The efficiency of an ideal heat engine is 30%. The gas received 6 kJ of heat from the heater.

The efficiency of an ideal heat engine is 30%. The gas received 6 kJ of heat from the heater. How much heat is transferred to the cooler?

Since the efficiency is 30%, it took the following to complete useful work:

6 * 30/100 = 1.8 kJ.

Therefore, the cooler is given:

6 – 1.8 = 4.2 kJ.



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