# Refrigerator problem!

by sportsrules
Tags: refrigerator
 P: 9 The question reads.... A heat pump is a refrigerator that uses the inside of a building as a hot-temp reservoir in the winter and the outside of the building as a hot-temp reservoir in the summer. So..... For a carnot refrigerator engine operating between 25C and 40C in the summer, what is the coefficient of performance?And, for each joule of heat trasnfer from the cooler reservoir per cycle, how many joules of work are done by the refrigerator engine? Is this work positive or negative....and, what additional information is needed to determine an appropriate heat pump for a building? Ok, so I figured out the COP.... COP=1/ [(Th/Tc)-1] COP=1/ [(313k/298K)-1] COP=19.9 But thenn I get confused on the second part I know that COP=Qc/W W=Qc/COP so is the value for Qc just 1 J? THat is the part i am having trouble with, and the work is negative correct? because it is work done by the system? And, any thouhts on the last part?
 Sci Advisor HW Helper P: 6,684 This appears to be identical to your earlier post which has already been responded to: http://www.physicsforums.com/showthr...338#post523338 AM
 P: 9 yes it was......but how about the last part...what else would I need to know? Just the size of the building?
HW Helper
P: 6,684
Refrigerator problem!

 Quote by sportsrules yes it was......but how about the last part...what else would I need to know? Just the size of the building?
First of all, the co-efficient of performance for a heat pump is defined a little differently from the co-efficient of performance for a refrigerator. For a refrigerator:

$$cop = \frac{Q_c}{W}$$

For a heat pump,

$$cop = \frac{Q_h}{W}$$

So for a heat pump:

$$W = \frac{Q_h}{cop}$$

That is all you need to determine how much work you consume in delivering 1 joule of heat. It has nothing to do with building size.

AM

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