An oven consumes 425 kWh of energy in order to provide 386 kWh of useful energy. What is its percent efficiency?

asked Sep 15, 2023 83.0k views

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Answer:

Work Input = 425kWh

Work Output = 386 kWh

Efficiency =

$(work \: output)/(work \: input) \\ = (386 \: kwh)/(425 \: kwh) = 0.90$

Efficiency % = 0.90 × 100 = 90%

Olson answered Sep 20, 2023

by Olson

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