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A kettle is found to output 65J when its input energy is 100J. What is the efficiency of the kettle, and what happens to the rest of the energy?

Efficiency=useful energy /input energy x 100, so for the kettle, efficiency = 65/100 x 100 = 65% efficient. The rest of the energy is wasted as heat energy in the kettle/sound energy.

Answered by Joshua L. • Physics tutor

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A kettle is found to output 65J when its input energy is 100J. What is the efficiency of the kettle, and what happens to the rest of the energy?

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