Question

Consider three engines that each use 2160 J of heat from a hot reservoir (temperature = 540 K). These three engines reject heat to a cold reservoir (temperature = 335 K). Engine I rejects 1497 J of heat. Engine II rejects 655 J of heat. Engine III rejects 1340 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. For each of the engines determine the total entropy change of the universe, which is the sum of the entropy changes of the hot and cold reservoirs. On the basis of your calculations, identify which engine operates reversibly, which operates irreversibly and could exist, and which operates irreversibly and could not exist.

Answer 1

In all cases, the hot reservoir transfers 2160J of thermal energy to the engine at a temperature of 540K. The change in thermal energy of the hot reservoir, ?q_hot, is -2160J, and the change in entropy of the hot reservoir is ?S = ?q/T = (-2160J)/(540K) = -4 J/K.

The engines add thermal energy to the cool reservoir at at temperature of 335K. The entropy changes of the cool reservoir due to the three engines is therefore:

Engine 1: ?q/T = (1497J)/(335K) = +4.468656 J/K
Engine 2: ?q/T = (655J)/(335K) = +1.9552 J/K
Engine 3: ?q/T = (1340J)/(335K) = +4 J/K

The change in entropy of the universe is the sum of the changes in the hot and cold reservoirs:
Engine 1: ?S = (4.46 - 4.00)J/K = 0.46 J/K
Engine 2: ?S = (1.95 - 4.00)J/K = -2.05 J/K
Engine 3: ?S = (4 - 4.00)J/K = 0 J/K

The Third engine operates reversibly (the change in entropy of the universe is zero). The Second engine violates the 2nd law because the entropy of the universe would decrease. It is therefore not a physically realizable engine.