BB My Courses - Blackboard L. LON-CAPA Entropy & Rev... C Physics question | Che
ID: 1515766 • Letter: B
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BB My Courses - Blackboard L. LON-CAPA Entropy & Rev... C Physics question | Chegg.c... x + triton physics ndsu nodak.edu/res/ndsu/demmelash/Giordano/Chap16/Prob1646 problem?symb= uploaded? 2fndsu%2f8A2361183b98e Makenzie Reutter (Student section: 01) Main Menu ContentsGrades Syllabus Feeds Q Search Physics 211 - College Physics I - Spring 2016 New Messages Courses Help Logout Course Contents . »Homework 14 - Extra Credit »Entropy & Reversibility Timer·Notes ,Feedback- Evaluate Print Info Consider a reversible heat engine that employs a hot reservoir at a temperature of 760 K and a cold reservoir at 210 K (a) What is the entropy change of the hot reservoir during a period in which 5670 is extracted from the hot reservoir? Submit Answer Tries 0/5 (b) What is the change in the entropy of the cold reservoir? Submit Answer Tries 0/5 (c) Find the change in the entropy of the engine itself during this time Submit Answer Tries 0/5 Post Discussion Send Feedback |1.4) 5:34 PM 5/2/2016Explanation / Answer
a) Entropy of the hot reservoir is assumed to have no irreversiblilites present so there is no entropy production involved. Therefore, entropy change is solely do to heat transfer. ie.
Change in entropy = -Q/T(h) = -5670/(760) = -7.4605 J/K
b) This answer is +7.4605 J/K. Because this is a REVERSIBLE heat engine, there can be no entropy generation in the engine. All entropy in = all entropy out.
C) Zero. Because this is a REVERSIBLE heat engine, there can be no entropy generation in the engine. All entropy in = all entropy out.
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