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If a heat engine takes in 4565 kJ and gives up 2955 kJ during one cycle, what is the engine’s efficiency?

Respuesta :

bhuvna789456

The engine efficiency is 64.73 %

Explanation:

Given data

To find the engine’s efficiency we have the formula,

Energy input- 4565 KJ

Energy output - 2955KJ

Efficiency= energy output/ energy input ×100%

 =2955/4565

  =0.6473 ×100

η               =64.73 %

The engine efficiency is 64.73 %

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