grid with axes labelled change in energy and change in temperature
Fig 6.2 — Plot your own results here.

Reference — Specific Heat Capacities

Accepted values for some metals commonly used in this experiment:

MetalAluminiumCopperIronLeadZinc
Specific heat capacity (J/kg°C)900385450128387

Questions

Q8
Explain why only the straight-line part of the graph should be used to find the gradient. (2 marks)
Model AnswerAt the start, some of the energy heats the heater itself before the block’s temperature responds, curving the graph; only the straight part shows a constant rate of energy transfer with temperature change.
Q9
Using your results:(a) Find the gradient of your line.(b) Calculate the percentage difference between your value and the accepted value, using: percentage difference = |experimental − accepted|accepted × 100 (6 marks)
Model Answer(a) Read ΔE and ΔT between two points on the straight-line part of your graph and calculate the gradient.
(b) Percentage difference = |your value of c − the table value|the table value × 100.
Q10
Suggest one way to improve the accuracy of this experiment. (2 marks)
Model AnswerAny sensible improvement, e.g. use more/better insulation to reduce energy loss, or use a data logger and temperature probe instead of reading a thermometer by eye.

Exam Question — SHC Investigation

A student investigates the specific heat capacity of a metal block of mass 1 kg. The graph of their results (energy supplied against temperature change) is a straight line with gradient 410 J/°C, once the curved section near the origin is excluded.
Total: 6 marks · mark allocations shown as (n) following AQA convention.
(a)
State the value of the specific heat capacity found by the student, and explain how you know. (2)
Mark Schemec = 410 J/kg°C, because with m = 1 kg the gradient of ΔE against ΔT is equal to c.
(b)
The accepted value for the metal is 385 J/kg°C. Calculate the percentage difference between the student’s value and the accepted value. (2)
Mark Scheme% difference = |410 − 385|385 × 100 ≈ 6.5%.
(c)
Suggest one reason why the student’s value is higher than the accepted value. (2)
Mark SchemeHeat was lost to the surroundings despite insulation, so more energy was needed than expected for the temperature rise, making the calculated c too high.

Lesson 7 · Power

Do Now

Specific Heat Capacity: ΔE = m × c × ΔT
Q1
Why is a metal block with a mass of exactly 1 kg normally used in the SHC investigation?
Model AnswerIt simplifies the equation ΔE = mcΔT, so the gradient of the graph is equal to c directly.
Q2
Why do we plot ΔE on the y-axis and ΔT on the x-axis, rather than the other way around?
Model AnswerThis is an exception made to find the gradient directly as c, since ΔE = mcΔT has the same form as y = mx when m = 1 kg.
Q3
Give one way to improve the accuracy of the SHC investigation.
Model AnswerAny valid: better insulation/lagging, more accurate thermometer, repeat and average.
Q4
A student heats a 0.5 kg aluminium block. Energy supplied = 4500 J; ΔT = 10°C. Calculate c.
Model Answerc = 45000.5 × 10 = 900 J/kg°C.

Part 1 of 4 · What is Power?

Work done is energy transferred.

Rate means how much each second.

Power is the rate at which energy is transferred.

The Watt (W) is the unit of power.

Questions

Q1
What is work done? (1 mark)
Model AnswerWork done is energy transferred.
Q2
What does rate mean? (1 mark)
Model AnswerRate means how much per second.
Q3
Define power. (1 mark)
Model AnswerPower is the rate at which energy is transferred.
Q4
What is the unit of power? (1 mark)
Model AnswerThe Watt (W).

Part 2 of 4 · Calculating Power

P = Et   ·   P = Wt

P = power (W); E = energy transferred (J); t = time (s).

Example 1: a device transfers 600 J of energy in 12 s. Calculate its power.

Worked Example
V
P = ?   E = 600 J   t = 12 s
E
P = Et
S
P = 60012
S
P = 50
U
Watts (W)

1 Watt means 1 Joule per second.  ·  1 kJ = 1000 J  ·  1 min = 60 s

Example 2: a machine transfers 1.2 kJ in 2 minutes. Calculate its power.

Worked Example
V
P = ?   E = 1.2 × 1000 = 1200 J   t = 2 × 60 = 120 s
E
P = Et
S
P = 1200120
S
P = 10
U
Watts (W)

Questions

Q5
A lamp transfers 200 J in 4 s. Calculate its power. (2 marks)
Model AnswerP = 2004 = 50 W.
Q6
A heater transfers 6000 J in 60 s. Calculate its power. (2 marks)
Model AnswerP = 600060 = 100 W.
Q7
An electric motor transfers 2400 J in 8 s. Calculate its power. (2 marks)
Model AnswerP = 24008 = 300 W.
Q8
A car engine transfers 84 000 J in 60 s. Calculate its power. (3 marks)
Model AnswerP = 84 00060 = 1400 W.
Q9
A device transfers 3 kJ of energy in 30 s. Calculate its power. (3 marks)
Model AnswerConvert: E = 3 × 1000 = 3000 J. P = 300030 = 100 W.
Q10
An appliance transfers 1800 J of energy in 3 minutes. Calculate its power. (3 marks)
Model AnswerConvert: t = 3 × 60 = 180 s. P = 1800180 = 10 W.
Q11
A motor transfers 4.8 kJ of energy in 60 s. Calculate its power. (3 marks)
Model AnswerConvert: E = 4.8 × 1000 = 4800 J. P = 480060 = 80 W.
Q12
A lamp transfers 3.6 kJ of energy in 3 minutes. Calculate its power. (3 marks)
Model AnswerConvert: E = 3.6 × 1000 = 3600 J; t = 3 × 60 = 180 s. P = 3600180 = 20 W.
Q13
Two motors both lift the same weight the same height. Motor A takes 5 s; Motor B takes 20 s. Which is more powerful? Explain. (2 marks)
Model AnswerMotor A — same energy transferred in less time, so power (Et) is greater.

Part 3 of 4 · Finding Energy or Time

Example 3 (finding energy): a device rated 250 W runs for 2 minutes. Calculate the energy transferred.

Worked Example — Finding E
V
E = ?   P = 250 W   t = 2 min = 120 s
E
P = Et
S
250 = E120
S
250 × 120 = E  →  E = 30 000
U
Joules (J)

Example 4 (finding time): a 500 W device transfers 15 000 J. Calculate the time taken.

Worked Example — Finding t
V
t = ?   P = 500 W   E = 15 000 J
E
P = Et
S
500 = 15 000t
S
500t = 15 000  →  t = 15 000500 = 30
U
Seconds (s)

Questions

Q14
A 40 W fan runs for 5 s. How much energy does it transfer? (2 marks)
Model AnswerE = P × t = 40 × 5 = 200 J.
Q15
A 250 W lamp runs for 2 minutes. Calculate the energy transferred. (3 marks)
Model AnswerE = P × t = 250 × 120 = 30 000 J.
Q16
A 3 kW kettle heats water for 4 minutes. Calculate the energy transferred in kJ. (3 marks)
Model AnswerE = P × t = 3000 × 240 = 720 000 J = 720 kJ.
Q17
A 60 W electric fan runs for 30 minutes. Calculate the energy transferred in kJ. (3 marks)
Model AnswerE = P × t = 60 × 1800 = 108 000 J = 108 kJ.
Q18
A device transfers 18 000 J at a power of 300 W. How long does it run for? (3 marks)
Model Answert = EP = 18 000300 = 60 s.
Q19
A battery stores 900 000 J. A 50 W lamp is connected. How many hours will it run? (3 marks)
Model Answert = EP = 900 00050 = 18 000 s; 18 0003600 = 5 hours.
Q20
A 2 kW motor transfers 480 000 J. How long does it take? Give your answer in minutes. (3 marks)
Model Answert = EP = 480 0002000 = 240 s; 24060 = 4 minutes.

Exam Question — Crane Lift

A crane lifts a 150 kg load through a height of 8 m in 12 s. g = 9.8 N/kg.
Total: 5 marks · mark allocations shown as (n) following AQA convention.
(a)
Calculate the gravitational potential energy gained by the load. (2)
Mark SchemeSubstitution: E = 150 × 9.8 × 8 (1 mark). Answer: E = 11 760 J (1 mark).
(b)
Using your answer to (a), calculate the power of the crane. (2)
Mark SchemeSubstitution: P = 11 76012 (1 mark). Answer: P = 980 W (1 mark).
(c)
State the unit of power. (1)
Mark SchemeWatt (W) (1 mark).

Exam Question — Water Pump

A pump raises 60 kg of water through a height of 5 m in 6 s. g = 9.8 N/kg.
Total: 5 marks · mark allocations shown as (n) following AQA convention.
(a)
Calculate the gravitational potential energy gained by the water. (2)
Mark SchemeSubstitution: E = 60 × 9.8 × 5 (1 mark). Answer: E = 2940 J (1 mark).
(b)
Calculate the power of the pump. (2)
Mark SchemeSubstitution: P = 29406 (1 mark). Answer: P = 490 W (1 mark).
(c)
State the unit of power. (1)
Mark SchemeWatt (W) (1 mark).

Part 4 of 4 · Two-Step Calculations

A motor lifts a 50 kg rock through a height of 4 m in 8 s. g = 9.8 N/kg. Calculate the power of the motor.

Step 1 — Calculate GPE
V
Ep = ?   m = 50 kg   h = 4 m   g = 9.8 N/kg
E
Ep = m × g × h
S
Ep = 50 × 9.8 × 4
S
Ep = 1960
U
Joules (J)
Step 2 — Calculate Power
V
P = ?   E = 1960 J   t = 8 s
E
P = Et
S
P = 19608
S
P = 245
U
Watts (W)

Questions

Q21
A 75 kg person climbs stairs of height 4 m in 8 s. g = 9.8 N/kg. Calculate power. (3 marks)
Model AnswerE = 75 × 9.8 × 4 = 2940 J; P = 29408 = 367.5 W.
Q22
A 90 kg weightlifter raises a bar 1.8 m in 2 s. g = 9.8 N/kg. Calculate power. (3 marks)
Model AnswerE = 90 × 9.8 × 1.8 = 1587.6 J; P = 1587.62794 W.
Q23
A motor transfers 540 000 J in 15 minutes. Calculate its power in W and in kW. (3 marks)
Model AnswerP = 540 000900 = 600 W = 0.6 kW.

Exam Question — Steel Beam

A crane lifts a 400 kg steel beam through a height of 15 m. g = 9.8 N/kg.
Total: 6 marks · mark allocations shown as (n) following AQA convention.
(a)
Calculate the gravitational potential energy gained by the steel beam. (2)
Mark SchemeE = mgh = 400 × 9.8 × 15 = 58 800 J.
(b)
The crane takes 40 seconds to lift the beam. Calculate the power of the crane. (2)
Mark SchemeP = Et = 58 80040 = 1470 W.
(c)
A second crane lifts the same beam through the same height but takes only 25 seconds. Calculate the power of the second crane. (2)
Mark SchemeP = Et = 58 80025 = 2352 W.

Lesson 8 · Energy Stores & Transfers

Do Now

Power: P = Et
Q1
What is the formula for power?
Model AnswerP = Et.
Q2
A lamp transfers 200 J in 4 s. Calculate its power.
Model AnswerP = 2004 = 50 W.
Q3
What is meant by the “rate” of energy transfer?
Model AnswerHow quickly energy is transferred (power, in Watts).
Q4
A 250 W lamp runs for 2 minutes. Calculate the energy transferred.
Model AnswerE = P × t = 250 × 120 = 30 000 J.

Part 1 of 3 · Energy Stores Recap

The four main stores studied so far: Gravitational (E = mgh), Kinetic (E = ½mv²), Elastic (E = ½ke²), Thermal (E = mcΔT).

Other stores: Chemical (food, fuel, batteries), Nuclear (uranium), Electrostatic (charged objects), Magnetic (magnets).

Energy is transferred when something happens. Transfers can be shown as bar models or Sankey diagrams.

In a bar model, the total height of bars stays constant (conservation) while the energy distributes between stores.

Three-stage bar model of an archer drawing a bow, the arrow in flight and the arrow landing
Fig 8.1 — Energy store bar models for an archer firing an arrow: elastic → kinetic as the arrow travels.

Questions

Q1
Name the eight energy stores. (4 marks)
Model AnswerKinetic, gravitational, elastic, thermal, chemical, nuclear, electrostatic, magnetic.
Q2
What energy store does a charged capacitor have? (1 mark)
Model AnswerElectrostatic.
Q3
Describe the energy stores at each stage as an archer fires an arrow and it lands in a target. (4 marks)
Model AnswerPulling back: chemical → elastic. Arrow in air: elastic → kinetic. Landing: kinetic → thermal + sound.
Q4
Why do the bars in a bar model always add up to the same total? (2 marks)
Model AnswerConservation of energy — energy is never created or destroyed.

Part 2 of 3 · Energy Transfer Mechanisms

Energy is transferred between stores by: Mechanical working (forces), Electrical working (current), Heating (conduction/convection/radiation), Radiation (light, sound, etc.).

The rate of energy transfer is power (Watts).

Example — electric motor: electrical → (electrical working) → kinetic + thermal.

Example — burning fuel: chemical → (heating) → thermal + (radiation) → light.

Questions

Q5
Name the four mechanisms by which energy can be transferred. (2 marks)
Model AnswerMechanical working, electrical working, heating, radiation.
Q6
A boy falls on a trampoline. Describe the energy transfers. (4 marks)
Model AnswerGPE → KE (falling) → elastic PE (trampoline deforms) → KE (bouncing) + thermal + sound.
Q7
A jack-in-the-box pops open. Describe the energy transfers. (3 marks)
Model AnswerChemical (spring compressed) → elastic PE → KE + sound (when released).
Q8
A weight hangs on a spring and oscillates. Describe the energy transfers. (3 marks)
Model AnswerKE ↔ EPE continually, with some thermal dissipated each cycle.

Part 3 of 3 · Energy Transfer Diagrams

A Sankey diagram shows energy transfers with arrow widths proportional to energy values.

Useful energy goes forwards (horizontal); wasted energy goes downward (usually thermal).

For a car engine: 1000 J input → 250 J kinetic (forward) + 750 J thermal (down).

Efficiency can be read from a Sankey diagram:

useful outputtotal input

Questions

Q9
What does the width of an arrow in a Sankey diagram represent? (1 mark)
Model AnswerThe amount of energy being transferred.
Q10
Which direction does wasted energy usually point in a Sankey diagram? (1 mark)
Model AnswerDownward.
Q11
A light bulb uses 100 J and emits 10 J as light. Sketch/describe its Sankey diagram. (3 marks)
Model Answer100 J input → 10 J light (forward, narrow arrow) + 90 J thermal (down, wide arrow).
Q12
Describe the energy transfers in a bungee jumper from jump to lowest point. (4 marks)
Model AnswerGPE → KE (falling freely) → elastic PE + KE (cord stretching) → elastic PE only (at lowest point).

Exam Question — Bungee Energy Transfers

A student jumps off a bridge on a bungee cord. The cord has an unstretched length of 20 m.
Total: 7 marks · mark allocations shown as (n) following AQA convention.
(a)
Give two reasons why it is important that the cord is appropriate for the student’s weight. (2)
Mark SchemeToo weak: won’t stop the student in time/hits ground. Too stiff: gives dangerous deceleration.
(b)
The student falls. Describe the energy transfers before the cord starts to stretch. (2)
Mark SchemeGPE → KE as student falls (before cord tenses).
(c)
When the cord is stretched, state the energy store in the cord. (1)
Mark SchemeElastic potential energy.
(d)
The student’s GPE decreases by 29 400 J. KE increases by 18 000 J. What has happened to the rest of the energy? (2)
Mark Scheme29 400 − 18 000 = 11 400 J stored as elastic PE in the cord.

Lesson 9 · Reducing Wasted Energy

Do Now

Q1
Name the four mechanisms by which energy can be transferred.
Model AnswerMechanical working, electrical working, heating, radiation.
Q2
What does the width of an arrow in a Sankey diagram represent?
Model AnswerThe amount of energy being transferred.
Q3
A boy falls on a trampoline. Describe the energy transfers.
Model AnswerGPE → KE (falling) → elastic PE (trampoline deforms) → KE (bouncing) + thermal + sound.
Q4
A light bulb uses 100 J and emits 10 J as light. Describe its Sankey diagram.
Model Answer100 J input → 10 J light (forward, narrow arrow) + 90 J thermal (down, wide arrow).

Part 1 of 3 · Thermal Insulation

Thermal insulation reduces the rate of energy transfer to the surroundings.

Methods: cavity wall insulation, loft insulation (foil or foam), double glazing, draught excluders.

Best insulating materials have low thermal conductivity (e.g. foam, wool, air gaps).

High thermal conductivity = faster energy transfer. Low thermal conductivity = slower transfer.

Metal has high thermal conductivity; foam has low thermal conductivity.

Side-by-side cross sections of an uninsulated and an insulated floor
Fig 9.1 — Heat transfer through an uninsulated floor (energy lost downward) vs. insulated floor (energy reflected/reduced).

Questions

Q1
Why do people want to reduce unwanted energy transfers? (2 marks)
Model AnswerTo save money/energy, improve efficiency, reduce environmental impact.
Q2
Give an example of an unwanted energy transfer in a sewing machine. (1 mark)
Model AnswerKE → thermal due to friction between moving parts.
Q3
Explain how unwanted energy transfer is reduced in heated buildings. (2 marks)
Model AnswerCavity wall insulation, double glazing, loft insulation — all reduce conduction/convection.
Q4
Relate thermal conductivity to rate of energy transfer. (2 marks)
Model AnswerHigher thermal conductivity → faster energy transfer through the material.
Q5
How does the thickness of a wall affect a building’s rate of cooling? (2 marks)
Model AnswerThicker walls → slower rate of cooling (more insulation).
Q6
How does the thermal conductivity of walls affect a building’s rate of cooling? (2 marks)
Model AnswerLower thermal conductivity → slower rate of cooling.

Part 2 of 3 · Reducing Friction & Other Losses

Lubrication (oil/grease) reduces friction between moving surfaces, reducing thermal energy waste.

Streamlining reduces air resistance, reducing thermal energy wasted in vehicles.

Insulation around pipes and tanks reduces thermal energy loss by conduction.

In electrical devices, thicker wires reduce resistance and therefore reduce heating losses.

Questions

Q7
Explain how lubrication reduces energy waste in engines. (2 marks)
Model AnswerLubrication reduces friction between moving parts, reducing energy dissipated as heat.
Q8
Why are racing cars designed to be streamlined? (2 marks)
Model AnswerStreamlining reduces air resistance, reducing thermal energy wasted overcoming drag.
Q9
How does insulating a hot water pipe reduce energy waste? (2 marks)
Model AnswerIt reduces the rate of thermal energy loss by conduction from the pipe to the surroundings.
Q10
A house loses 20% of heat through the roof. Suggest one way to reduce this. (1 mark)
Model AnswerAdd loft insulation.

Part 3 of 3 · Reducing Energy Waste — Hot Water Tanks

A copper hot water tank loses energy quickly because copper has high thermal conductivity.

Insulation (foam jacket) around the tank reduces the rate of energy transfer to the room.

An insulated tank keeps water hot for longer, reducing the need to reheat it.

The electric immersion heater inside the tank converts electrical energy to thermal energy.

Cutaway of a hot water tank with insulation and an immersion heater
Fig 9.2 — Insulated hot water tank showing tank, insulation layer and electric immersion heater.

Questions

Q11
Why does a copper hot water tank lose heat quickly? (1 mark)
Model AnswerCopper has high thermal conductivity so conducts heat to surroundings rapidly.
Q12
How does a foam jacket reduce heat loss from the tank? (2 marks)
Model AnswerFoam has low thermal conductivity so slows conduction from tank to air.
Q13
Compared to an uninsulated tank, how does the rate of energy transfer change? (1 mark)
Model AnswerThe rate of energy transfer is reduced (slower cooling).
Q14
Explain why insulating a hot water tank saves money. (2 marks)
Model AnswerWater stays hot longer → immersion heater needed less often → lower electricity bill.

Exam Question — Hot Water Tank

Figure 9.2 shows a copper hot water tank with insulation.
Total: 5 marks · mark allocations shown as (n) following AQA convention.
(a)
Copper has higher thermal conductivity than most metals. How does the rate of energy transfer compare to most metals? Tick one box: Higher / Lower / The same. (1)
Mark SchemeHigher.
(b)
The tank is insulated. The water is heated, then the heater switches off. Compare the rate of cooling with and without insulation. (2)
Mark SchemeWith insulation, the rate of energy transfer to the room is reduced so the water stays hotter for longer.
(c)
During one morning, 4 070 000 J is transferred from the heater. 4 030 000 J goes to the water. Calculate the proportion of energy transferred to the water. (2)
Mark Scheme4 030 0004 070 000 = 0.990 (99.0%).

Lesson 10 · Efficiency

Do Now

Q1
What method is used to reduce heat loss through a house’s walls?
Model AnswerCavity wall insulation (or loft insulation / double glazing).
Q2
Explain how lubrication reduces energy waste in engines.
Model AnswerLubrication reduces friction between moving parts, reducing energy dissipated as heat.
Q3
Why does a copper hot water tank lose heat quickly?
Model AnswerCopper has high thermal conductivity, so it conducts heat to the surroundings rapidly.
Q4
How does a foam jacket reduce heat loss from a hot water tank?
Model AnswerFoam has low thermal conductivity, so it slows conduction from the tank to the air.

Part 1 of 3 · What is Efficiency?

Efficiency is the fraction of input energy that is transferred to a useful output.

Formula 1: Efficiency = useful output energytotal input energy
Formula 2: Efficiency = useful output powertotal input power

Efficiency has no units. It is between 0 and 1 (or expressed as a percentage 0–100%).

No machine is 100% efficient — some energy is always dissipated as thermal or sound.

Two power stations with input fuel arrows and useful and wasted output arrows
Fig 10.1 — Power Station A transfers 400 J usefully from 600 J of input energy (efficiency 0.67); Power Station B transfers 380 J usefully from 540 J of input energy (efficiency 0.70). Station B is more efficient.

Questions

Q1
What does the efficiency of an appliance tell us? (2 marks)
Model AnswerWhat fraction of the total input energy is transferred to a useful output.
Q2
Write both equations for efficiency. (2 marks)
Model AnswerEff = useful output Etotal input E   ·   Eff = useful output Ptotal input P
Q3
Why would a more efficient motor give a car a higher top speed? (2 marks)
Model AnswerMore KE output per unit of fuel → reaches higher speed before air resistance balances thrust.
Q4
A loudspeaker uses 400 W and produces 325 W of sound. Find its efficiency. (3 marks)
Model AnswerEff = 325400 = 0.81 = 81%.
Q5
A car uses 3890 J of fuel to generate 2650 J of KE. Find its efficiency. (3 marks)
Model AnswerEff = 26503890 = 0.68 = 68%.

Part 2 of 3 · Calculating and Applying Efficiency

Example: A lightbulb uses 470 J, emits 180 J as heat and 290 J as light.

Efficiency = 290470 = 0.617

To find wasted energy: wasted = input − useful output.

Increasing efficiency: lubrication, streamlining, better insulation, using waste heat (CHP).

CHP (Combined Heat and Power) stations recapture waste steam to heat homes, greatly increasing overall efficiency.

Questions

Q6
A lightbulb uses 470 J; 180 J is heat and 290 J is light. Find the efficiency. (3 marks)
Model AnswerEff = 290470 = 0.617.
Q7
A power station uses 2000 J of fuel to generate 600 J of electricity. How much energy is wasted? (2 marks)
Model Answer2000 − 600 = 1400 J wasted.
Q8
An electric motor uses 2.4×108 J to give a weight 1.8×107 J of GPE. Find efficiency. (3 marks)
Model AnswerEff = 1.8×1072.4×108 = 0.075 = 7.5%.
Q9
A lightbulb (efficiency 0.4) emits 3000 J of light in 20 s. Find the power input. (3 marks)
Model AnswerUseful power = 300020 = 150 W. Input P = 1500.4 = 375 W.

Part 3 of 3 · Efficiency as a Percentage & Increasing It

To express as percentage: multiply the decimal by 100 (e.g. 0.67 → 67%).

Wasted energy is usually thermal. Reducing thermal losses increases efficiency.

Environmental argument: more efficient appliances use less fuel → lower CO2 emissions.

Economic argument: more efficient appliances cost less to run.

Questions

Q10
Convert efficiency 0.82 to a percentage. (1 mark)
Model Answer82%.
Q11
A power station uses 2000 J to produce 600 J electricity. For every 600 J of electricity, how much fuel is used? (2 marks)
Model Answer2000 J of fuel needed for every 600 J of electricity.
Q12
State two reasons why improving efficiency is desirable. (2 marks)
Model AnswerSaves money; reduces fuel consumption and environmental impact.
Q13
A fan uses 180 W and delivers 150 W of useful power to the air. Calculate its efficiency. (3 marks)
Model AnswerEff = 150180 = 0.833 = 83.3%.

Exam Question — Power Station Efficiency

A fuel burning power station uses 2000 J of fuel energy to generate 600 J of electrical energy.
Total: 6 marks · mark allocations shown as (n) following AQA convention.
(a)
Calculate the efficiency of the power station. (2)
Mark SchemeEff = 6002000 = 0.3 = 30%.
(b)
State where the rest of the energy goes. (1)
Mark SchemeThermal energy transferred to the surroundings (wasted heat).
(c)
A second power station has efficiency 0.70. For every 1000 J of fuel, how much electricity does it produce? (2)
Mark Scheme0.70 × 1000 = 700 J.
(d)
Give one way the second power station could increase its efficiency. (1)
Mark SchemeUse CHP to capture waste steam heat; reduce friction in turbines; better insulation.

Lesson 11 · Non-Renewable Energy Resources

Do Now

Efficiency = useful output energytotal input energy
Q1
Give one way to increase the efficiency of a machine.
Model AnswerAny valid: lubrication, streamlining, better insulation, using waste heat (CHP).
Q2
A loudspeaker uses 400 W and produces 325 W of sound. Find its efficiency.
Model AnswerEff = 325400 = 0.81 = 81%.
Q3
A power station uses 2000 J of fuel to generate 600 J of electricity. How much energy is wasted?
Model Answer2000 − 600 = 1400 J.
Q4
Convert an efficiency of 0.82 to a percentage.
Model Answer82%.

Part 1 of 3 · Fossil Fuels

Fossil fuels (coal, oil, natural gas) formed from ancient organisms over millions of years. They are non-renewable.

Fossil fuels store chemical energy. When burned, chemical energy → thermal → electrical (via turbine/generator).

Advantages: reliable, high energy density, existing infrastructure.

Disadvantages: non-renewable (will run out ~50 years); produce CO2 (greenhouse gas/climate change); coal also produces SO2 (acid rain).

They provide a consistent, controllable supply — useful for meeting peak demand.

Questions

Q1
State three things humans use energy resources for. (3 marks)
Model AnswerElectricity generation, heating, transport.
Q2
Define “non-renewable”. (2 marks)
Model AnswerA resource that cannot be replenished as fast as it is consumed; it will eventually run out.
Q3
Name three common non-renewable energy resources. (3 marks)
Model AnswerCoal, oil, natural gas (and nuclear counts as non-renewable).
Q4
What are the benefits of fossil fuels? (2 marks)
Model AnswerReliable, high energy density, existing infrastructure, controllable output.
Q5
When are fossil fuels likely to run out? (1 mark)
Model AnswerApproximately 50 years.
Q6
Name two things formed when fossil fuels are burned and explain the problems they cause. (4 marks)
Model AnswerCO2 → greenhouse effect/climate change; SO2 (from coal) → acid rain/respiratory problems.

Part 2 of 3 · Nuclear Fuels

Nuclear fuels (uranium, plutonium) are non-renewable but will last ~80 years.

Nuclear fission: heavy nucleus absorbs a neutron and splits, releasing large amounts of energy as heat.

Advantages: no CO2 emissions during operation; very high energy density; reliable.

Disadvantages: expensive to build; produces radioactive waste (difficult to dispose of safely); risk of contamination if accident occurs.

Scientists research nuclear fusion (joining light nuclei) which would be cleaner but is not yet commercially viable.

Questions

Q7
Explain why nuclear energy is non-renewable. (2 marks)
Model AnswerUranium and plutonium are finite resources; they will eventually run out.
Q8
What are the risks with using nuclear power? (3 marks)
Model AnswerRadioactive waste is toxic and hard to dispose of; risk of accidental contamination of surroundings.
Q9
What are the benefits of nuclear power compared to fossil fuels? (3 marks)
Model AnswerNo CO2 produced; much longer-lasting fuel; much higher energy density per kg.
Q10
What is nuclear fusion and why is it not yet used commercially? (3 marks)
Model AnswerFusion joins light nuclei releasing enormous energy; not yet commercially viable because containing the plasma requires more energy than it produces currently.

Part 3 of 3 · UK Energy Mix

The UK uses a mix of energy resources. Gas and wind together now provide the majority of UK electricity, with nuclear, biomass, solar and imports making up the rest.

The UK stopped using coal-fired power stations entirely in 2024, to reduce CO2 emissions.

Electricity demand varies throughout the day — non-renewables are used to meet peak demand because they are controllable.

Renewables are increasingly replacing non-renewables as technology improves and costs fall.

Exam Question — UK Energy Resources

Figure 11.1 shows how different energy resources generated electricity in the UK across 2025. Figure 11.2 shows how GB electricity demand varies over the course of a day.
Total: 8 marks · mark allocations shown as (n) following AQA convention.
Pie chart of UK electricity generation by source in 2025
Fig 11.1 — UK electricity generation by source, 2025. Source: Carbon Brief analysis of NESO/DESNZ data.
Line graph of GB electricity demand across 24 hours
Fig 11.2 — GB electricity demand over 24 hours. Source: Electricity Maps, Electricity Grid Review 2025.
(a)
The UK stopped using coal-fired power stations in 2024. Explain one environmental problem caused when electricity is generated by burning coal. (2)
Mark SchemeCO2 is a greenhouse gas contributing to climate change; SO2 causes acid rain/respiratory problems.
(b)
Use Figure 11.1 to determine the percentage of electricity generated by nuclear power in 2025. (2)
Mark SchemeNuclear = 11% (from Fig 11.1).
(c)
Figure 11.2 shows electricity demand varying with time of day. Identify the maximum and minimum demand and calculate the difference. (2)
Mark SchemeMax ≈ 36.5 GW (17:00); min ≈ 23 GW (04:00); difference ≈ 13.5 GW.
(d)
Explain why non-renewable sources are used to meet peak electricity demand. (2)
Mark SchemeNon-renewables can be turned up quickly to match demand; renewables depend on weather.

Lesson 12 · Renewable Energy Resources

Do Now

Q1
What is meant by a “non-renewable” energy resource?
Model AnswerA resource that cannot be replenished at the rate at which it is used.
Q2
Name three fossil fuels.
Model AnswerCoal, oil, natural gas.
Q3
Approximately how long will fossil fuels last at current usage?
Model AnswerApproximately 50 years.
Q4
What is nuclear fission?
Model AnswerThe splitting of a heavy atomic nucleus (e.g. uranium) releasing large amounts of energy.

Part 1 of 3 · Wind, Solar & Geothermal

Wind power: Wind turns turbine blades → generator → electricity. Zero emissions, renewable. Disadvantage: intermittent; visual impact; harms birds.

Solar power: Photovoltaic cells convert sunlight to electricity. Renewable, no emissions. Disadvantage: no output at night or in cloudy weather; requires large area.

Geothermal: Water pumped into hot rocks underground, returns as steam → drives turbine. Renewable, no emissions. Disadvantage: only available in geologically active areas.

Wind turbine cutaway showing blades, gearbox and generator
Fig 12.1 — Wind turbine: (1) wind rotates blades, (2) gearbox speeds up rotation, (3) generator produces electricity.
Diagram of a photovoltaic panel, inverter and household supply
Fig 12.2 — How a solar panel works: sunlight excites electrons in photovoltaic cells (negative layer → diode → positive layer), generating direct current (DC); an inverter converts DC to alternating current (AC) for use in the home.
Cross section of a geothermal power plant
Fig 12.3 — Geothermal power plant: (1) cold water injected, (2) steam rises, (3) turbine, (4) cooling tower, (5) injection well.

Questions

Q1
Explain how wind power generates electricity. (3 marks)
Model AnswerWind turns turbine blades attached to a generator; rotation of the generator produces electricity.
Q2
Give one advantage and one disadvantage of solar power. (2 marks)
Model AnswerAdv: renewable/no emissions. Disadv: only works in sunlight/weather-dependent.
Q3
Why is geothermal power only available in certain locations? (2 marks)
Model AnswerIt requires high underground temperatures close to the surface (volcanic/tectonic regions).
Q4
A wind turbine generates 2 MW of power. How much energy does it produce in 1 hour? (3 marks)
Model AnswerE = 2×106 × 3600 = 7.2×109 J = 7200 MJ.

Part 2 of 3 · Hydroelectric, Tidal & Wave

Hydroelectric: Water in reservoir flows through turbines. Renewable, zero emissions, reliable. Disadvantage: habitat destruction; needs suitable geography.

Tidal power: Turbines in sea turn with incoming/outgoing tides. Renewable, predictable. Disadvantage: few suitable sites; affects marine life.

Wave power: Floating devices harness wave motion to drive generators. Renewable. Disadvantage: unreliable, easily damaged by storms.

Cross section of a hydroelectric dam
Fig 12.4 — Hydroelectric dam: reservoir → penstock → turbine → generator → power lines.
Tidal barrage with turbine between high and low tide water levels
Fig 12.5 — Tidal power: water flows from high tide to low tide, passing through a turbine connected to a generator; the direction of water flow reverses with each tide, so the turbine generates electricity on both the incoming and outgoing tides.
Wave power buoys on the sea surface with cables to shore
Fig 12.6 — Wave power: surface buoys rise and fall with ocean waves, driving a linear generator submerged beneath each device; the electricity produced is carried to shore via undersea cables.

Questions

Q5
Describe how a hydroelectric power station generates electricity. (3 marks)
Model AnswerWater flows from reservoir through penstocks → turbines → generator → electricity.
Q6
Give one advantage of tidal power over wind power. (1 mark)
Model AnswerTides are predictable (unlike wind), so output can be planned in advance.
Q7
Why might wave power be unreliable? (2 marks)
Model AnswerWave height varies with weather; storms can damage equipment.
Q8
What environmental problem is associated with hydroelectric dams? (2 marks)
Model AnswerDamming floods land/habitat; disrupts river ecosystems and fish migration.

Part 3 of 3 · Biofuels & Global Energy Trends

Biofuels (wood, bioethanol, manure) are produced from living things and are considered renewable.

Biofuels are carbon-neutral in theory (CO2 absorbed growing = CO2 released burning), but in practice still contribute to climate change.

Disadvantage: land used for biofuel crops cannot grow food — a concern where food shortages exist.

Global primary energy consumption has risen sharply since 1800, driven by industrialisation and population growth.

Renewables are a small but growing fraction; fossil fuels still dominate globally.

Stacked area chart of global primary energy consumption since 1800
Fig 12.7 — Global primary energy consumption (TWh/year) since 1800. Source: Our World in Data (Smil 2017; Energy Institute Statistical Review 2025). Coal, oil and gas dominate; renewables are rising rapidly.

Questions

Q9
Why are biofuels considered renewable? (2 marks)
Model AnswerThey come from living organisms that can be regrown/replenished.
Q10
Give one reason why biofuels are not completely carbon-neutral in practice. (2 marks)
Model AnswerLand-use changes (deforestation) and transport/processing release extra CO2.
Q11
From Fig 12.7, when did global energy use begin to increase most rapidly? (1 mark)
Model AnswerAfter approximately 1950.
Q12
What was the main cause of the rapid energy use increase after 1950? (2 marks)
Model AnswerIndustrialisation and population growth dramatically increased energy demand.
Q13
How has the use of renewables compared to non-renewables changed? Give two differences. (4 marks)
Model AnswerNon-renewables still far larger in total; renewables growing faster proportionally; non-renewables dominated for most of history.

Exam Question — Electric vs Diesel Car

An electric car has a motor powered by a battery. A diesel car has an engine powered by diesel fuel.
Total: 6 marks · mark allocations shown as (n) following AQA convention.
(a)
The table compares the two cars. State one way the electric car is better for the environment. (1)
Mark SchemeThe electric car produces no direct CO2/pollution emissions during use.
(b)
Electricity is generated in a gas power station (efficiency 0.35) and transmitted to the electric car (transmission efficiency 0.90). Calculate the overall efficiency of generating and transmitting the electricity. (3)
Mark SchemeOverall efficiency = 0.35 × 0.90 = 0.315 = 31.5%.
(c)
Give one reason why renewable energy sources are important for the future. (2)
Mark SchemeFossil fuels will run out / to reduce CO2 emissions and combat climate change.
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