A petrol-powered go-kart engine transfers 12 000 J of chemical energy every second. Only 3 600 J per second is transferred as useful kinetic energy to move the kart forward. Explain why the go-kart engine cannot be 100% efficient, and describe what happens to the energy that is not usefully transferred.
Written & reviewed by James Millett — Biology (Imperial College London), PGCE Science (University of Cambridge).
A petrol-powered go-kart engine transfers 12 000 J of chemical energy every second. Only 3 600 J per second is transferred as useful kinetic energy to move the kart forward.
Model answer (5 marks)
The engine cannot be 100 % efficient because energy is conserved – all 12 000 J of chemical energy must be accounted for. In practice some of this energy is lost as heat through friction in the engine and moving parts, and is transferred to the surroundings. Therefore only 3 600 J can be converted into useful kinetic energy, giving an efficiency of 3 600/12 000 = 0.30 or 30 %. The remaining 70 % of the input energy is dissipated as thermal energy, spreading into the surroundings and becoming unavailable for useful work.
Examiner tips
- Show the conservation of energy first, then explain heat loss, calculate efficiency, and state the fate of wasted energy
Common mistakes
- Claiming the engine is 100 % efficient, ignoring heat loss; mis‑calculating efficiency (e.g. 3 600/12 000 = 0.3 instead of 30 %); not mentioning that the wasted energy becomes heat in the surroundings
Mark scheme (5 marks)
- Energy cannot be created or destroyed (conservation of energy) so all input energy must be accounted for
- Some energy is dissipated / transferred to the surroundings as heat (thermal energy) due to friction in the engine / moving parts
- This means not all input energy is converted to useful kinetic energy, so efficiency must be less than 1 (100%)
- The wasted energy is spread out into the surroundings (dissipation), making it unavailable for further useful energy transfer
- The efficiency of the go-kart engine is 0.3 / 30%, meaning 70% of the input energy is wasted
Key terms in this question
Related
- All Edexcel A-Level Physics (9PH0) revision notes →
- How to answer a "Explain" question →
- Decode the mark scheme abbreviations →
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