alleryStart learning →A heat engine takes in 6.0 kJ of heat and produces 4.8 kJ of waste heat. What is its efficiency?
A20 percent
B25 percent
C48 percent
D80 percent
Answer & Solution
Correct answer: A. 20 percent
1. Work output is W = Qh minus Qc = 6.0 kJ minus 4.8 kJ = 1.2 kJ.
2. Efficiency is Eff = W divided by Qh = 1.2 divided by 6.0.
3. Dividing gives Eff = 0.20, or 20 percent.
4. 80 percent is the fraction of heat wasted, not the efficiency, and 48 or 25 percent do not match this division.
5. So 20 percent is correct.
_Source: OpenStax Physics (CC BY 4.0), Ch 12 "Thermodynamics", section 12.4 Applications of Thermodynamics: Heat Engines, Heat Pumps, and Refrigerators_
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