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4.1.3Energy Resources and Energy Transfers

Efficiency

5 min read 5 parts Lesson 4 of 15 1 quiz

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Section 1 of 5Defining Efficiency

4.1.3 Efficiency

Efficiency tells us what fraction of the energy put into a device becomes useful output.

Efficiency
  • The proportion of the total energy input that is converted into useful energy output.
Percentage
  • Efficiency is found as a fraction and then multiplied by 100 to give a percentage.
Maximum value
  • Efficiency can never be greater than 100%, because no device can create energy.
Ideal device
  • A device with 100% efficiency would transfer all of its energy input into useful output with no waste.

Section 2 of 5Improving Efficiency

Engineers improve efficiency by reducing the energy wasted in each process.

Improving Efficiency
Insulation
  • Reduces the energy wasted by conduction and convection from a hot object to its surroundings.
Lubrication
  • Reduces friction between moving parts, so less kinetic energy is transferred to the thermal store.
Streamlining
  • Reduces air resistance on moving vehicles, so less energy is wasted by heating the air.
Better components
  • Using more efficient motors or bulbs increases the proportion of input energy that becomes useful output.

Section 3 of 5Calculating Efficiency

The same method works for energy values in joules or power values in watts.

1
Write the equation as useful energy output divided by total energy input, multiplied by 100 percent.
2
Work out the useful energy output first if only the wasted energy is given, using useful = total input − wasted.
3
Substitute the values in the same units, then divide and multiply by 100 to convert the answer into a percentage.
4
Check the answer lies between 0% and 100%, because a larger value shows an error in the calculation.

Section 4 of 5Worked Example

Question

  • A device takes in 300 J of energy and wastes 85 J.
  • Calculate its efficiency.

  1. Useful energy output = total energy input − wasted energy = 300 J − 85 J = 215 J.
  2. Efficiency = useful energy output ÷ total energy input × 100%.
  3. Efficiency = 215 J ÷ 300 J × 100% = 71.7%.

Answer The efficiency is 71.7%, which is about 72% to two significant figures.

Section 5 of 5Exam Checklist & Tips

4.1.3 – Key Points for the Exam

  • Efficiency Useful energy output divided by total energy input, multiplied by 100 percent.
  • Useful output The energy or power that does the job the device is intended to do.
  • Insulation A material with low thermal conductivity used to slow the transfer of energy by heating.
  • Friction A force between surfaces that transfers kinetic energy to the thermal store.

Common Mistakes

  • Giving an efficiency above 100%, which shows that the useful output has been calculated incorrectly.
  • Dividing the wasted energy by the total input instead of dividing the useful energy output by the total input.
  • Forgetting to multiply by 100 when the question asks for a percentage.
  • Treating efficiency as having a unit; it is a ratio, so it is given as a percentage with no unit.
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