ENERGY

AQA 8463 & 8464 - All Lessons
Student Booklet Teacher Booklet

Energy Stores & Gravitational Potential Energy

Do Now

Q1
What is the unit of energy?
Model Answer Joule (J).
Q2
Name two types of energy store.
Model Answer Any two of: kinetic, gravitational, elastic, thermal, chemical, nuclear, electrostatic, magnetic.
Q3
What does "g" represent in physics equations?
Model Answer Gravitational field strength (N/kg). On Earth, g = 9.8 N/kg.
Q4
If an object is lifted higher, what happens to its gravitational potential energy?
Model Answer It increases.

Part 1 · 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). Larger quantities are expressed in kJ (1 kJ = 1000 J) or MJ.

Questions

Q1
Define energy. (1)
Model Answer The ability to do work / capacity to cause change.
Q2
Name four energy stores. (2)
Model Answer Any four from: gravitational, kinetic, elastic, thermal, chemical, nuclear, electrostatic, magnetic.
Q3
State the Law of Conservation of Energy. (2)
Model Answer Energy cannot be created or destroyed, only transferred between stores.
Q4
A ball is held above the ground. Which energy store does it have? (1)
Model Answer Gravitational potential energy store.
Q5
Convert 4500 J into kJ. (1)
Model Answer 4.5 kJ

Part 2 · 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 (J); m = mass (kg); g = gravitational field strength (N/kg); h = height (m).

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

Example: A rock of mass 75 kg is lifted 4 m. Ep = 75 × 9.8 × 4 = 2940 J.

Questions

Q1
Define gravitational potential energy store. (1)
Model Answer Energy stored due to an object's position in a gravitational field.
Q2
What does g measure? Give its value on Earth. (2)
Model Answer Gravitational field strength; 9.8 N/kg on Earth.
Q3
A 5 kg book is placed on a shelf 1.5 m high. g = 9.8 N/kg. Calculate its GPE. (3)
Model Answer E = 5 × 9.8 × 1.5 = 73.5 J.
Q4
If the height of an object doubles, what happens to its GPE? Explain. (2)
Model Answer GPE doubles, because GPE ∝ h.
Q5
A 60 kg person stands on a 3 m diving board. g = 9.8 N/kg. Calculate GPE. (3)
Model Answer E = 60 × 9.8 × 3 = 1764 J.

Part 3 · Rearranging E = mgh

To find mass: m = Ep ÷ (g × h)

To find height: h = Ep ÷ (m × g)

To find g: g = Ep ÷ (m × h)

Use the VESSS method: Values → Equation → Substitute → Solve → State units.

Questions

Q1
A ball has a GPE of 196 J and a mass of 2 kg. g = 9.8 N/kg. Find the height. (3)
Model Answer h = 196 ÷ (2 × 9.8) = 10 m.
Q2
A buzzard stores 5403 J of GPE at a height of 98 m. g = 9.8 N/kg. Find its mass. (3)
Model Answer m = 5403 ÷ (9.8 × 98) = 5.63 kg.
Q3
An object of mass 12 kg is at a height of 5 m and stores 588 J of GPE. Find g. (3)
Model Answer g = 588 ÷ (12 × 5) = 9.8 N/kg.
Q4
A 3 kg ball is dropped from 10 m. g = 9.8 N/kg. Calculate the GPE lost. (3)
Model Answer E = 3 × 9.8 × 10 = 294 J.
Q5
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)
Model Answer E = 20 × 1.6 × 8 = 256 J.

Exam Question — Zip Wire (7 marks)

A person slides down a zip wire from height h. Mass = 60 kg, g = 9.8 N/kg.

Exam Questions

Part 1
(a) The change in GPE is 1.47 kJ. Calculate the change in vertical height. (3)
Model Answer 1.47 kJ = 1470 J. h = 1470 ÷ (60 × 9.8) = 2.5 m.
Part 2
(b) As the person moves down the zip wire, the increase in KE is less than the decrease in GPE. Explain why. (2)
Model Answer Some GPE is transferred to thermal energy due to friction / air resistance.
Part 3
(c) Different people reach different speeds at the bottom. Explain why. (2)
Model Answer People have different masses, so the same GPE change gives different velocities.

Kinetic Energy

Do Now

Q1
What is the formula for gravitational potential energy?
Model Answer E = mgh.
Q2
What is the unit of energy?
Model Answer Joule (J).
Q3
A 10 kg object is 5 m high. g = 9.8 N/kg. Calculate its GPE.
Model Answer E = 10 × 9.8 × 5 = 490 J.
Q4
What does "velocity" mean and how does it differ from speed?
Model Answer Velocity is speed in a given direction (it is a vector quantity).

Part 1 · Kinetic Energy

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

Ek = ½ × m × v²

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

Velocity is used (not speed) because KE depends on both magnitude and direction of motion.

Tip: calculate v² first, then multiply by ½m. Using 0.5 for ½ on a calculator avoids errors.

Questions

Q1
Define kinetic energy store. (1)
Model Answer Energy stored in a moving object.
Q2
Write the formula for kinetic energy. (1)
Model Answer Ek = ½mv².
Q3
Why does a train waiting at a station have zero kinetic energy store? (1)
Model Answer Because it is stationary; v = 0.
Q4
A 4 kg object moves at 3 m/s. Calculate its KE. (3)
Model Answer E = 0.5 × 4 × 9 = 18 J.
Q5
A 0.2 kg ball moves at 5 m/s. Calculate its KE. (3)
Model Answer E = 0.5 × 0.2 × 25 = 2.5 J.

Part 2 · Rearranging E_k = ½mv²

To find mass: m = 2Ek ÷ v²

To find velocity: v = √(2Ek ÷ m)

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 velocity — doubling speed quadruples KE.

Questions

Q1
Why does a lorry at 30 mph have more KE than a car at the same speed? (1)
Model Answer The lorry has a much greater mass, so even at the same velocity KE = ½mv² is much larger.
Q2
A car of mass 1200 kg travels at 20 m/s. Calculate its KE. (3)
Model Answer E = 0.5 × 1200 × 400 = 240 000 J.
Q3
An object has KE = 200 J and mass 25 kg. Find its velocity. (3)
Model Answer v = √(2 × 200 ÷ 25) = √16 = 4 m/s.
Q4
An object has KE = 450 J and velocity = 6 m/s. Find its mass. (3)
Model Answer m = 2 × 450 ÷ 36 = 25 kg.
Q5
How does doubling the velocity of an object affect its KE? Explain. (2)
Model Answer KE is quadrupled because KE ∝ v².

Part 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

Q1
A trampolinist has KE = 8500 J and mass 65 kg. Find their velocity. (3)
Model Answer v = √(2 × 8500 ÷ 65) = √261.5 ≈ 16.2 m/s.
Q2
A 0.05 kg pancake is tossed with initial KE = 0.6 J. Calculate its initial velocity. (3)
Model Answer v = √(2 × 0.6 ÷ 0.05) = √24 ≈ 4.9 m/s.
Q3
A 2 kg ball is dropped from 10 m. g = 9.8 N/kg. What is its KE just before hitting the ground? (3)
Model Answer GPE = 2 × 9.8 × 10 = 196 J = KE.
Q4
A cyclist of mass 70 kg (including bike) reaches 15 m/s. Calculate their KE. (3)
Model Answer E = 0.5 × 70 × 225 = 7875 J.

Exam Question — Falling Ball (6 marks)

A ball of mass 0.5 kg is dropped from a height of 8 m. g = 9.8 N/kg. Ignore air resistance.

Exam Questions

Part 1
(a) Calculate the gravitational potential energy of the ball before it is dropped. (2)
Model Answer E = 0.5 × 9.8 × 8 = 39.2 J.
Part 2
(b) State the kinetic energy of the ball just before it hits the ground. (1)
Model Answer 39.2 J (GPE fully converted to KE).
Part 3
(c) Calculate the velocity of the ball just before it hits the ground. (3)
Model Answer v = √(2 × 39.2 ÷ 0.5) = √156.8 ≈ 12.5 m/s.

Conservation of Energy & Dissipation

Do Now

Q1
State the Law of Conservation of Energy.
Model Answer Energy cannot be created or destroyed, only transferred between stores.
Q2
What is a closed system?
Model Answer A closed system is one where no energy enters or leaves.
Q3
A 2 kg ball moves at 4 m/s. Calculate its KE.
Model Answer E = 0.5 × 2 × 16 = 16 J.
Q4
A 3 kg ball is dropped from 5 m. g = 9.8 N/kg. What is its KE just before hitting the ground?
Model Answer E = 3 × 9.8 × 5 = 147 J.

Part 1 · 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.

Questions

Q1
What is a closed system? (1)
Model Answer A system in which no energy enters or leaves.
Q2
State the law of conservation of energy. (2)
Model Answer Energy 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)
Model Answer 20 J — all KE converts to GPE at the highest point.
Q4
Why is no real system truly closed? (2)
Model Answer Some 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)
Model Answer KE = ½ × 25 × 64 = 800 J. h = 800 ÷ (25 × 9.8) ≈ 3.27 m.

Part 2 · 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

Q1
What does "dissipation" mean in physics? (2)
Model Answer Energy transferred to surroundings in a less useful (often thermal) form.
Q2
Give two examples of energy dissipation in everyday life. (2)
Model Answer Any two: friction in engines, air resistance on vehicles, sound in brakes, heat from light bulbs.
Q3
A bullet is shot upward with KE = 32 J. At its highest point its GPE = 31.8 J. Explain why the values differ. (2)
Model Answer 0.2 J was transferred to thermal energy / sound due to air resistance.
Q4
Name two ways to reduce energy dissipation in machines. (2)
Model Answer Lubrication (to reduce friction) and streamlining (to reduce air resistance).
Q5
Why is dissipated energy considered "wasted"? (2)
Model Answer Because it cannot easily be recovered and put back to useful work.

Part 3 · Energy Transfers in Scenarios

When describing energy transfers, state: 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

Q1
Describe the energy transfer as a ball falls from a height (ignore air resistance). (2)
Model Answer GPE → KE.
Q2
Describe the energy transfers for a person braking on a bicycle. (3)
Model Answer KE in cyclist → thermal energy in brakes (friction) + sound.
Q3
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)
Model Answer 1470 − 1100 = 370 J dissipated as thermal energy / sound.
Q4
Explain why a bouncing ball never bounces back to its original height. (2)
Model Answer Each bounce transfers some KE to thermal / sound, so less energy remains as KE / GPE.

Exam Question — Zip Wire Energy Loss (6 marks)

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.

Exam Questions

Part 1
(a) Calculate the decrease in GPE. (2)
Model Answer E = 60 × 9.8 × 2.5 = 1470 J.
Part 2
(b) The increase in KE is 1100 J. Calculate the energy transferred to thermal energy. (2)
Model Answer 1470 − 1100 = 370 J.
Part 3
(c) Explain why different people reach different speeds at the bottom. (2)
Model Answer Different masses → same GPE loss gives different KE → different velocities.

Elastic Potential Energy

Do Now

Q1
What is meant by the term "energy store"?
Model Answer A way energy can be held in an object.
Q2
What is the unit of the spring constant?
Model Answer N/m (Newtons per metre).
Q3
What is the difference between compression and extension?
Model Answer Compression = pushing together; extension = pulling apart.
Q4
If a spring with k = 5 N/m is stretched 2 m, what is its elastic potential energy?
Model Answer E = 0.5 × 5 × 4 = 10 J.

Part 1 · 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.

Questions

Q1
Define elastic potential energy store. (1)
Model Answer Energy stored in a stretched or compressed elastic object.
Q2
What two factors affect the elastic potential energy stored in a spring? (2)
Model Answer Spring constant (k) and extension (e).
Q3
Do you need the total length of a spring to calculate its EPE? Explain. (2)
Model Answer No — 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)
Model Answer E = 0.5 × 2 × 9 = 9 J.
Q5
Which spring is easier to stretch: k = 0.5 N/m or k = 40 N/m? Why? (2)
Model Answer k = 0.5 N/m is easier — it has a lower spring constant so less force is needed.

Part 2 · Rearranging E = ½ke²

To find spring constant: k = 2Ee ÷ e²

To find extension: e = √(2Ee ÷ k)

Example: A spring stores 18 J with k = 4 N/m. Find the extension. e = √(2 × 18 ÷ 4) = √9 = 3 m.

Questions

Q1
A spring (k = 75 N/m) is extended 40 cm. Calculate its EPE. (3)
Model Answer 0.4 m extension. E = 0.5 × 75 × 0.16 = 6 J.
Q2
A spring stores 400 J with k = 12 N/m. Find the extension. (3)
Model Answer e = √(2 × 400 ÷ 12) = √66.7 ≈ 8.16 m.
Q3
A spring stores 50 J when extended 5 m. Calculate the spring constant. (3)
Model Answer k = 2 × 50 ÷ 25 = 4 N/m.
Q4
A spring (k = 200 N/m) has 180 J of EPE. Find the extension. (3)
Model Answer e = √(2 × 180 ÷ 200) = √1.8 ≈ 1.34 m.

Part 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.

Questions

Q1
Describe the energy transfers in a bow-and-arrow as the arrow is fired. (3)
Model Answer Chemical energy in archer → EPE in string → KE in arrow (+ thermal from friction).
Q2
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 velocity. (4)
Model Answer EPE = 0.5 × 50 × 0.04 = 1 J. v = √(2 × 1 ÷ 0.1) = √20 ≈ 4.47 m/s.
Q3
Give two examples of objects that store elastic potential energy. (2)
Model Answer Any two: spring, rubber band, trampoline, bungee cord.

Exam Question — Bungee Cord (6 marks)

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.

Exam Questions

Part 1
(a) Give two reasons why the cord must be appropriate for the student's weight. (2)
Model Answer If too short / stiff the student won't reach maximum stretch; if too long / weak they may hit the ground.
Part 2
(b) The cord stretches 18 m. Calculate the elastic PE stored. (3)
Model Answer E = 0.5 × 15 × 18² = 0.5 × 15 × 324 = 2430 J.
Part 3
(c) The student's KE at the lowest point is zero. Explain why. (1)
Model Answer At the lowest point, velocity = 0 so KE = 0.

Specific Heat Capacity

Do Now

Q1
What is thermal energy?
Model Answer Energy stored due to the movement of particles in a substance.
Q2
What is the difference between temperature and thermal energy?
Model Answer Temperature is the average KE per particle; thermal energy is the total energy stored.
Q3
If 500 J is added to two objects of different materials but equal mass, will they heat up equally?
Model Answer No — different materials require different amounts of energy to raise temperature.
Q4
What unit is temperature measured in?
Model Answer Degrees Celsius (°C) or Kelvin (K).

Part 1 · Temperature and Thermal Energy

Temperature is the average kinetic energy store of the particles in a system.

Two blocks of equal mass but different materials will heat up by different amounts when given the same energy.

Copper has SHC = 386 J/kg°C; gold has SHC = 126 J/kg°C. Gold heats up faster with the same energy input.

The material that is "easier to heat up" has the lower specific heat capacity.

Questions

Q1
What does temperature measure at a particle level? (1)
Model Answer The average kinetic energy store of the particles.
Q2
Two blocks of equal mass are given the same energy. One heats up more than the other. What can you conclude? (2)
Model Answer The one that heated up more has a lower specific heat capacity.
Q3
Which metal requires more energy to raise its temperature by 1°C per kg: copper or gold? Why? (2)
Model Answer Copper — it has a higher SHC (386 vs 126 J/kg°C) so needs more energy per kg per °C.
Q4
Define the term "specific heat capacity". (2)
Model Answer The amount of energy required to raise the temperature of 1 kg of a substance by 1°C.

Part 2 · The SHC Equation

ΔE = m × c × ΔT

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

Example: Water (c = 4200 J/kg°C), mass 2 kg, heated by 10°C. ΔE = 2 × 4200 × 10 = 84 000 J.

The higher the SHC, the more energy needed to raise the temperature — water has one of the highest SHCs.

This equation is on the AQA formula sheet.

Questions

Q1
Write the equation for specific heat capacity. (1)
Model Answer ΔE = mcΔT.
Q2
A 1 kg block of iron (c = 450 J/kg°C) is heated from 20°C to 70°C. Calculate the energy transferred. (3)
Model Answer ΔE = 1 × 450 × 50 = 22 500 J.
Q3
A block of silver (m = 12.5 kg) is heated by 37°C. c = 233 J/kg°C. Calculate the thermal energy. (3)
Model Answer ΔE = 12.5 × 233 × 37 = 107 763 J.
Q4
A 3 kg copper block absorbs 5790 J. c = 386 J/kg°C. Find the temperature change. (3)
Model Answer ΔT = 5790 ÷ (3 × 386) = 5 °C.

Part 3 · Rearranging E = mcΔT

To find mass: m = ΔE ÷ (c × ΔT)

To find SHC: c = ΔE ÷ (m × ΔT)

To find temperature change: ΔT = ΔE ÷ (m × c)

Water has c = 4200 J/kg°C — this is why it is used in central heating systems (stores a lot of energy).

Questions

Q1
Find the mass of zinc needed to absorb 2300 J heated from 50°C to 61°C. c = 387 J/kg°C. (4)
Model Answer ΔT = 11 °C. m = 2300 ÷ (387 × 11) ≈ 0.54 kg.
Q2
A 3.4 kg mass of water (c = 4186 J/kg°C) gains 6800 J. Find the temperature change. (3)
Model Answer ΔT = 6800 ÷ (3.4 × 4186) ≈ 0.48 °C.
Q3
Water in a kettle (c = 4200 J/kg°C, m = 0.5 kg) is heated by 80°C. Calculate the energy used. (3)
Model Answer ΔE = 0.5 × 4200 × 80 = 168 000 J.
Q4
A material with c = 500 J/kg°C absorbs 12 500 J with a 5°C rise. Find the mass. (3)
Model Answer m = 12 500 ÷ (500 × 5) = 5 kg.

Exam Question — Heating Water (6 marks)

A student heats 0.5 kg of water using a 100 W immersion heater for 5 minutes. The specific heat capacity of water is 4200 J/kg°C.

Exam Questions

Part 1
(a) Calculate the energy supplied by the heater in 5 minutes. (2)
Model Answer E = P × t = 100 × 300 = 30 000 J.
Part 2
(b) Calculate the expected temperature rise of the water. (3)
Model Answer ΔT = 30 000 ÷ (0.5 × 4200) ≈ 14.3 °C.
Part 3
(c) The actual temperature rise is less than calculated. Give one reason why. (1)
Model Answer Some energy is transferred to the surroundings / container (heat loss).

Investigating Specific Heat Capacity

Do Now

Q1
Write the equation for specific heat capacity.
Model Answer ΔE = mcΔT.
Q2
What does the specific heat capacity of a substance tell us?
Model Answer How much energy is needed to raise the temperature of 1 kg of it by 1°C.
Q3
A 2 kg block is heated by 20°C. c = 400 J/kg°C. Calculate the energy transferred.
Model Answer ΔE = 2 × 400 × 20 = 16 000 J.
Q4
Why is water used in central heating systems?
Model Answer Water has a very high SHC so it stores a lot of energy per kg.

Part 1 · The SHC Investigation

Aim: to find the specific heat capacity of a metal block by measuring energy transferred and temperature change.

Equipment: metal block, immersion heater, thermometer, joulemeter (or ammeter + voltmeter + stopwatch), digital scales.

Method

  • Step 1: Measure the mass of the block using digital scales.
  • Step 2: Apply glycerine to the immersion heater and thermometer to ensure good thermal contact. Insert both.
  • Step 3: Record the starting temperature and joulemeter reading.
  • Step 4: Switch on the heater. Record temperature and energy at regular intervals for 10 minutes.
  • Step 5: Use ΔE = mcΔT to calculate c.

Questions

Q1
Name the key pieces of equipment for the SHC investigation. (3)
Model Answer Metal block, immersion heater, thermometer, joulemeter (or ammeter + voltmeter), digital scales.
Q2
Why is glycerine applied to the heater and thermometer? (2)
Model Answer To ensure good thermal contact (transfers heat more efficiently).
Q3
What measurements must be taken during the experiment? (3)
Model Answer Mass of block, starting temperature, energy supplied (joulemeter), temperature at regular intervals.
Q4
Why is it important to insulate (lag) the metal block? (2)
Model Answer To reduce heat loss to the surroundings and improve accuracy.

Part 2 · Analysing Results

Calculate c using: c = ΔE ÷ (m × ΔT)

Plot a graph of temperature (y-axis) against energy supplied (x-axis). The gradient = 1 ÷ (mc).

Possible errors: heat loss to surroundings (c appears too high); poor thermal contact (uneven heating).

To improve: use lagging; allow system to reach thermal equilibrium; repeat and average.

Compare calculated c with the known value (e.g. aluminium: 900 J/kg°C). Discuss percentage error.

Questions

Q1
A student heats a 0.5 kg aluminium block. Energy supplied = 4500 J; ΔT = 10°C. Calculate c. (3)
Model Answer c = 4500 ÷ (0.5 × 10) = 900 J/kg°C.
Q2
The known SHC of aluminium is 900 J/kg°C. Calculate the percentage error. (2)
Model Answer % error = 0% — matches exactly.
Q3
A student carried out an experiment with 100 g (0.1 kg) of water in sunlight for 30 minutes. The temperature rose by 3°C. c = 4200 J/kg°C. Find the energy absorbed. (2)
Model Answer ΔE = 0.1 × 4200 × 3 = 1260 J.
Q4
Suggest one improvement to increase the accuracy of the SHC experiment. (2)
Model Answer Any valid: better insulation, more accurate thermometer, repeat measurements.

Part 3 · Evaluating the SHC Investigation

Sources of error: heat loss to surroundings; thermal lag (thermometer not reaching equilibrium); poor contact.

Systematic errors: if the joulemeter is not zeroed; if mass measurement is incorrect.

Reliability: repeat the experiment and average values; use a data logger for precision.

Variables: independent = energy supplied; dependent = temperature change; controlled = mass, material, insulation.

Questions

Q1
What is the independent variable in the SHC investigation? (1)
Model Answer Energy supplied (joulemeter reading).
Q2
What is the dependent variable? (1)
Model Answer Temperature change.
Q3
Give two sources of error that could make the calculated c too high. (2)
Model Answer Heat loss to surroundings; poor thermal contact between heater and block.
Q4
How could you check if your result is repeatable? (1)
Model Answer Repeat the experiment and compare values.
Q5
Why should you record temperature over time rather than just at the end? (2)
Model Answer Allows you to identify when temperature stabilises and spot anomalies.

Exam Question — SHC Investigation (5 marks)

A student investigates the specific heat capacity of a metal block of mass 0.8 kg. The immersion heater supplies 3600 J of energy. The temperature rises from 20°C to 25°C.

Exam Questions

Part 1
(a) Calculate the specific heat capacity of the metal. (3)
Model Answer ΔT = 5 °C. c = 3600 ÷ (0.8 × 5) = 900 J/kg°C.
Part 2
(b) The student's value is higher than the accepted value. Give one reason why. (1)
Model Answer Heat was lost to the surroundings so less energy went into the block.
Part 3
(c) Suggest one way to reduce this source of error. (1)
Model Answer Insulate / lag the block to reduce heat loss.

Power

Do Now

Gravitational Potential Energy: EGP = m × g × h Kinetic Energy: Ek = ½ × m × v²
Q1
What is the unit of energy?
Model Answer Joule (J).
Q2
Calculate the GPE of a 3 kg object at a height of 5 m. g = 9.8 N/kg.
Model Answer EGP = 3 × 9.8 × 5 = 147 J.
Q3
A 20 kg object is 8 m high. g = 9.8 N/kg. Calculate its GPE.
Model Answer EGP = 20 × 9.8 × 8 = 1568 J.
Q4
A 4 kg ball moves at 6 m/s. Calculate its kinetic energy.
Model Answer Ek = 0.5 × 4 × 6² = 0.5 × 4 × 36 = 72 J.
Q5
A ball has a kinetic energy of 72 J and a mass of 4 kg. Calculate its speed.
Model Answer v² = (2 × 72) ÷ 4 = 36; v = √36 = 6 m/s.

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

Part 2 · Calculating Power

P = E ÷ t   |   P = W ÷ t
  • P = power (W)
  • E = Energy Transferred (J)
  • t = time (s)
Example 1
Values
P = ? E = 600 J t = 12 s
Equation
P = E ÷ t
Substitute
P = 600 ÷ 12
Solve
P = 50
Units
Watts (W)

Questions

Q5
A lamp transfers 200 J in 4 s. Calculate its power. (2)
Model AnswerP = 200 ÷ 4 = 50 W.
Q6
A heater transfers 6000 J in 60 s. Calculate its power. (2)
Model AnswerP = 6000 ÷ 60 = 100 W.
Q7
An electric motor transfers 2400 J in 8 s. Calculate its power. (2)
Model AnswerP = 2400 ÷ 8 = 300 W.
Q8
A car engine transfers 84 000 J in 60 s. Calculate its power. (3)
Model AnswerP = 84 000 ÷ 60 = 1400 W.

Unit Conversions

  • 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
V
P = ? E = 1.2 × 1000 = 1200 J t = 2 × 60 = 120 s
E
P = E ÷ t
S
P = 1200 ÷ 120
S
P = 10
U
Watts (W)

Questions

Q9
A device transfers 3 kJ of energy in 30 s. Calculate its power. (3)
Model AnswerConvert: E = 3 × 1000 = 3000 J. P = 3000 ÷ 30 = 100 W.
Q10
An appliance transfers 1800 J of energy in 3 minutes. Calculate its power. (3)
Model AnswerConvert: t = 3 × 60 = 180 s. P = 1800 ÷ 180 = 10 W.
Q11
A motor transfers 4.8 kJ of energy in 60 s. Calculate its power. (3)
Model AnswerConvert: E = 4.8 × 1000 = 4800 J. P = 4800 ÷ 60 = 80 W.
Q12
A lamp transfers 3.6 kJ of energy in 3 minutes. Calculate its power. (3)
Model AnswerConvert: E = 3.6 × 1000 = 3600 J; t = 3 × 60 = 180 s. P = 3600 ÷ 180 = 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)
Model AnswerMotor A — same energy transferred in less time, so power (E ÷ t) is greater.

Part 3 · Finding Energy or Time

Rearranging P = E ÷ t gives: E = P × t and t = E ÷ P

Example 3 — finding energy
V
E = ? P = 250 W t = 2 min = 120 s
E
P = E ÷ t
S
250 = E ÷ 120
S
250 × 120 = E = 30 000
U
Joules (J)
Example 4 — finding time
V
t = ? P = 500 W E = 15 000 J
E
P = E ÷ t
S
500 = 15 000 ÷ t
S
500t = 15 000
S
t = 15 000 ÷ 500 = 30
U
Seconds (s)

Questions

Q14
A 40 W fan runs for 5 s. How much energy does it transfer? (2)
Model AnswerE = P × t = 40 × 5 = 200 J.
Q15
A 250 W lamp runs for 2 minutes. Calculate the energy transferred. (3)
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)
Model AnswerE = 3000 × 240 = 720 000 J = 720 kJ.
Q17
A 60 W electric fan runs for 30 minutes. Calculate the energy transferred in kJ. (3)
Model AnswerE = 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)
Model Answert = E ÷ P = 18 000 ÷ 300 = 60 s.
Q19
A battery stores 900 000 J. A 50 W lamp is connected. How many hours will it run? (3)
Model Answert = 900 000 ÷ 50 = 18 000 s; 18 000 ÷ 3600 = 5 hours.
Q20
A 2 kW motor transfers 480 000 J. How long does it take? Give your answer in minutes. (3)
Model Answert = 480 000 ÷ 2000 = 240 s; 240 ÷ 60 = 4 minutes.

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

Two-Step Question [5 marks]

A crane lifts a 150 kg load through a height of 8 m in 12 s. g = 9.8 N/kg.

Exam Questions

Part a
(a) Calculate the gravitational potential energy gained by the load. (2)
Model AnswerE = m × g × h = 150 × 9.8 × 8; E = 11 760 J.
Part b
(b) Using your answer to (a), calculate the power of the crane. (2)
Model AnswerP = E ÷ t = 11 760 ÷ 12; P = 980 W.
Part c
(c) State the unit of power. (1)
Model AnswerWatt (W).

Two-Step Question [5 marks]

A pump raises 60 kg of water through a height of 5 m in 6 s. g = 9.8 N/kg.

Exam Questions

Part a
(a) Calculate the gravitational potential energy gained by the water. (2)
Model AnswerE = 60 × 9.8 × 5; E = 2940 J.
Part b
(b) Calculate the power of the pump. (2)
Model AnswerP = 2940 ÷ 6; P = 490 W.
Part c
(c) State the unit of power. (1)
Model AnswerWatt (W).

Questions

Q21
A 75 kg person climbs stairs of height 4 m in 8 s. g = 9.8 N/kg. Calculate the power. (3)
Model AnswerE = 75 × 9.8 × 4 = 2940 J; P = 2940 ÷ 8 = 367.5 W.
Q22
A 90 kg weightlifter raises a bar 1.8 m in 2 s. g = 9.8 N/kg. Calculate the power. (3)
Model AnswerE = 90 × 9.8 × 1.8 = 1587.6 J; P = 1587.6 ÷ 2 ≈ 794 W.
Q23
A motor transfers 540 000 J in 15 minutes. Calculate its power in W and in kW. (3)
Model Answert = 15 × 60 = 900 s; P = 540 000 ÷ 900 = 600 W = 0.6 kW.

Exam Question [6 marks]

A crane lifts a 400 kg steel beam through a height of 15 m. g = 9.8 N/kg.

Exam Questions

Part a
(a) Calculate the gravitational potential energy gained by the steel beam. (2)
Model AnswerE = mgh = 400 × 9.8 × 15 = 58 800 J.
Part b
(b) The crane takes 40 seconds to lift the beam. Calculate the power of the crane. (2)
Model AnswerP = E ÷ t = 58 800 ÷ 40 = 1470 W.
Part c
(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)
Model AnswerP = E ÷ t = 58 800 ÷ 25 = 2352 W.

Energy Stores & Transfers

Do Now

Q1
Name four energy stores.
Model Answer Any four: kinetic, gravitational, elastic, thermal, chemical, nuclear, electrostatic, magnetic.
Q2
Name four ways energy can be transferred.
Model Answer Any four: mechanically, electrically, by radiation (light/sound), by heating.
Q3
Describe the energy transfer when a candle burns.
Model Answer Chemical store → thermal + light (by heating and radiation).
Q4
What is meant by the "rate" of energy transfer?
Model Answer How quickly energy is transferred (power in Watts).

Part 1 · 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.

Questions

Q1
Name the eight energy stores. (4)
Model Answer Kinetic, gravitational, elastic, thermal, chemical, nuclear, electrostatic, magnetic.
Q2
What energy store does a charged capacitor have? (1)
Model Answer Electrostatic.
Q3
Describe the energy stores at each stage as an archer fires an arrow and it lands in a target. (4)
Model Answer Pulling 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)
Model Answer Conservation of energy — energy is never created or destroyed.

Part 2 · 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

Q1
Name the four mechanisms by which energy can be transferred. (2)
Model Answer Mechanical working, electrical working, heating, radiation.
Q2
A boy falls on a trampoline. Describe the energy transfers. (4)
Model Answer GPE → KE (falling) → elastic PE (trampoline deforms) → KE (bouncing) + thermal + sound.
Q3
A jack-in-the-box pops open. Describe the energy transfers. (3)
Model Answer Chemical (spring compressed) → elastic PE → KE + sound (when released).
Q4
A weight hangs on a spring and oscillates. Describe the energy transfers. (3)
Model Answer KE ↔ EPE continually, with some thermal dissipated each cycle.

Part 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 output ÷ total input.

Questions

Q1
What does the width of an arrow in a Sankey diagram represent? (1)
Model Answer The amount of energy being transferred.
Q2
Which direction does wasted energy usually point in a Sankey diagram? (1)
Model Answer Downward.
Q3
A light bulb uses 100 J and emits 10 J as light. Sketch/describe its Sankey diagram. (3)
Model Answer 100 J input → 10 J light (forward, narrow arrow) + 90 J thermal (down, wide arrow).
Q4
Describe the energy transfers in a bungee jumper from jump to lowest point. (4)
Model Answer GPE → KE (falling freely) → elastic PE + KE (cord stretching) → elastic PE only (at lowest point).

Exam Question — Bungee Jump Transfers (7 marks)

A student jumps off a bridge on a bungee cord. The cord has an unstretched length of 20 m.

Exam Questions

Part 1
(a) Give two reasons why it is important that the cord is appropriate for the student's weight. (2)
Model Answer Too weak: won't stop the student in time / hits ground. Too stiff: gives dangerous deceleration.
Part 2
(b) The student falls. Describe the energy transfers before the cord starts to stretch. (2)
Model Answer GPE → KE as student falls (before cord tenses).
Part 3
(c) When the cord is stretched, state the energy store in the cord. (1)
Model Answer Elastic potential energy.
Part 4
(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)
Model Answer 29 400 − 18 000 = 11 400 J stored as elastic PE in the cord.

Reducing Wasted Energy

Do Now

Q1
What is dissipation?
Model Answer Energy transferred to the surroundings in a less useful form.
Q2
Give two examples of wasted energy in a car engine.
Model Answer Thermal energy from friction; sound from the engine.
Q3
What is thermal conductivity?
Model Answer How quickly a material allows thermal energy to pass through it.
Q4
Why does a metal spoon feel colder than a wooden spoon at the same temperature?
Model Answer Metal has higher thermal conductivity so it conducts heat away from your hand faster.

Part 1 · 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.

Questions

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

Part 2 · 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

Q1
Explain how lubrication reduces energy waste in engines. (2)
Model Answer Oil reduces friction between metal surfaces, so less KE is converted to thermal energy.
Q2
Why are racing cars designed to be streamlined? (2)
Model Answer Streamlining reduces air resistance, so less KE is converted to thermal energy at high speed.
Q3
How does insulating a hot water pipe reduce energy waste? (2)
Model Answer Insulation reduces conduction from the hot pipe to cooler surroundings.
Q4
A house loses 20% of heat through the roof. Suggest one way to reduce this. (1)
Model Answer Loft insulation / fibre glass in the roof space.

Part 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.

Questions

Q1
Why does a copper hot water tank lose heat quickly? (1)
Model Answer Copper has high thermal conductivity so conducts heat to surroundings rapidly.
Q2
How does a foam jacket reduce heat loss from the tank? (2)
Model Answer Foam has low thermal conductivity so slows conduction from tank to air.
Q3
Compared to an uninsulated tank, how does the rate of energy transfer change? (1)
Model Answer The rate of energy transfer is reduced (slower cooling).
Q4
Explain why insulating a hot water tank saves money. (2)
Model Answer Water stays hot longer → immersion heater needed less often → lower electricity bill.

Exam Question — Hot Water Tank (5 marks)

A copper hot water tank with insulation is shown.

Exam Questions

Part 1
(a) Copper has higher thermal conductivity than most metals. How does the rate of energy transfer compare to most metals? Tick one: Higher / Lower / The same. (1)
Model Answer Higher.
Part 2
(b) The tank is insulated. The water is heated, then the heater switches off. Compare the rate of cooling with and without insulation. (2)
Model Answer With insulation, the rate of energy transfer to the room is reduced so the water stays hotter for longer.
Part 3
(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)
Model Answer 4 030 000 ÷ 4 070 000 = 0.990 (99.0%).

Efficiency

Do Now

Q1
What is meant by "wasted" energy?
Model Answer Energy transferred to the surroundings in a less useful form.
Q2
Give two examples of useful energy outputs from a car engine.
Model Answer Kinetic energy (motion) and possibly electrical energy (alternator).
Q3
What is the formula for power?
Model Answer P = E ÷ t.
Q4
A 500 W machine runs for 20 s. How much energy does it transfer?
Model Answer E = 500 × 20 = 10 000 J.

Part 1 · What is Efficiency?

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

Efficiency = useful output energy ÷ total input energy
Efficiency = useful output power ÷ total 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.

Questions

Q1
What does the efficiency of an appliance tell us? (2)
Model Answer What fraction of the total input energy is transferred to a useful output.
Q2
Write both equations for efficiency. (2)
Model Answer Eff = useful output E ÷ total input E; Eff = useful output P ÷ total input P.
Q3
Why would a more efficient motor give a car a higher top speed? (2)
Model Answer More 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)
Model Answer Eff = 325 ÷ 400 = 0.81 = 81%.
Q5
A car uses 3890 J of fuel to generate 2650 J of KE. Find its efficiency. (3)
Model Answer Eff = 2650 ÷ 3890 = 0.68 = 68%.

Part 2 · Calculating and Applying Efficiency

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

Q1
A lightbulb uses 470 J; 180 J is heat and 290 J is light. Find the efficiency. (3)
Model Answer Eff = 290 ÷ 470 = 0.617.
Q2
A power station uses 2000 J of fuel to generate 600 J of electricity. How much energy is wasted? (2)
Model Answer 2000 − 600 = 1400 J wasted.
Q3
An electric motor uses 2.4 × 10⁵ J to give a weight 1.8 × 10⁴ J of GPE. Find efficiency. (3)
Model Answer Eff = 1.8 × 10⁴ ÷ 2.4 × 10⁵ = 0.075 = 7.5%.
Q4
A lightbulb (efficiency 0.4) emits 3000 J of light in 20 s. Find the power input. (3)
Model Answer Useful power = 3000 ÷ 20 = 150 W. Input P = 150 ÷ 0.4 = 375 W.

Part 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

Q1
Convert efficiency 0.82 to a percentage. (1)
Model Answer 82%.
Q2
A power station uses 2000 J to produce 600 J electricity. For every 600 J of electricity, how much fuel is used? (2)
Model Answer 2000 J of fuel needed for every 600 J of electricity.
Q3
State two reasons why improving efficiency is desirable. (2)
Model Answer Saves money; reduces fuel consumption and environmental impact.
Q4
A fan uses 180 W and delivers 150 W of useful power to the air. Calculate its efficiency. (3)
Model Answer Eff = 150 ÷ 180 = 0.833 = 83.3%.

Exam Question — Power Station Efficiency (6 marks)

A fuel burning power station uses 2000 J of fuel energy to generate 600 J of electrical energy.

Exam Questions

Part 1
(a) Calculate the efficiency of the power station. (2)
Model Answer Eff = 600 ÷ 2000 = 0.3 = 30%.
Part 2
(b) State where the rest of the energy goes. (1)
Model Answer Thermal energy transferred to the surroundings (wasted heat).
Part 3
(c) A second power station has efficiency 0.70. For every 1000 J of fuel, how much electricity does it produce? (2)
Model Answer 0.70 × 1000 = 700 J.
Part 4
(d) Give one way the second power station could increase its efficiency. (1)
Model Answer Use CHP to capture waste steam heat; reduce friction in turbines; better insulation.

Non-Renewable Energy Resources

Do Now

Q1
What is meant by a "non-renewable" energy resource?
Model Answer A resource that cannot be replenished at the rate at which it is used.
Q2
Name three fossil fuels.
Model Answer Coal, oil, natural gas.
Q3
What gas is released when fossil fuels are burned?
Model Answer Carbon dioxide (CO₂) — a greenhouse gas.
Q4
What is nuclear fission?
Model Answer The splitting of a heavy atomic nucleus (e.g. uranium) releasing large amounts of energy.

Part 1 · 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 → 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)
Model Answer Electricity generation, heating, transport.
Q2
Define "non-renewable". (2)
Model Answer A 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)
Model Answer Coal, oil, natural gas (and nuclear counts as non-renewable).
Q4
What are the benefits of fossil fuels? (2)
Model Answer Reliable, high energy density, existing infrastructure, controllable output.
Q5
When are fossil fuels likely to run out? (1)
Model Answer Approximately 50 years.
Q6
Name two things formed when fossil fuels are burned and explain the problems they cause. (4)
Model Answer CO₂ → greenhouse effect / climate change; SO₂ (from coal) → acid rain / respiratory problems.

Part 2 · 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

Q1
Explain why nuclear energy is non-renewable. (2)
Model Answer Uranium and plutonium are finite resources; they will eventually run out.
Q2
What are the risks with using nuclear power? (3)
Model Answer Radioactive waste is toxic and hard to dispose of; risk of accidental contamination of surroundings.
Q3
What are the benefits of nuclear power compared to fossil fuels? (3)
Model Answer No CO₂ produced; much longer-lasting fuel; much higher energy density per kg.
Q4
What is nuclear fusion and why is it not yet used commercially? (3)
Model Answer Fusion joins light nuclei releasing enormous energy; not yet commercially viable because containing the plasma requires more energy than it produces currently.

Part 3 · UK Energy Mix

The UK uses a mix of energy resources. In 2018, gas and nuclear provided the majority of electricity.

The UK government plans to phase out coal-fired power stations 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 Mix (8 marks)

Figures show how different energy resources generated electricity in the UK on one day in June 2018, and how electricity demand varies with time of day.

Exam Questions

Part 1
(a) The UK government plans to stop using coal-fired power stations by 2025. Explain one environmental problem caused when electricity is generated by burning coal. (2)
Model Answer CO₂ is a greenhouse gas contributing to climate change; SO₂ causes acid rain / respiratory problems.
Part 2
(b) Use Figure 11.1 to determine the percentage of electricity generated by nuclear power that day. (2)
Model Answer Nuclear ≈ 22% (approximately; from the pie chart, the hatched sector).
Part 3
(c) Figure 11.2 shows electricity demand varies with time of day. Identify the maximum and minimum demand and calculate the difference. (2)
Model Answer Max ≈ 37 500 MW; min ≈ 22 000 MW; difference ≈ 15 500 MW.
Part 4
(d) Explain why non-renewable sources are used to meet peak electricity demand. (2)
Model Answer Non-renewables can be turned up quickly to match demand; renewables depend on weather.

Renewable Energy Resources

Do Now

Q1
What is a renewable energy resource?
Model Answer One that is naturally replenished and will not run out.
Q2
Name three renewable energy resources.
Model Answer Any three: solar, wind, hydroelectric, tidal, wave, geothermal, biofuels.
Q3
What is one disadvantage of wind power?
Model Answer Only works when wind is blowing; unreliable.
Q4
Approximately how long will fossil fuels last at current usage?
Model Answer Approximately 50 years.

Part 1 · 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.

Questions

Q1
Explain how wind power generates electricity. (3)
Model Answer Wind turns turbine blades attached to a generator; rotation of the generator produces electricity.
Q2
Give one advantage and one disadvantage of solar power. (2)
Model Answer Adv: renewable / no emissions. Disadv: only works in sunlight / weather-dependent.
Q3
Why is geothermal power only available in certain locations? (2)
Model Answer It 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)
Model Answer E = 2 × 10⁶ × 3600 = 7.2 × 10⁹ J = 7200 MJ.

Part 2 · 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.

Questions

Q1
Describe how a hydroelectric power station generates electricity. (3)
Model Answer Water flows from reservoir through penstocks → turbines → generator → electricity.
Q2
Give one advantage of tidal power over wind power. (1)
Model Answer Tides are predictable (unlike wind), so output can be planned in advance.
Q3
Why might wave power be unreliable? (2)
Model Answer Wave height varies with weather; storms can damage equipment.
Q4
What environmental problem is associated with hydroelectric dams? (2)
Model Answer Damming floods land / habitat; disrupts river ecosystems and fish migration.

Part 3 · Biofuels & Global Energy Trends

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

Biofuels are theoretically carbon-neutral (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.

Questions

Q1
Why are biofuels considered renewable? (2)
Model Answer They come from living organisms that can be regrown / replenished.
Q2
Give one reason why biofuels are not completely carbon-neutral in practice. (2)
Model Answer Land-use changes (deforestation) and transport / processing release extra CO₂.
Q3
From Fig 12.4, when did global energy use begin to increase most rapidly? (1)
Model Answer After approximately 1950.
Q4
What was the main cause of the rapid energy use increase after 1950? (2)
Model Answer Industrialisation and population growth dramatically increased energy demand.
Q5
How has the use of renewables compared to non-renewables changed? Give two differences. (4)
Model Answer Non-renewables still far larger in total; renewables growing faster proportionally; non-renewables dominated for most of history.

Exam Question — Electric vs Diesel Cars (6 marks)

An electric car has a motor powered by a battery. A diesel car has an engine powered by diesel fuel.

Exam Questions

Part 1
(a) State one way the electric car is better for the environment. (1)
Model Answer The electric car produces no direct CO₂ / pollution emissions during use.
Part 2
(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)
Model Answer Overall efficiency = 0.35 × 0.90 = 0.315 = 31.5%.
Part 3
(c) Give one reason why renewable energy sources are important for the future. (2)
Model Answer Fossil fuels will run out / to reduce CO₂ emissions and combat climate change.
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