How to calculate the entropy for two sources

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Homework Statement:
Find entropy variation due to irreversibility when heat sources at two different temperatures (T_h>T_c) exchange heat (Q)
Relevant Equations:
T_c=temperature of cold source
T_h=temperature of hot source
Q= exchanged heat
ΔS= entropy variation
ΔS_irr= entropy variation due to irreversibility
Hello to everyone, I'm studying thermodinamics and I would like to understand better the meaning of entropy and how to calculate it.
I know that if A and B are two possible states of a system, the equation whcih defines variation of entropy from A to B is:
ΔS=∫(dQ/T)_rev (1)
So I have to find a generic reversible transformation, which go from A to B, and resolve the integral along this transformation. In this case system would evolve through infinite sources which are step by step at the same temperature of system.
I have thought to this example where two heat sources at different temperatures (T_h>T_c) exchange an amount of heat (Q).
As reversible transformations I have choosed isotermal transformations at the two different temperatures of the sources (you can see my solution in the pic below).
Now I start to have some problems. Let's consider the equation (1), that integral can be calculated also on the real (irreversible) transformation which system follows and we have that:
ΔS=∫(dQ/T)_rev≥∫(dQ/T)_irr (2)
In general it can be written:
∫(dQ/T)_rev=∫(dQ/T)_irr+ΔS_irr
where the term ΔS_irr is a contribution due to irreversible process.
What I have tried to do is to calculate ΔS_irr for each source and for the universe. While I have calculated it for the universe I couldn't for the sources.

This is a simple example to help me to understand better the problem. I have other doubts which I'm going to specify below:

1) In equation (1) is T the temperature of the system or od the environment? (I think it's system temperature but I also read it refers to envirnoment)

2) Equation (1) can be also written as:
T*dS=dQ (2)
In this equation I think I'm considering an infinitesimal step of a trasformation where system are exchanging with a source at temperature T (wichi is the same temperature of system). If I consider a irreversible process equation (2) becomes:
T*dS=dQ+dS_irr(q)+dS_irr(L_f) (3)
dS_irr(q)=contribution of heat
dS_irr(L_f)= contribution of friction
However I really don't understand equation (3). First of all, if it's an irresversible transformation how can I consider an infinitesimal process? During an irreversible transformation I can't know the state of system because thermodinamics properties are not defined.
Also, since it's an irreverisble transformation and source and system could be different temperatures, what does T refers to?
Then, I have read that the term [dQ+dS_irr(q)] is equivalent to the heat of the reversible transformation.
Indeed I have read:
(dQ/T_source)+dS_irr(q)=(dQ/T)
where (dQ/T) refers to the equivalent reversible transformation and T should be the temperature of system.
I know that some passages could be not so clear but I did my best especially I don't speak English very well. I hope you can help me especially with equation (3) because I don't know its meaning at all.
thank you
 

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Answers and Replies

  • #2
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You have two ideal reservoirs, one at TA and one at TB, and your irreversible process involves a transfer of a finite amount of heat Q between the two ideal reservoirs, correct? And you are wondering, if you use the reversible equation to calculate the entropy change for each (since they are ideal reservoirs) where the irreversible entropy generation comes in, correct?
 
  • #3
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yes because also for the irreversible trasnformation I can write (cosidering for example hot reservoir)
∫(dQ/T)_irr=(1/T_h)*∫dQ
In fact also for irreversible transformation the temperatures of resevoirs are constat.
Instead the other questions regard some specific equations, which I didn't understand.
 
  • #4
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Imagine a heat conducting medium sandwiched between the two reservoirs, of thermal conductivity k. Heat enters the medium at the higher temperature and exits the medium at the lower temperature. Over the entire process, its temperature returns to its original value of ##(T_h+T_c)/2##, so its entropy doesn't change. But the amount of entropy generated within the conductive medium during the process is ##Q\left(\frac{1}{T_c}-\frac{1}{T_h}\right)##. This entropy gets transferred to the combination of the two reservoirs. So the net result over the entire process is that the entropy of the combination of the two reservoirs changes by this amount.
 
  • #5
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Thanks for your answer. However can you explain me the full passages to arrive to Q(1/T_c-1/t_h)? Because I didn't understand how to get it.
 
  • #6
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Thanks for your answer. However can you explain me the full passages to arrive to Q(1/T_c-1/t_h)? Because I didn't understand how to get it.
The change in entropy of an ideal isothermal reservoir is equal to the amount of heat it receives Q divided by its absolute temperature T. This is the case irrespective of the irreversibility of the overall process. Since the reservoirs are ideal, there is no entropy generated within them during the process we are considering. Therefore, all the entropy that is generated in the process we are considering must be generated within the conductive medium connecting them during the process. So, the entropy balance on the hot reservoir is $$\Delta S_h=-\frac{Q}{T_h}$$
The entropy balance on the cold reservoir is $$\Delta S_c=+\frac{Q}{T_c}$$and the entropy balance on the intervening conductive medium whose entropy change is zero (because it had negligible heat capacity and returns to its original state after the process) is: $$\Delta S_m=+\frac{Q}{T_h}-\frac{Q}{T_c}+\sigma=0$$where ##\sigma## is the entropy generated in the conductive medium during the process. If we solve this balance equation for ##\sigma##, we obtain: $$\sigma=\frac{Q}{T_c}-\frac{Q}{T_h}$$And, if we add the three entropy equations together, we get the entropy change of the combination: $$\Delta S_h+\Delta S_c+\Delta S_m=\sigma=\frac{Q}{T_c}-\frac{Q}{T_h}$$

The entropy balance on the conductive medium says that the amount of entropy transferred to the conductive medium from the hot reservoir is ##\frac{Q}{T_h}##, the amount of entropy transferred from the conductive medium to the cold reservoir is ##\frac{Q}{T_c}##, and the amount of entropy generated within the conductive medium during the process is ##\sigma##.
 

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