C Physics question I Chegg LON-CAPA Non-Reversibl... Bb My Courses Blackboard L.
ID: 1515911 • Letter: C
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C Physics question I Chegg LON-CAPA Non-Reversibl... Bb My Courses Blackboard L. triton physi dak,ed Giordano 16/Prob1638 problem? symb uploaded%2fndsu%2f8A2361183b98e C a search /Chap Physics 211 College Physics I Spring 2016 New Messages Courses Help Logout Makenzie Reutter v Student section: 01 Main Menu Contents Grades syllabus Feeds Timer Notes Evaluate Feedback Print Info Homework 14 Extra Credit Non-Reversible Pump Course Contents A heat pump operates between reservoirs at 303 K (inside the house and 266 K (outside). It is not a reversible heat pump; it only pumps 67% of the maximum possible heat into the house. How much work must be done to "pump" 5590 J of heat energy into the house? Submit Answer Tries 0/10 Threaded View Chronological View Other Views My general preferences on what is marked as NEW Mark NEW posts no longer new Export NEW Help Anonymous 1 Reply Thu Apr 28 01:05: 14 am 2016 (CDT)) Has anyone figured out how to do this problem? It seems straightforward, but everything I try is wrong. I've tried finding the ma efficiency then the specified percent of that efficiency, followed by the amount of heat energy divided by that level of efficiency...? NEW Re: Help Tasha Hahka Reply (Fri Apr 29 04:44:23 pm 2016 (CDT amount of J needed to pump)(Hot-Cold) all divided by percent hot temp My settings for this discussion 1. Display All posts 2. Not new Once marked not NEW Change Threaded View Chronological View Other Views My general preferences on what is marked as NEW Mark NEW posts no longer new Export Send Feedback Post Discussion 8:05 PM 5/2/2016Explanation / Answer
here,
Tc = 266 k
Th = 303 k
heat tranfered to the house , Q = 0.67 * 5590
Q = 3745.3 J
the work done , W = Q * ( 1 - Tc/Th)
W = 3745.3 * ( 1 - 266/303)
W = 457.35 J
the work done by the pump is 457.35 J
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