Calculate efficiency using useful energy output
Divide the useful share by the full input and keep consistent units, boundaries and intervals.
Section P1.6.4 · Energy resources
Cross-referenced from the related Cambridge IGCSE Chemistry 0620 or Physics 0625 teaching to the 0652 syllabus. Exact 0652 Core/Supplement objective coverage and this final-series course’s examiner fit still need specialist review. Written practice is self-checked and does not predict a Cambridge grade or demonstrate supervised laboratory competence.
Before you begin
- Identify the intended useful output.
- Convert kilojoules to joules and use percentages.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which fraction represents 40%?
What you will learn
- Recall efficiency = useful energy output / total energy input.
- Convert between fraction and percentage and find missing amounts.
The denominator is the full input
Efficiency as a fraction is useful energy output divided by total energy input. Multiply by 100% for percentage efficiency. For 200 J input and 120 J useful output, the fraction is 0.60 and efficiency is 60%. The remaining 80 J has other conserved destinations; it is not deleted from the input denominator.
Use matching units and the same event, boundary and interval. A 2.0 kJ input is 2000 J, so 500 J useful output gives fraction 0.25 or 25%. Dividing 500 by 2 without converting units gives a meaningless result. If the account includes an additional energy source or an initial store decrease, include it in the full relevant input rather than claiming a conversion exceeds its complete input.
Rearrange a complete account
Useful output = efficiency fraction × total input. At 40% efficiency and 300 J input, useful output is 120 J and other transfers are 180 J in an exhaustive account. If 150 J useful output is required at 30%, total input is 150/0.30 = 500 J. A percentage value of 30 means fraction 0.30, not thirty times the input.
At the same useful task and consistent full boundary, a larger useful fraction means higher efficiency. It need not mean a larger useful amount if input quantities differ. A reported ordinary conversion value above 100% requires checking units, boundaries, omitted inputs or measurements. Heating-system ratios using additional environmental energy cannot be called this complete-input conversion efficiency by ignoring that energy source.
Sources for this lesson
Pause and explain
What denominator belongs in efficiency for 200 J input with 120 J useful and 80 J other output?
Worked example
A device receives 200 J and supplies 120 J to its stated useful task. Find fraction, percentage and the exhaustive other share.
Show the worked solution
- Fraction = 120/200 = 0.60.
- Percentage = 0.60 × 100% = 60%.
- Other destinations receive 200 − 120 = 80 J.
Answer 0.60 or 60%; other share 80 J.
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Find efficiency and the other share for 200 J input and 120 J useful output.
Give me a hint
Use the whole input and then an exhaustive difference.
Compare my reasoning
- Fraction = 120/200 = 0.60.
- Percentage = 60%.
- Other share = 80 J.
60%; 80 J other output.
Look for these in your work
- Use the full denominator.
- Account for the remaining conserved energy.
Find efficiency for 500 J useful from 2.0 kJ input, and useful output from 300 J input at 40%.
Give me a hint
Convert input units and percentage fractions before calculating.
Compare my reasoning
- 2.0 kJ = 2000 J, so 500/2000 = 25%.
- 40% is fraction 0.40.
- Useful output = 0.40 × 300 = 120 J.
25%; 120 J.
Look for these in your work
- Use matching units.
- Multiply by the fraction, not forty.
A device is claimed to supply 150 J usefully at 30% using only 300 J complete input. Check the numbers and find the input actually required by that efficiency.
Give me a hint
Compare the claimed ratio and the rearranged input.
Compare my reasoning
- 150/300 = 50%, not 30%.
- At 30%, input must be 150/0.30.
- The required input is 500 J; the original combined claim is inconsistent.
The claim gives 50%; 30% would require 500 J input.
Look for these in your work
- Check rather than accept inconsistent labels.
- Keep a complete input boundary.
Common mistakes
- Divide by full input, not wasted share.
- Convert units before the ratio.
- Use percentage as a fraction when rearranging.
What can you explain now?
What input supplies 150 J usefully at 30%?
Compare with the explanation
500 J.
150/0.30 = 500 J for the consistent supplied account.
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Make it stick
Recall energy efficiency, convert units, find missing input or output and identify a boundary or percentage error.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.