What is the efficiency (in percent) of a cyclical heat engine in which 25.0 kJ of heat transfer occurs to the environment for every 40.0 kJ of heat transfer into the engine?

62.5%

Explanation:

Efficiency of a machine is expressed as;

Efficiency = Work output/Work input * 100%

work input is the quantity of heat transferred to the engine

Work output is the heat transferred to the surrounding.

Given

Work input = 40.0kJ

Work output = 25.0kJ

Get the efficiency of the engine;

Efficiency = 25/40 * 100

Efficiency = 5/8 * 100

Efficiency = 500/8

Efficiency = 62.5%

Hence the efficiency (in percent) of the cyclical heat engine is 62.5%

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