A coal power station operates at 40.0 percent efficiency and takes in 1.20 x 10^12 J by heat. How much work does it do?

1. Efficiency is Eff = W divided by Qh, so work output is W = Eff multiplied by Qh. 2. Substitute Eff = 0.400 and Qh = 1.20 x 10^12 J. 3. Multiplying gives W = 0.400 x 1.20 x 10^12 = 4.8 x 10^11 J. 4. 3.0 x 10^12 J and 4.8 x 10^13 J are far too large for a 40 percent share of the input heat. 5. 3.0 x 10^10 J is too small by more than an order of magnitude. 6. So 4.8 x 10^11 J is correct. _Source: OpenStax Physics (CC BY 4.0), Ch 12 "Thermodynamics", section 12.4 Applications of Thermodynamics: Heat Engines, Heat Pumps, and Refrigerators_