alleryStart learning →A coal power station transfers 3.0 x 10^12 J by heat from burning coal and releases 1.5 x 10^12 J by heat into the environment. What is its efficiency?
A0.33
B0.50
C0.66
D1.00
Answer & Solution
Correct answer: B. 0.50
1. Efficiency is Eff = W divided by Qh, where W = Qh minus Qc.
2. First find the work: W = 3.0 x 10^12 J minus 1.5 x 10^12 J = 1.5 x 10^12 J.
3. Then Eff = 1.5 x 10^12 divided by 3.0 x 10^12.
4. Dividing gives Eff = 0.50.
5. 0.33 and 0.66 do not match this division, and 1.00 would mean no heat at all was rejected to the environment.
6. So 0.50 is correct.
_Source: OpenStax Physics (CC BY 4.0), Ch 12 "Thermodynamics", section 12.4 Applications of Thermodynamics: Heat Engines, Heat Pumps, and Refrigerators_
Keep practising Thermodynamics
Browse all Thermodynamics questions for GCE O-Level (Singapore), or explore more Physics topics.
Related questions
What condition would be required for a heat engine to reach 100 percent thermal efficiencyIn the heat pump described, which component raises the temperature and pressure of the worA coal power station operates at 40.0 percent efficiency and takes in 1.20 x 10^12 J by heA heat engine takes in 6.0 kJ of heat and produces 4.8 kJ of waste heat. What is its efficA heat engine is given 120 J of heat and releases 20 J of heat to the environment. How mucWhat is a cyclical process, as used to describe heat engines?What does a heat engine do?Can the entropy of one part of the universe, such as a system built by doing work, ever de