Consider three engines that each use 1680 J of heat from a hot reservoir (temper
ID: 2215629 • Letter: C
Question
Consider three engines that each use 1680 J of heat from a hot reservoir (temperature = 560 K). These three engines reject heat to a cold reservoir (temperature = 335 K). Engine I rejects 1005 J of heat. Engine II rejects 1203 J of heat. Engine III rejects 655 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. (Round off to two decimal places.) delta S Engine I: delta S Engine II: delta S Engine III: Type of Operation for- Engine I: Engine II: Engine III:Explanation / Answer
Entropy change = 1680/560 - H/335
for 1st engine s = 3 - 1005/335 = 0
for 2nd engine s = 3 - 1203/335 = -.59
for 3rd engine s = 3 - 655/335 = 1.044
for second engine total entropy change is -ve, which is not possible so it could not exist.
for 1st engine s = 0, it is irreversible engine
and 3rd engine is reversible...