AQA GCSE Combined Science (8464), Higher tier · Physics › Energy › Conservation and dissipation of energy
Practise Efficiency. 14 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
Efficiency tells you how much of the energy supplied to a device is transferred usefully. You need to calculate it from energy or power values, as a decimal or a percentage, and rearrange the equation. Higher tier students must also describe ways to increase efficiency. Most questions are 2 to 3 mark calculations.
Grade by grade
What you need to be able to do, from the first marks up to the top grade.
3
Identify useful and wasted energy transfersE.g. for a lamp, light is the useful output and heating the surroundings is wasted.
4
Calculate efficiency from energy valuesDivide the useful output energy by the total input energy.
5
Give efficiency as a decimal or percentageMultiply the decimal by 100 to get a percentage; efficiency is never more than 1 (100%).
5
Calculate efficiency from power valuesDivide the useful power output by the total power input.
6
Find the wasted or useful energyWasted energy = total input − useful output, so useful output = total input − wasted energy.
7
Rearrange to find the input or outputUseful output = efficiency × total input; total input = useful output ÷ efficiency.
7
Describe ways to increase efficiencyReduce the wasted transfers, e.g. lubricate moving parts, streamline, insulate (Higher tier only).
Notes
What efficiency means
Energy supplied to a device is either transferred usefully (to the store you want) or wasted (dissipated, usually to the thermal store of the surroundings).
total input energy = useful output energy + wasted energy
Efficiency = useful output energy transfer ÷ total input energy transfer
You can also use power: efficiency = useful power output ÷ total power input.
Efficiency has no unit. Give it as a decimal (e.g. 0.35) or as a percentage (× 100, e.g. 35%).
No real device is 100% efficient, because some energy is always dissipated. Efficiency can never be more than 1 (100%).
Using the equation
Both energies (or both powers) must be in the same unit, e.g. both in J or both in kJ.
If you are given a percentage, divide it by 100 before you use it in the equation.
To find the useful output: useful output = efficiency × total input.
To find the input: total input = useful output ÷ efficiency. grade 7+
Increasing efficiency (Higher tier only) grade 7+
Reduce the wasted energy transfers, so more of the input energy is transferred usefully.
Lubricate moving parts to reduce friction.
Streamline vehicles to reduce air resistance.
Thermally insulate devices that heat things (e.g. an oven), so less energy is dissipated to the surroundings.
Use wires with a low resistance, and LED lamps instead of filament lamps, so less energy is wasted by heating.
Cheatsheet
efficiency = useful output energy transfer ÷ total input energy transfer
efficiency = useful power output ÷ total power input
percentage efficiency = efficiency × 100
total input = useful output + wasted
Efficiency has no unit and is never more than 1 (100%)
Calculate the efficiency when you are given the wasted energy
3 marks6
Find the useful energy: total input − wasted energy.
Divide the useful energy by the total input.
Convert to a percentage if the question asks.
Example. A filament lamp is supplied with 500 J of energy. 425 J of this energy is wasted heating the surroundings. Calculate the efficiency of the lamp as a percentage.
Rearrange: total input = useful output ÷ efficiency.
Example. An electric motor is 80% efficient. Its useful power output is 1.2 kW. Calculate the total power input to the motor in watts.
Show the model answer
0.80 = 1200 ÷ total power input (1) total power input = 1200 ÷ 0.80 (1) total power input = 1500 W (1)
Describe how to increase efficiency (Higher tier only)
2 to 4 marks7
Name the wasted energy transfer, e.g. friction heating the moving parts.
Give a change that reduces it, e.g. lubrication.
Explain the effect: less energy is dissipated, so more is transferred usefully.
Example. An electric motor gets hot when it is used. Describe two ways the efficiency of the motor could be increased. Give a reason for each.
Show the model answer
Lubricate the moving parts (bearings) (1), to reduce friction so less energy is dissipated by heating (1). Use wires with a lower resistance in the motor (1), so less energy is dissipated by heating when there is a current in them (1).
Shortcuts and memory tricks
'Useful over total': the smaller number goes on top.
Sense check: if your efficiency is more than 1 (or 100%), you have divided the wrong way round.
Decimal to percentage: multiply by 100, so 0.35 becomes 35%.
An electric heater is almost 100% efficient, because heating the room is its useful output.
Where marks are lost
Dividing the total input by the useful output.
Putting the wasted energy on top instead of the useful energy.
Using one value in J and the other in kJ.
Giving efficiency a unit (such as J), or forgetting the % sign.
Putting a percentage (e.g. 80) into the equation instead of the decimal (0.80).
Exam technique
Check whether the question wants a decimal or a percentage.
Show the subtraction when you have to find the useful energy first; it can earn a mark on its own.
For 'describe how to increase efficiency', link each change to the wasted transfer it reduces.
Quick recall
Cover the answers and test yourself. The app has these as flashcards that come back just before you'd forget them.
Calculate the total energy dissipated from the moment the ball is released until the ball reaches the top of its rise after the 4th bounce.
0.86 J
Write down the equation that links efficiency, useful output energy transfer and total input energy transfer.
efficiency = useful output energy transfer ÷ total input energy transfer
Sample questions
Written for this site in the style of AQA exam questions. They are not taken from real past papers.
Question 1Easy4 marks
An electric motor is used in a toy car.
(a) The motor is supplied with 500 J of energy. 350 J is transferred usefully to the kinetic energy store of the car. Calculate the efficiency of the motor. Use the equation: efficiency = useful output energy transfer ÷ total input energy transfer Give your answer as a decimal.[2]
(b) Calculate the energy wasted by the motor.[1]
(c) What happens to the energy wasted by the motor? Tick (✓) one box.[1]
It is destroyed.
It is dissipated to the surroundings.
It is stored in the battery.
It is transferred to the kinetic energy store of the car.
Show the answer and mark scheme
(a)Answer: 0.70
efficiency = 350 ÷ 500
efficiency = 0.70
(b)Answer: 150 J
150 (J)
(c)Answer: It is dissipated to the surroundings.
Question 2Medium6 marks
An electric motor in a fan has an efficiency of 0.65.
(a) The useful power output of the motor is 390 W. Calculate the total power input to the motor.[3]
(b) Calculate the power wasted by the motor.[1]
(c) The fan runs for 5.0 minutes. Calculate the energy wasted by the motor in this time.[2]
Show the answer and mark scheme
(a)Answer: 600 W
0.65 = 390 ÷ Pin
Pin = 390 ÷ 0.65
Pin = 600 (W)
(b)Answer: 210 W
600 − 390 = 210 (W)
(c)Answer: 63 000 J
E = 210 × 300
E = 63 000 (J)
Question 3Hard8 marks
A factory uses an electric winch to lift crates. The winch lifts a crate of mass 45 kg vertically through 8.0 m. The electrical energy supplied to the winch during the lift is 6000 J. Gravitational field strength = 9.8 N/kg
(a) Calculate the efficiency of the winch. Give your answer as a decimal to 2 significant figures.[3]
(b) Suggest two ways the efficiency of the winch could be increased. Explain how each change increases the efficiency.[4]
(c) The factory manager says that the winch could be improved until it is 100% efficient. Explain why this is not possible.[1]
Show the answer and mark scheme
(a)Answer: 0.59
useful energy = 45 × 9.8 × 8.0 (= 3528 J)
efficiency = 3528 ÷ 6000 (= 0.588)
efficiency = 0.59 given to 2 significant figures
(b)Answer: Lubricate the gears and pulley to reduce friction, so less energy is dissipated by heating; use lower-resistance (thicker) wires in the motor so less energy is dissipated in the wires; use a lighter hook and cable so less energy is transferred to them.
lubricate the moving parts / gears / pulley
reduces friction so less energy is dissipated (to the thermal store of the surroundings)
use wires with a lower resistance / thicker wires in the motor
less energy is dissipated by heating in the wires
use a lighter hook / cable
less energy is (wastefully) transferred to the gravitational potential store of the hook and cable
(c)Answer: There will always be some friction / resistance, so some energy is always dissipated to the surroundings.
there will always be some friction / electrical resistance so some energy is always dissipated (to the surroundings)