Work out the entropy change

  • #1

Homework Statement



a fridge at temp 2 C is in a kitchen of temp 22 C, over 1 hour 75 Kj of energy is supplied,
calculate the change in entropy of the fridge & kitchen


Homework Equations



S = Qrev / T

Fridge Temp in Kelvin 275.15 K
Kitchen Temp in Kelvin 295.15 K

The Attempt at a Solution



so S1 = 75 Kj / 275.15
and S2 = 75 Kj / 295.15

But this doesnt seem to make sense as i would expect the entropy in the fridge to go down and the kitchen to go up but these are both positive values
 

Answers and Replies

  • #2
Andrew Mason
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Homework Statement



a fridge at temp 2 C is in a kitchen of temp 22 C, over 1 hour 75 Kj of energy is supplied,
calculate the change in entropy of the fridge & kitchen


Homework Equations



S = Qrev / T

Fridge Temp in Kelvin 275.15 K
Kitchen Temp in Kelvin 295.15 K

The Attempt at a Solution



so S1 = 75 Kj / 275.15
and S2 = 75 Kj / 295.15

But this doesnt seem to make sense as i would expect the entropy in the fridge to go down and the kitchen to go up but these are both positive values
When calculating entropy, you have to determine Qc and get the sign right. Since heat is being removed from the cold reservoir Qc is negative. Since heat is being delivered to the hot reservoir, Qh is positive.

Are you given the COP of the fridge? You will need that.

The fridge pumps heat from the cold reservoir (inside) to the hot reservoir (outside). The amount of heat removed from the cold reservoir is not 75kJ. That is the amount of work done, not Qc. The amount of heat removed: Qc = Work x COP

AM
 
  • #3
i am not sure what you mean by COP, if this is what my text book describes as "performance - k" then it states the following equation

k = Qc / W = Tc / Th-Tc

so i would need
Tc - 275.15
Th - 295.15

275.15 / 295.15-275.15

k or COP = 13.7575

then Qc = 75kj x 13.76

do i do the same for the hot reservior & how do i go from this to get the entropy change
 
  • #4
135
2
When calculating entropy, you have to determine Qc and get the sign right. Since heat is being removed from the cold reservoir Qc is negative. Since heat is being delivered to the hot reservoir, Qh is positive.

Are you given the COP of the fridge? You will need that.

The fridge pumps heat from the cold reservoir (inside) to the hot reservoir (outside). The amount of heat removed from the cold reservoir is not 75kJ. That is the amount of work done, not Qc. The amount of heat removed: Qc = Work x COP

AM

I don't agree with this:
If you are calculating entropy change, there are two different equations. If you can assume from your question that the temperature of the fridge and the kitchen are not going to change significantly, then you can use the simplest:

DeltaS = Qrev/T

But as stated above, you have to be sure about the sign of the heat transfer. Heat transfer can be measured in Joules (or kiloJoules of course) giving the entropy in J K^-1.
Work is not a measure of heat transfer, but is measured in Nm, and doesn't come into it in my understanding of the question.

It's important to remember that the fridge is losing energy (transferring heat), so the sign will be negative for Qfridge, giving a decline in entropy.
 
  • #5
vela
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i am not sure what you mean by COP, if this is what my text book describes as "performance - k" then it states the following equation

k = Qc / W = Tc / Th-Tc
COP stands for coefficient of performance, and it's the ratio of the heat removed from the cold reservoir and the work needed to accomplish that. For an ideal refrigerator, the performance can be calculated from the temperatures of the hot and cold reservoirs. Did the problem say you have an ideal refrigerator?

Do I do the same for the hot reservior & how do I go from this to get the entropy change?
Use conservation of energy to calculate [itex]Q_h[/itex]. You already stated the equation for calculating entropy.
 
  • #6
Andrew Mason
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I don't agree with this:
If you are calculating entropy change, there are two different equations. If you can assume from your question that the temperature of the fridge and the kitchen are not going to change significantly, then you can use the simplest:

DeltaS = Qrev/T

But as stated above, you have to be sure about the sign of the heat transfer. Heat transfer can be measured in Joules (or kiloJoules of course) giving the entropy in J K^-1.
Work is not a measure of heat transfer, but is measured in Nm, and doesn't come into it in my understanding of the question.
I am not sure what it is that you do not agree with.

Entropy is [itex]\int dQ_{rev}/T[/itex]. But there are two reservoirs here: the fridge and the room. The change in entropy is the sum of the changes in entropy of the two reservoirs:

[tex]\Delta S = \int dQ_c/T_c + \int dQ_h/T_h = -|Q_c|/T_c + Q_h/T_h[/tex]

(Note: one assumes that the heat transfer from the cold reservoir occurs at constant temperature and the heat transfer to the hot reservoir also occurs isothermally- ie. reversibly).

You can determine Qc from the coefficient of performance (COP). In order to calculate the heat delivered to the hot register, you have to add the work done to the heat removed: Qh = Qc + W.

I don't understand your objection to the use of W. I agree that W is not a measure of heat flow. That was my point. But it can be used to determine heat flow (Qc = COP x W and Qh = Qc + W.).

AM
 
  • #7
135
2
Sorry Andrew, I didn't mean to criticise. I agree with everything in your last post. The only part of your explanation I disagreed with originally was:

"The amount of heat removed from the cold reservoir is not 75kJ. That is the amount of work done, not Qc."

I took this to mean you were saying the work done was 75kJ, which you obviously weren't.
As the question makes no mention of COP, I would answer it using the assumption that this is an ideal fridge, that the process is reversible, and that the temperature remains unchanged. If you make those assumptions, then entropy change for the fridge can be -75kJ / temperature of the fridge and entropy change for the room can be +75kg / temperature of the room, no? Then add them together for total entropy change, if needed.
 
  • #8
Andrew Mason
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Sorry Andrew, I didn't mean to criticise. I agree with everything in your last post. The only part of your explanation I disagreed with originally was:
I don't want to discourage criticism - ever. But I was not sure what it was you were objecting to.

"The amount of heat removed from the cold reservoir is not 75kJ. That is the amount of work done, not Qc."

I took this to mean you were saying the work done was 75kJ, which you obviously weren't.
I was, actually. 75kJ is the input work/hour (eg voltage x current x 3600 of electric motor running the compressor). The Coefficient of Performance is the heat removed from the fridge divided by the input work. The input work does contribute to the heat flowing to the hot reservoir but it is only part of it.
As the question makes no mention of COP, I would answer it using the assumption that this is an ideal fridge, that the process is reversible, and that the temperature remains unchanged. If you make those assumptions, then entropy change for the fridge can be -75kJ / temperature of the fridge and entropy change for the room can be +75kg / temperature of the room, no? Then add them together for total entropy change, if needed.
No. The heat flow to the room is Qc + 75kJ. Qc = W x COP

If it was an ideal fridge, the entropy change would be 0. So a problem asking what the entropy change of the fridge and surroundings of an ideal refrigerator is not a very challenging one. You can see this if you work it out:

COP = Qc/W = Tc/(Th-Tc) = 275/(20) = 13.75

The heat flow from the cold reservoir would be:

Qc = 13.75 x 75kJ = 1031 kJ
Qh = Qc + W = 1106 kJ

[tex]\Delta S = -1031/275 + 1106/295 = -3.75 + 3.75 = 0[/tex]

So there is something missing from the original question as posted. Perhaps the OP could give us the entire question verbatim.

AM
 
  • #9
135
2
You're right Andrew, my mistake.

I misread the original question, actually, and presumed that the 75kJ was the heat being transferred from the fridge to the kitchen, but rereading it I see that it is actually the energy being supplied (presumably to the fridge).
 
  • #10
Andrew Mason
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I misread the original question, actually, and presumed that the 75kJ was the heat being transferred from the fridge to the kitchen, but rereading it I see that it is actually the energy being supplied (presumably to the fridge).
Ok. Now I understand what it was you were saying. Looking back at the question I can see why you might interpret it that way. I was hoping victoriafello would post the complete question though.

AM
 
  • #11
thanks for your help so far. here is my homework question verbatim.


The temperature inside the refrigerator is 2.0 °C. It stands in a kitchen whose
temperature is 22 °C. In the course of an hour, 75 kJ of heat are transferred
from the interior of the refrigerator to the kitchen.
As a result of this heat transfer, what is the entropy change of (i) the contents
of the refrigerator, and (ii) the kitchen? (Assume that the heat transfer is
reversible.)

was i going in the correct direction calculating the COP or is it as simple as Qrev/T ?

Thanks so much
 
  • #12
Andrew Mason
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thanks for your help so far. here is my homework question verbatim.


The temperature inside the refrigerator is 2.0 °C. It stands in a kitchen whose
temperature is 22 °C. In the course of an hour, 75 kJ of heat are transferred
from the interior of the refrigerator to the kitchen.
As a result of this heat transfer, what is the entropy change of (i) the contents
of the refrigerator, and (ii) the kitchen? (Assume that the heat transfer is
reversible.)

was i going in the correct direction calculating the COP or is it as simple as Qrev/T ?

Thanks so much
Ok. This is a totally different question than the one originally posted. It now makes sense. Tomwilliam was correct in thinking that the 75kJ/h was Qc.

The COP is 13.75 (COP = Qc/W = Qc/(Qh-Qc) = Tc/(Th-Tc))
Qc is given
W = Qc/COP
Qh = Qc+W

So with Qc and Qh the entropy calculation is very straightforward.

AM
 
  • #13
Sorry I put the original question from memory & didn’t realize that I had made a mistake in the wording, I will be more careful !!

So then this is my answer

Qc – 75Kj
W = Qc / COP – 75kj / 13.75
W = 5.45
Qh = Qc+W = 75Kj + 5.45 = 80.45

DeltaSc = Qc/T = -75Kj/275.15= -0.27
DeltaSh = Qh/T – 80.45/295.15 = 0.27

So overall Change in Entropy DeltaS = 0

Is this correct?
 
  • #14
vela
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Yup, you got it. Don't forget the units though.
 
  • #15
135
2
Now I'm confused.

The question states that the 75kJ of heat is transferred from the refrigerator to the kitchen.

So if Qc is the heat absorbed in the refrigerator compartment
and Qh is the heat absorbed by the room

Why are we assuming that -75kJ = Qc
Why not 75kJ = Qh?

Do you know what I mean?

Could we not interpret this question as: 75kJ of heat are released into the room due to the cooling action of the fridge (Qh)?
 
  • #16
vela
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It takes work to move heat from a cold reservoir to a hot reservoir. Qh is not only contains Qc but the work required to move the heat as well, so it will always be larger than Qc. If you used Qh=75 kJ, then Qc would have to be less than 75 kJ, and you won't have transferred 75 kJ from the compartment to the room.
 
  • #17
135
2
Thanks - very clear.
 

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