How Does an Air Conditioner Affect Entropy and Heat Transfer?

AI Thread Summary
An air conditioner operating at 800W with a performance coefficient of 2.80 is analyzed for its heat transfer and entropy effects. The discussion focuses on calculating the rate of heat removal and heat discharge to the outside air, along with the total entropy changes for both the room and the outside environment over one hour. Key equations include the coefficient of performance and entropy formulas, which help in determining the necessary values. Participants seek guidance on finding the rate of heat removal and the final temperature needed for entropy calculations. The conversation emphasizes the importance of converting power into a rate for accurate calculations.
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Homework Statement


An air conditioner operates on 800W of power and has a performance coefficient of 2.80 with a room temperature of 21oC, and an outside temperature of 35oC.
a) Calculate the rate of heat removal for this unit.
b) calculate the rate at which heat is discharged to the outside air.
c) calculate the total entropy change in the room if the air conditioner runs for 1 hour. Calculate the total entropy change in the outside air for the same time period.
d) What is the net change in entropy for the system(room+outside air)?


Homework Equations


\DeltaQ=cm\DeltaT
I think that that is the equation of heat lost
Coefficient of performance = Qcold, input/Winput
C.o.P=Qcold,input/(Qhot,output-Qcold,input)

The Attempt at a Solution



a) I'm not sure how to find the rate of heat which is removed

b) I'm not sure either

c) I know that Entropy=Q/T, and
\DeltaS=mcln(Th/Tc)
c is the specific heat, m is mass

but I'm not sure how to find the final temperature from what's given in the problem, or the mass...


d) once I know the answer to part C, its fairly self explanatory

Can someone guide in the right direction for solving this
 
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for part a, the useful equations are CoP= Qci/(Qho-Qci and CoP = Qci/W. However, since we only have a value for power, W/t, to use the latter equation, you have to turn everything into a rate. This isn't hard to do, just multiply the numerator and denominator by 1/t, you'll get CoP = (Qci/t)/P. Now plug in the known values for P and CoP, can you take it from there? (hint, you'll have to use the same trick for the equation involving Qho
 
Ok thanks that's makes a little more sense now, thanks
 
No problem, that should give you the answer for a and b. For part c, remember,
5058da7e100f84df3d05e7ba922259c8.png

This should help you find the temperature change.
 
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