Calculate the ideal (Carnot) efficiency of a heat engine operating between 24.0deg C and 660.0deg C.?

Question

How much heat would be rejected by the engine if 1.00 x 10^6 calories were taken from the high temperature reservoir?

Answer 1

Carnot efficiency is directly related to the temperature (in Kelvin degrees--which has the same interval as Celcius but starts at absolute zero.) K = C + 273.15. So...

24.0 deg C = 297.15 deg K
660.0 deg C = 933.15 deg K
Max theoretical efficiency is (933.15 - 297.15)/933.15 or 68.2%

For the amount of heat rejected if 1.00 x 10^6 calories are taken from the heat source (the reservoir), that 1.00 x 10^6 amount is the input energy. Since you now know from above that it's ideally 68.2% efficient, the rest of that energy must be rejected. 100%-68.2% or 31.8% of that energy is therefore rejected to the heat sink. 1.00 x 10^6 * 31.8% is 3.18 x 10^5 or 318,000 calories rejected.

Answer 2

http://www.engineersedge.com/thermodynamics/power_plant_components.htm


Look at the very bottom of the screen, its correct. use the formulas he talked about above... it works.

ttyl