Energy & Efficiency
The forms energy takes, how it is transferred in engineered systems, and the calculations the exam asks for most.
Forms of energy & energy systems Exam
Energy is the ability to do work or cause change. Engineers must recognise the common forms and follow how energy changes form as a system works.
In the exam Naming forms of energy, and identifying useful and wasted energy, is tested most years. 2024 Q7(b), Q9(c) · 2025 Q10(d)
Measured in joules (J), the common forms are: kinetic (moving), potential (raised up), electrical (a current), heat (hot objects), chemical (fuels & batteries), light and sound. As a system works, energy changes from one form to another (a motor: electrical → kinetic + heat).
Input, useful and wasted energy
Almost every engineered product moves, heats, lights or sounds something — and all of these need energy. Engineers track what goes in, what useful energy comes out, and how much is wasted.
Input energy is the total supplied to a system. The useful output does the job it was designed for; the rest is wasted, usually as heat or sound. Efficiency measures how much of the input becomes useful output.
1. Identify the useful output energy from an electric kettle. 1 mark
Answer
2. Identify one form of wasted energy from an electric motor. 1 mark
Answer
3. A hairdryer takes in electrical energy. Describe two forms of energy that come out. 2 marks
Answer
4. Explain why no real system gives out only useful energy. 2 marks
Answer
Conservation of energy & energy transfers Exam
Energy cannot be created or destroyed. It can only be transferred from place to place or transformed from one form to another. So the total energy in a system always stays the same.
In the exam The conservation idea underpins the energy-audit and efficiency questions. every year
Block diagrams
Engineers draw a block diagram to show how energy changes form — the input form on the left, the device in the middle, the useful output and the wasted output on the right.
1. State the law of conservation of energy. 1 mark
Answer
2. An electric motor is supplied with 1200 J. It produces 900 J of useful kinetic energy. Calculate the wasted energy. 2 marks
Answer
3. A lift motor takes in 5000 J and does 3800 J of useful work raising the cage. State the wasted energy and one form it takes. 2 marks
Answer
Work done Exam
The exam gives you the data booklet with this relationship. Your marks come from showing the working: substitute the numbers, rearrange if needed, then give the answer with a unit, rounded to 2 significant figures.
In the exam A direct work-done question does come up — a workbench pushed 12 m by a 2200 N force (2 marks). 2022 Q3
Worked example — find the work done
Worked example — find the force
1. A pump pushes water with a force of 220 N. The water moves 6 m. Calculate the work done. 3 marks
Answer
Kinetic & potential energy Exam
Both formulae are in the data booklet. Remember to square the speed for kinetic energy, and use g = 9.8 ms−2 for potential energy — that is how the data booklet writes it, and it means the same as the 9.8 N/kg you meet in Physics.
In the exam Kinetic and potential energy calculations appear every year. Ek: 2021 Q11(e) · 2022 Q13(c) · 2024 Q13(b) | Ep: SQP Q17(b)(i) · 2023 Q11(b) · 2025 Q10(c)
Kinetic energy — Ek = ½ m v2
Worked example — find Ek
Worked example — find the speed
Potential energy — Ep = m g h
Worked example — find Ep
Worked example — find the mass
1. A delivery vehicle of mass 1800 kg travels at 12 m/s. Calculate the kinetic energy. 3 marks
Answer
2. A goods lift carries a 320 kg load up a height of 12 m. Calculate the gain in potential energy. (g = 9.8 ms−2) 3 marks
Answer
Electrical & heat energy Exam
Both formulae are in the data booklet. Two rules the markers always check: convert time to seconds first (×60 for minutes); and use ΔT (the temperature change), not the final temperature. The booklet lists water as c = 4180 J kg−1K−1 — a rise of 1 K is a rise of 1 °C, so J/kg°C is the same number.
In the exam Electrical and heat energy calculations appear every year. Ee: 2021 Q10(b)(i) · 2024 Q12(b) | Eh: 2021 Q10(a) · 2023 Q12(e) · 2025 Q12(b)
Electrical energy — Ee = V I t
Worked example — find Ee (time in minutes)
Worked example — find the current
Heat energy — Eh = c m ΔT
Worked example — find Eh (work out ΔT first)
Worked example — find the temperature rise
1. A workshop motor runs at 240 V with a current of 5 A for 8 minutes. Calculate the electrical energy supplied. 4 marks
Answer
2. An industrial heater warms 25 kg of water from 18 °C to 80 °C. Calculate the heat energy required. (c = 4180 J kg−1K−1) 4 marks
Answer
Power & choosing the right formula Exam
The exam gives you the data booklet with all six energy and power relationships — Ew, Ek, Ep, Ee, Eh and P. Your job is to spot which one fits the numbers you've been given — match the symbols in the question to the symbols in the formula.
In the exam Power calculations, and rearranging Ee = V I t for time, are tested every year. SQP Q17(b)(ii)
Worked example — find the power
Worked example — find the energy used
Which formula? Match the symbols in the question
Every one of these is printed in your data booklet. The skill is reading the question for the quantities you have been given, then finding the relationship that uses those symbols.
| If the question gives you… | Use | The thing that costs marks |
|---|---|---|
| a force (N) and a distance (m) | Ew = F d | d is the distance moved in the direction of the force |
| a mass (kg) and a speed (m/s) | Ek = ½ m v2 | square the speed before you multiply — not at the end |
| a mass (kg) and a height (m) | Ep = m g h | g = 9.8 ms−2; h is the height gained |
| a voltage (V), a current (A) and a time | Ee = V I t | time in seconds — minutes × 60 first |
| a mass (kg), a specific heat capacity and two temperatures | Eh = c m ΔT | subtract to get ΔT — never substitute the final temperature |
| an energy (J) and a time, or a power (W) | P = Et | watts need joules and seconds; 1.5 kW = 1500 W |
| an input and a useful output (energy or power) | η = EoutEin | × 100 for a percentage — the booklet only gives you the ratio |
1. A 1.5 kW heater is switched on for 4 minutes. Calculate the energy supplied. Give your answer in kJ. 4 marks
Answer
Efficiency Exam Assignment
Efficiency measures how much of the input energy becomes useful output energy. A more efficient system wastes less. No real system is 100% efficient.
In the exam Efficiency is examined every year. 2021 Q10(b)(ii) · 2022 Q14(d) · SQP Q11(b) · 2025 Q10(d)(i)
Worked example — find the percentage efficiency
Worked example — find the useful energy
1. Describe what is meant by efficiency. 1 mark
Answer
2. Explain why no real engineering system can be 100% efficient. 2 marks
Answer
3. A motor uses 1500 J of electrical energy and produces 1050 J of useful kinetic energy. Calculate the percentage efficiency. 3 marks
Answer
4. A lighting system uses 240 J of electrical energy and is 25% efficient. Calculate the useful light energy produced. 3 marks
Answer
5. A factory uses an old motor that wastes a lot of energy as heat. Suggest one engineering change that would reduce the waste and explain how it works. 2 marks
Answer
Energy audits Exam
An energy audit shows where the input energy goes: the total in, the useful out and the wasted energy. It is just the conservation rule drawn as a diagram, with real numbers.
In the exam Completing an energy audit diagram is tested directly — the lift: 44 kJ in, 32 kJ useful Ep, find the losses (3 marks). 2022 Q11(c)
1. A lift motor is supplied with 44 kJ of electrical energy. The lift cage gains 32 kJ of potential energy. Calculate the energy wasted, and state one form it takes. 3 marks
Answer
2. A pump is supplied with 1500 J of electrical energy and is 60% efficient. Complete the audit: state the useful energy and the wasted energy. 3 marks
Answer
3. An industrial heater takes in 50 000 J and wastes 5000 J. Calculate its percentage efficiency. 3 marks
Answer
Engineering decision task Exam Assignment
Engineers decide with numbers. Calculate the efficiency for each option, then back your recommendation with the calculation and another reason — cost, lifetime or environmental impact.
In the exam Compare-and-recommend, backed by an efficiency calculation, is a common extended question. most years
A common exam task gives you two options and asks you to recommend one. Calculate each efficiency, then back your choice with the numbers and another reason (cost, lifetime, environment).
Worked example — choose a motor
Motor A: η = (700 ÷ 1000) × 100 = 70%.
Motor B: η = (640 ÷ 800) × 100 = 80%.
Recommendation: choose Motor B. It is more efficient (80% vs 70%), so it wastes less energy and costs less to run; it also lasts twice as long, so it needs replacing less often.
1. A workshop needs a new compressor. Compressor A is supplied with 2000 W and delivers 1400 W of useful power. Compressor B is supplied with 1600 W and delivers 1200 W. B costs more to buy but lasts twice as long. Recommend one, supporting your choice with a calculation and one other reason. 4 marks
Answer
Booklet answer keys
Final answers to the TRY THIS tasks in your printed Topic 2 booklet — work each one out first, then open the matching key. The booklet's exam-style Practice sets now sit with the lesson they belong to, further up this page.
Task — Match the energy form p.3
- A car driving along a road — kinetic
- A book held high above the floor — potential
- Current flowing through a heater element — electrical
- A hot mug of tea — heat
- Petrol in a fuel tank — chemical
- Light shining from an LED — light
- A buzzer sounding in a circuit — sound
Task — Identify input, useful and wasted energy p.4
- Electric heater: input electrical → useful heat → wasted: light/sound.
- Electric motor: input electrical → useful kinetic (movement) → wasted: heat and sound.
- Battery torch: input chemical → useful light → wasted: heat.
- Filament light bulb: input electrical → useful light → wasted: heat.
- Petrol car engine: input chemical → useful kinetic → wasted: heat and sound.
- Hairdryer: input electrical → useful heat and kinetic → wasted: sound.
Task — Energy transfer diagrams p.5
- Electric motor: electrical → kinetic (+ heat, sound wasted).
- Generator: kinetic → electrical (+ heat wasted).
- Battery-powered torch: chemical → electrical → light (+ heat wasted).
- Hairdryer: electrical → heat + kinetic (+ sound wasted).
Task — Account for the energy p.15
wasted = input − useful:
- Electric motor: 1200 − 900 = 300 J
- Lift cage: 5000 − 3800 = 1200 J
- Light bulb: 60 − 12 = 48 J
- Electric kettle: 200 000 − 168 000 = 32 000 J
- Hairdryer: 8000 − 1500 = 6500 J
Task — Complete the audit p.17
wasted = input − useful; efficiency = (useful ÷ input) × 100:
- Hairdryer: input 8000, useful 1500 → wasted 6500 J, efficiency 18.75% ≈ 19%.
- LED bulb: input 10, useful 8 → wasted 2 J, efficiency 80%.
- Industrial heater: input 50 000, wasted 5000 → useful 45 000 J, efficiency 90%.
- Crane motor: input 20 000, wasted 4000 → useful 16 000 J, efficiency 80%.
- Pump: efficiency 60% → useful = 0.60 × 1500 = 900 J, wasted 600 J.
Task — Choose a new motor p.18
- Motor A: (700 ÷ 1000) × 100 = 70%.
- Motor B: (640 ÷ 800) × 100 = 80%.
- Motor B is more efficient. Economic reason for B: cheaper to run and lasts twice as long, so fewer replacements. Environmental reason: wastes less energy, so less electricity has to be generated.
- Recommendation: Motor B — higher efficiency plus the longer lifetime outweighs its higher purchase price.
Everything on one screen
The six relationships below are all printed in your data booklet — you do not have to memorise them. What you do have to remember is the working: substitute, rearrange if you need to, and finish with a unit.
The six things that lose marks
- No substitution line. Write the relationship, then the relationship with your numbers in, then the answer. Two of the three marks are usually for those first two lines.
- Time left in minutes. Ee = V I t and P = E ÷ t both need seconds: × 60 first.
- The final temperature instead of ΔT. Heated from 18 °C to 80 °C, subtract: ΔT = 80 − 18 = 62 °C, not 80.
- The speed not squared. In Ek = ½ m v2, square v before multiplying.
- The × 100 forgotten. An efficiency of 0.75 is not an answer to "calculate the percentage efficiency".
- No unit. J, W, N, %, kg — an answer without one is an answer short of a mark.
| You are given | Substitute | Because |
|---|---|---|
| 5 minutes | 300 s | × 60 — the joule is defined with seconds |
| 1.5 kW | 1500 W | kilo = × 1000 |
| 44 kJ | 44 000 J | kilo = × 1000 |
| heated from 18 °C to 80 °C | ΔT = 62 °C | the change, so subtract |
| 85% efficient | η = 0.85 | ÷ 100 to use the ratio form |
Check yourself
Where this comes up next
Sources & credits: The Topic 2 booklet © R Stewart, 2026. The Past Paper Finder is compiled by Mr McDonald, 2024; past-paper questions © Qualifications Scotland (SQA). The N4/N5 data booklet and the Unit 1 test slides' past-paper material are reproduced for educational use, © Qualifications Scotland (SQA).