A heat engine takes in 6.0 kJ of heat and produces 4.8 kJ of waste heat. What is its efficiency?
- 20 percent
- 25 percent
- 48 percent
- 80 percent
Show answer and 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_