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Efficiency and power

Calculate efficiency and compare useful energy transfers.

Start here

The key idea

Efficiency compares the useful output with the total input.

Equation to know

efficiency = useful output / total input

Efficiency And Power
efficiency = useful outputtotal inputpower = energy / time

Use the labels to explain the scientific relationship shown.

Revision notes

The bit that matters

Short notes first. Learn the idea, then use the worked example and questions to check it properly.

1

What efficiency means

Efficiency describes how much of the energy supplied to a device is transferred usefully rather than wasted.No device is 100% efficient because some energy is always dissipated to the surroundings, usually as thermal energy through friction or electrical heating.The more energy transferred usefully, the higher the efficiency.

2

Calculating efficiency

Efficiency = useful output energy transfer / total input energy transfer.You can also use efficiency = useful power output / total power input.The answer is a decimal between 0 and 1, or a percentage when multiplied by 100.Efficiency can never be greater than 1 (100%).

3

Power

Power is the rate of energy transfer, or the rate of doing work.Power = energy transferred / time (P = E / t), and also power = work done / time (P = W / t).Power is measured in watts (W), where 1 watt = 1 joule per second.A more powerful device transfers the same energy in less time.

4

Reducing wasted energy

Wasted energy is usually dissipated to the surroundings as thermal energy.Lubrication reduces friction in moving parts so less energy is wasted by heating.Thermal insulation, such as loft insulation or cavity wall foam, reduces the rate of energy transfer out of a home, raising overall efficiency.

Key terms

Definitions to learn

Efficiency

The proportion of input energy transferred usefully.

Power

The rate of energy transfer or rate of doing work.

Watt

The unit of power, equal to one joule per second.

Dissipated energy

Energy that spreads to the surroundings and is wasted.

Useful energy

Energy transferred to the store the device is designed to fill.

Worked example

A motor receives 500 J and transfers 350 J usefully. Calculate its efficiency.

1

Divide useful output by total input.

2

Convert to a percentage if requested.

Final answer

0.70 or 70%

Exam habit

70) or percentage (70%) — read the question. Show the division clearly: useful ÷ total, not total ÷ useful.

Watch out

Do not divide the input by the useful output.

Examiner tips

How to score full marks

  • 1Efficiency has no units — it is a ratio or percentage.
  • 2Efficiency can never exceed 1 or 100%; if your answer does, you have flipped the fraction.
  • 3Check whether the question wants a decimal or a percentage and convert if needed.
Practice questions

Try these yourself

Start with the core skill, then open the answer only after you have attempted the full question.

1A lamp is 18% efficient and receives 1200 J. Calculate its useful output.
Mark scheme
  1. 1.Convert 18% to 0.18.
  2. 2.Multiply the input by the efficiency.
216 J
2A kettle transfers 168 kJ usefully from a 200 kJ input. Calculate its efficiency.
Mark scheme
  1. 1.Use efficiency = useful / total.
0.84 or 84%
3Explain one way to reduce unwanted energy transfer from a house.
Mark scheme
  1. 1.Name a suitable method.
  2. 2.Link it to reduced thermal transfer.
For example, loft insulation reduces energy transfer by conduction and convection through the roof.
4State the equation linking power, energy transferred and time.[1 mark]
Mark scheme
  1. 1.Power is the rate of energy transfer.
  2. 2.Divide energy by time.
power = energy transferred / time (P = E / t) (1)
5A motor receives 500 J and transfers 350 J usefully. Calculate its efficiency.[2 marks]
Mark scheme
  1. 1.Use efficiency = useful output / total input.
  2. 2.Substitute 350 / 500.
efficiency = 350 / 500 = 0.70 (1), or 70% (1)
6A lamp transfers 2400 J of energy in 60 s. Calculate its power output.[2 marks]
Mark scheme
  1. 1.Use power = energy transferred / time.
  2. 2.Substitute E = 2400 and t = 60.
P = E / t (1) = 2400 / 60 = 40 W (1)
7A kettle has a power of 2000 W and is used for 90 s. Calculate the energy transferred.[2 marks]
Mark scheme
  1. 1.Rearrange P = E / t to E = P x t.
  2. 2.Substitute P = 2000 and t = 90.
E = P x t (1) = 2000 × 90 = 180000 J (1) (180 kJ)
8An electric motor is supplied with 1500 J of energy. It does 900 J of useful work lifting a load, and the rest is wasted. Calculate the efficiency as a percentage and state where the wasted energy goes.[3 marks]
Mark scheme
  1. 1.Useful output = 900 J, total input = 1500 J.
  2. 2.Use efficiency = useful output / total input.
  3. 3.Multiply by 100 for a percentage.
  4. 4.Identify the wasted energy pathway.
efficiency = 900 / 1500 (1) = 0.60 = 60% (1); the wasted 600 J is transferred by heating to the surroundings, mainly due to friction in the moving parts (1)
9A cyclist exerts a force of 120 N over a distance of 50 m in 30 s. Calculate the power output of the cyclist.[3 marks]
Mark scheme
  1. 1.First calculate work done using W = F x d.
  2. 2.Then use power = work done / time.
  3. 3.Substitute values and evaluate.
work done = F x d = 120 × 50 = 6000 J (1); power = W / t = 6000 / 30 (1) = 200 W (1)
10A car engine has a power input of 45 kW and a useful power output of 27 kW. Calculate its efficiency and state the power wasted.[3 marks]
Mark scheme
  1. 1.Convert kW to W if needed (or keep consistent units).
  2. 2.Use efficiency = useful power output / total power input.
  3. 3.Wasted power = total input - useful output.
efficiency = 27 / 45 (1) = 0.60 = 60% (1); wasted power = 45 - 27 = 18 kW (1)
11A student investigates a device and states: 'The device transferred 900 J in 15 s but only 720 J were useful.' Calculate the efficiency and the power wasted.[3 marks]
Mark scheme
  1. 1.efficiency = useful / total = 720 / 900.
  2. 2.Wasted energy = 900 - 720 = 180 J.
  3. 3.Wasted power = wasted energy / time = 180 / 15.
efficiency = 720 / 900 = 0.80 = 80% (1); wasted energy = 180 J; wasted power = 180 / 15 (1) = 12 W (1)
12Explain two methods a building can use to reduce energy wastage and how each method reduces heat transfer.[3 marks]
Mark scheme
  1. 1.Name method 1 and its mechanism.
  2. 2.Name method 2 and its mechanism.
  3. 3.Relate each to reducing rate of thermal energy transfer.
Loft insulation traps air in the fibres, reducing convection and conduction through the roof so less thermal energy is lost to the cold air above (1); double-glazing traps a layer of air between the panes, which is a poor conductor and also reduces convection, so less energy is transferred through the windows (1); both reduce the rate of energy dissipation to the surroundings, lowering energy bills and improving the building's thermal efficiency (1)
13A pump of power 800 W lifts water from a well. In one minute it raises 120 kg of water by 4.0 m. Calculate the useful power output of the pump and its efficiency. Take g = 9.8 N/kg.[3 marks]
Mark scheme
  1. 1.Calculate gravitational potential energy gained by the water: Ep = m x g x h.
  2. 2.Useful power = Ep / time; convert 1 minute to 60 s.
  3. 3.Efficiency = useful power / input power.
Ep = 120 × 9.8 × 4.0 = 4704 J (1); useful power = 4704 / 60 = 78.4 W (1); efficiency = 78.4 / 800 = 0.098 = 9.8% (accept 10%) (1)
14Two lamps are tested: Lamp A has 40 W input and emits 8 W of light; Lamp B has 60 W input and emits 15 W of light. Which lamp is more efficient? Show your working clearly.[3 marks]
Mark scheme
  1. 1.Efficiency A = useful output / total input = 8 / 40.
  2. 2.Efficiency B = 15 / 60.
  3. 3.Compare efficiencies.
Efficiency A = 8 / 40 = 0.20 = 20% (1); Efficiency B = 15 / 60 = 0.25 = 25% (1); Lamp B is more efficient because it converts a greater proportion of its input energy to useful light (1)
15Evaluate the suggestion that a building should use cavity wall insulation AND loft insulation to maximise energy efficiency. Discuss the trade-off between installation cost and long-term energy savings, and explain why reducing dissipation improves efficiency.[4 marks]
Mark scheme
  1. 1.Explain the physics of why each insulation reduces energy transfer.
  2. 2.Refer to the rate of energy dissipation falling.
  3. 3.Consider cost vs saving argument.
  4. 4.Conclude whether the suggestion is sound.
Cavity wall insulation fills the air gap with foam or fibres, reducing convection and conduction between the inner and outer walls; loft insulation reduces conduction and convection through the ceiling (1); both reduce the rate at which thermal energy dissipates to the surroundings, meaning more of the input energy from heating appliances remains as useful thermal energy inside the building, increasing overall efficiency (1); installation has an upfront cost but over years the reduced energy bills save money, and if the payback time is shorter than the lifespan of the insulation the investment is worthwhile (1); using both together addresses the two largest areas of heat loss, so the combined effect is greater than using one alone, making the suggestion well-founded (1)
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