Entropy - show positive net entropy change...

  • #1

Homework Statement


Two systems that have the same heat capacity Cv but different initial temperatures T1 and T2 (with T2 > T1) are placed in thermal contact with each other for a brief time, so that some heat flows but the temperature of neither system changes appreciably. Show that there is a positive net entropy change associated with this heat flow.

Homework Equations


ΔS = ΔST2+ΔST1 > (1/T1 - 1/T2)δQ > 0
TdS = CvdT +PdV

The Attempt at a Solution


The answer seems intuitive, but I'm not really sure how to approach a proper solution. Can someone clarify what this question is asking, and point me in the right direction in terms of appropriate formulas or concepts? I'm relatively new to entropy, and still working on grasping the fundamentals.

Cheers
 

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Answers and Replies

  • #2
I presume this is an irreversible process, but in terms of calculations I'm not entirely sure how to expand. entropy change is positive for spontaneous reactions as far as I know, so I'm not sure what else this question wants me to show.
 
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  • #3
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If ##\delta Q## is the amount of heat transferred in the process, what is the final equilibrium temperature of system 1 (when the system finally equilibrates after being taken out of contact with system 2) and what is the final equilibrium temperature of system 2? Once you answer these questions, we can continue.
 
  • #4
"so that some heat flows but the temperature of neither system changes appreciably." so I suppose neither T1 nor T2 change (?) but some heat is exchanged. some volume change then?
 
  • #5
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"so that some heat flows but the temperature of neither system changes appreciably." so I suppose neither T1 nor T2 change (?) but some heat is exchanged. some volume change then?
No. Even though neither system changes appreciably, there is still some small change. That is what we will work with. So, again, please tell me what you would calculate for the final temperatures. Please bear with me. I will get you to where you need to be.
 
  • #6
ok, so for system 1 would it be -Q/Cv + T1 and for system 2: +Q/Cv + T2?
(I appreciate your patience; apologies for implying doubt)
 
  • #7
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ok, so for system 1 would it be -Q/Cv + T1 and for system 2: +Q/Cv + T2?
(I appreciate your patience; apologies for implying doubt)
Excellent. So, in the initial equilibrium state of system 1, the temperature is T1, and, in the final equilibrium state of system 1, the temperature is T1-Q/C. This establishes the two end points. Now, we need to calculate the entropy change of system 1. By considering a reversible path for system 1 between the two end states, we find that this entropy change can be determined by integrating the equation $$dS_1=\frac{CdT}{T}$$ between the two end temperatures. What do you get?
 
  • #8
I suppose: S1= Cvln( (T1-Q/Cv)/T1)

I take it I do the same thing for system 2 and then add?
S2= Cvln( (T2+Q/Cv)/T2)
 
  • #9
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I suppose: S1= Cvln( (T1-Q/Cv)/T1)

I take it I do the same thing for system 2 and then add?
S2= Cvln( (T2+Q/Cv)/T2)
Yes, but I would write them a little differently:

$$\Delta S_1=C\ln{\left(1-\frac{Q}{CT_1}\right)}$$

OK??
 
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  • #10
Ok, so ΔS = Cln(1-Q/CT1) + Cln(1+Q/T2) is where we're at?
 
  • #11
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Ok, so ΔS = Cln(1-Q/CT1) + Cln(1+Q/T2) is where we're at?
These are the results for arbitrary Q. We want the results for small finite Q. Do you know the series expansion of ln(1+x) for small x?
 
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  • #12
≈x-x2/2. so, Cln(1+Q/T2) ≈ C (Q/T2 - (Q/T2)2/2), where that second term would go to zero

so, I think we'd arrive at:
ΔS ≈ C (Q/T2 - Q/T1), or ≈ CQ (1/T2 - 1/T1)]
this seems incorrect... maybe I mixed up the positive and negative Q earlier
 
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  • #13
ok, so the Q should be a positive value in system 1 due to T2 > T1... so, adjusting for that and recalculating, I end up with:
ΔS ≈ CvQ (1/T1 - 1/T2).

One of the relevant equations from my text says ΔS = ΔST2+ΔST1 > (1/T1 - 1/T2)δQ > 0. Do I need to expand any further with more work? also, noob question: I'm not sure if I should treat "Q" the same as "δQ" in all these formulas... is it ok do to so in this case? sometimes I get a bit mixed up with δQ vs ΔQ vs Q, etc.. I understand that we're using δ_ because of infinitesimal change, but am I good as far as formula consistency so far?
 
  • #14
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≈x-x2/2. so, Cln(1+Q/T2) ≈ C (Q/T2 - (Q/T2)2/2), where that second term would go to zero
Actually, according to the specification of your problem, the second term would be considered negligible ("the temperature of neither system changes appreciably")

so, I think we'd arrive at:
ΔS ≈ C (Q/T2 - Q/T1), or ≈ CQ (1/T2 - 1/T1)]
this seems incorrect... maybe I mixed up the positive and negative Q earlier[/QUOTE]
No. I took Q as the heat transferred from T1 to T2. If T2>T1, Q is negative
 
  • #15
if initial T2 > T1 then heat would flow from T2 to T1, which means in system 2 the Q would be negative and it would be positive for 1... right? I'm saying that I think I mixed that up earlier
 
  • #16
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ok, so the Q should be a positive value in system 1 due to T2 > T1... so, adjusting for that and recalculating, I end up with:
ΔS ≈ CvQ (1/T1 - 1/T2).

One of the relevant equations from my text says ΔS = ΔST2+ΔST1 > (1/T1 - 1/T2)δQ > 0. Do I need to expand any further with more work?
No. What you have done is more than adequate, and, moreover, is exactly what they asked for.
also, noob question: I'm not sure if I should treat "Q" the same as "δQ" in all these formulas... is it ok do to so in this case? sometimes I get a bit mixed up with δQ vs ΔQ vs Q, etc.. I understand that we're using δ_ because of infinitesimal change, but am I good as far as formula consistency so far?
This is never presented very well in the textbooks, and there is a lot of ambiguity. The important thing is to understand exactly what you personally mean.
 
  • #17
ok, so just to back things up a bit and make sure I'm all squared away here...:
ΔS1= Cvln( (T1+δQ/Cv)/T1) (if δQ was positive in system 1)
ΔS2= Cvln( (T2-δQ/Cv)/T2) (if δQ was negative in system 2)

ΔS ≈ CvδQ (1/T1 - 1/T2), and knowing that (1/T1 - 1/T2)δQ > 0, ΔS is therefore > 0

... all good on my end? Is my assumption of the positive / negative Q correct?
 
  • #18
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if initial T2 > T1 then heat would flow from T2 to T1, which means in system 2 the Q would be negative and it would be positive for 1... right? I'm saying that I think I mixed that up earlier
Actually, it was I who mixed it up. I got it in my head that T1 was higher than T2.

One more question for you. Do you understand how the equation in post #7 was obtained, and, specifically, what reversible path was used to obtain ##dq_{rev}/T##? What reversible path was used here?

Here is a Physics Forums Insights article I wrote explaining how to determine the change of entropy for an irreversible process like this. One of the examples in the article solves your exact problem: https://www.physicsforums.com/insights/grandpa-chets-entropy-recipe/
 
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  • #19
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ok, so just to back things up a bit and make sure I'm all squared away here...:
ΔS1= Cvln( (T1+δQ/Cv)/T1) (if δQ was positive in system 1)
ΔS2= Cvln( (T2-δQ/Cv)/T2) (if δQ was negative in system 2)

ΔS ≈ CvδQ (1/T1 - 1/T2), and knowing that (1/T1 - 1/T2)δQ > 0, ΔS is therefore > 0

... all good on my end? Is my assumption of the positive / negative Q correct?
Prefect.
 
  • #20
Awesome, thanks so much for the help.
---
"One more question for you. Do you understand how the equation in post #7 was obtained, and, specifically, what reversible path was used to obtain dqrev/TdqrevT? What reversible path was used here?"
I... kinda. An applicable equation is mentioned in my class notes for a system that comes into contact with a reservoir and experiences an increase in temperature. We only recently introduced 'reversible paths' so I haven't quite wrapped my head around the concept. I suppose I'll check out that link you posted.

I assumed this question covered an irreversible process.
 
  • #21
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Awesome, thanks so much for the help.
---
"One more question for you. Do you understand how the equation in post #7 was obtained, and, specifically, what reversible path was used to obtain dqrev/TdqrevT? What reversible path was used here?"
I... kinda. An applicable equation is mentioned in my class notes for a system that comes into contact with a reservoir and experiences an increase in temperature. We only recently introduced 'reversible paths' so I haven't quite wrapped my head around the concept. I suppose I'll check out that link you posted.

I assumed this question covered an irreversible process.
I'm amazed that you are learning about entropy without first being introduced to reversible process paths, since the change in entropy from one thermodynamic equilibrium state to another can only be determined by calculating the integral of dq/T for a reversible process path between these two states. A reversible process is one for which the system under consideration passes through a continuous sequence of thermodynamic equilibrium states. So, in the case of your problem, to find the entropy change for system 1, you would first separate it in its initial state from system 2, and you would then bring it into contact with a sequence of ideal reservoirs running in temperature from T1 to the final temperature T1+Q/C. The contact with these reservoirs would be carried out very gradually. This would constitute a reversible process path from its initial thermodynamic equilibrium state to its final thermodynamic equilibrium state.
 
  • #22
gotcha. the wording of the question tripped me up, but I suppose to interpret it I would need to conceptualize it the way you're mentioning (?). they made it seem spontaneous and quick rather than gradual.
 
  • #23
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gotcha. the wording of the question tripped me up, but I suppose to interpret it I would need to conceptualize it the way you're mentioning (?). they made it seem spontaneous and quick rather than gradual.
Let's be clear about this. The actual process experienced by these systems is spontaneous, rapid, and irreversible. However, to calculate the entropy change for these systems, we must separately devise (dream up) reversible paths for the two systems that take each of them from their same starting state to their same final state as the irreversible path; and the entropy change is then the integral of dq/T for the reversible paths. The reversible paths do not have to bear any resemblance whatsoever to the actual irreversible path, other then to match the initial and final end points.
 
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  • #24
ok, I think I get it. cheers
 

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