Question

Consider three engines that each use 1650 J of heat from a hot resevoir (temperature = 550 K). These three engines reject heat to a cold resevoir (temperature 330 K). Engine I rejects 1380 J of heat. Engine II rejects 990 J of heat. Engine III rejects 785 J of heat. One of the engines operates reversibly, and two operate irreversibly. However, of the two irreversible engines, one violates the second law of thermodynamics and could not exist. What is the total entropy change of the universe for (a) engine I, (b) engine II, and (c) engine III? (Note: The total entropy change of the universe is the sum of the entropy changes of the hot and cold resevoirs.)

Answer 1

Given :-

Heat used by each engine, Qh = 1650 J

Temperture of hot reservoir, Th = 550 K

Heat rejected by engine I, Qc1 = 1380 J

Heat rejected by engine II, Qc2 = 990 J

Heat rejected by engine III, Qc3 = 785 J

Tempreture of cold reservoir, Tc = 330 K

change in entropy is

deltaS = Qc/Tc - Qh/Th

1)

engine I

deltaS1 = Qc/Tc - Qh/Th

deltaS1 = 1380/330 - 1650/550

deltaS1 = 1.182 J/k

1)

engine II

deltaS2 = Qc/Tc - Qh/Th

deltaS2 = 990/330 - 1650/550

deltaS2 = 0 J/k

3)

engine III

deltaS3 = Qc/Tc - Qh/Th

deltaS3 = 785/330 - 1650/550

deltaS3 = -0.62 J/k