A electric kettle is used to heat water. Explain why the useful energy transferred to the water is always less than the total electrical energy input to the kettle.

Eduqas GCSE Physics — 1.6 Distance, speed and acceleration · Explain · 4 marks · View as Markdown

Written & reviewed by James Millett — Biology (Imperial College London), PGCE Science (University of Cambridge).

An electric kettle has a power-rating of 3 kW. When it is used, some energy is transferred usefully to the water and some energy is wasted.

Model answer (4 marks)

The kettle’s power rating of 3 kW means 3 kJ of electrical energy is supplied every second. When the kettle operates, part of this energy is converted into heat that raises the temperature of the water – this is the useful energy. However, some of the electrical energy is also lost as heat to the surrounding air and to the kettle’s outer casing. This wasted heat does not contribute to heating the water.

Because energy cannot be created or destroyed (conservation of energy), the useful energy transferred to the water must be less than the total electrical energy input. Therefore the efficiency, defined as useful energy divided by total energy input, is always less than 1 (or 100 %).

Examiner tips

  • Use the word ‘wasted’ or ‘lost’ to describe heat to the surroundings; mention the kettle’s outer casing. Show the conservation‑of‑energy argument. State efficiency <1. Keep answer to 4 points.

Common mistakes

  • Saying the useful energy equals the total input, ignoring heat loss. Not mentioning the kettle’s outer casing or surrounding air as heat sinks. Using vague terms like ‘some energy’ without specifying heat loss.

Mark scheme (4 marks)

  1. Some energy is transferred to the surroundings / to the air / as heat from the outer casing of the kettle
  2. This is a waste energy transfer / this energy does not directly serve the purpose of the kettle
  3. Conservation of energy means energy cannot be created or destroyed, so the useful output energy transfer cannot exceed the total energy input
  4. Therefore efficiency is less than 1 (or 100%) because useful output energy transfer is less than total energy input

Key terms in this question

useful energy transfer

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