A demonstration on the necessary positive change in the entropy

In summary, Homework Equations state that if two bodies have different temperatures, then the entropy of the system will change. However, it is still unclear how to prove this inequality.
  • #1
Rulonegger
16
0

Homework Statement


Hello everyone. My problem is as follows: In a spontaneous process where two bodies at different temperatures [itex]T_{1}[/itex] and [itex]T_{2}[/itex], where [itex]T_{1}>T_{2}[/itex], are put together until they reach thermal equilibrium. The number of atoms or molecules of the first is [itex]N_{1}[/itex] and [itex]N_{2}[/itex] for the second one, with [itex]N_{1} \neq N_{2}[/itex], and they have heat capacities equal to [itex]C_{V_{1}}=aN_{1}k[/itex] and [itex]C_{V_{2}}=aN_{2}k[/itex], with [itex]a[/itex] given with the appropriate units. Past some sufficiently large time, the system reaches a temperature [itex]T[/itex], provided that [itex]T_{1}>T>T_{2}[/itex], which is in function of the initial temperatures and the number of atoms or molecules of the two bodies. The problem is that i can't demonstate that the change of the entropy of the system as a whole is positive, i.e. [tex]\bigtriangleup S>0[/tex]

Homework Equations


When i compute the change of the entropy for the i-th body, i get
[tex]\bigtriangleup S_{i}=\int_{T_{i}}^T \! \frac{1}{T} \, \mathrm{d} Q=\int_{T_{i}}^T \! \frac{aN_{i}k}{T} \, \mathrm{d} T=aN_{i}k\int_{T_{i}}^T \! \frac{1}{T} \, \mathrm{d} T=aN_{i}k\ln{\frac{T}{T_{i}}}[/tex]
With the hypothesis that the entropy is an extensive property, then [tex]\bigtriangleup S=\bigtriangleup S_{1}+\bigtriangleup S_{2}=aN_{1}k\ln{\frac{T}{T_{1}}}+aN_{2}k\ln{\frac{T}{T_{2}}}[/tex]
So i just have to prove that [tex]N_{1}\ln{\frac{T}{T_{1}}}+N_{2}\ln{\frac{T}{T_{2}}} > 0[/tex]

The Attempt at a Solution


I think that i have to use the two cases ([itex]N_{1}>N_{2}[/itex] and [itex]N_{1}<N_{2}[/itex]), and using the fact that [itex]T_{1}>T>T_{2}[/itex], to prove the inequality, but i have tried to do it in very different ways, and i get nothing, so i think there is some trick to demonstrating that, but I'm still a bit of an amateur in proving tricky inequalities.
 
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  • #2
Don't you need conservation of energy? Otherwise there's nothing to stop both ending arbitrarily close to T2.
 
  • #3
Actually I've used conservation of energy, and i can determine the temperature [itex]T[/itex] of equilibrium, which is [tex]T=\frac{N_{1}T_{1}+N_{2}T_{2}}{N_{1}+N_{2}}[/tex]but when i substitute that expression on the inequality, the later just complicates a little bit more. In despite of this, i think i must substitute [itex]T[/itex] in the inequality, but i get nothing again.
 
  • #4
I'm sure it can be done this way, but I feel that performing the integral just makes life harder. Consider a small dQ transferred. The hotter body loses dQ/T1, the cooler gains dQ/T2. Since T1 > T2, the sum has increased.
 
  • #5
Thanks haruspex. If i understand, you say that [tex]dS=dS_{1}+dS_{2}=\frac{dQ}{T_{1}}+\frac{dQ}{T_{2}}[/tex]With the fact that [itex]T_{1}>T_{2}[/itex], therefore a small dQ transferred between them would lead a change in the first entropy which is smaller than the change of the second one, without sayin anything about [itex]C_{V_{1}}[/itex] nor [itex]C_{V_{2}}[/itex]? But maybe i should take into account a small change in the temperature [itex]T_{1}[/itex] and in [itex]T_{2}[/itex] just to get a formal demonstration, don't you think?
 
  • #6
Rulonegger said:
[tex]dS=dS_{1}+dS_{2}=\frac{dQ}{T_{1}}+\frac{dQ}{T_{2}}[/tex]
[tex]dS=dS_{1}+dS_{2}=\frac{-dQ}{T_{1}}+\frac{dQ}{T_{2}}[/tex]
But maybe i should take into account a small change in the temperature [itex]T_{1}[/itex] and in [itex]T_{2}[/itex] just to get a formal demonstration, don't you think?
Yes, I suppose strictly you should say ΔS1 ≥ -dQ/(T1 + ΔT1) and ΔS2 ≥ dQ/(T2 + ΔT2) (ΔT1 being negative). You will probably also need to specify T1 + ΔT1 > T2 + ΔT2, i.e. the deltas are small enough that the temperatures do not cross over.
 
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Related to A demonstration on the necessary positive change in the entropy

1. What is entropy and why is it important?

Entropy is a measure of the disorder or randomness in a system. In other words, it measures the amount of energy in a system that is not available to do work. It is important because it is a fundamental concept in thermodynamics and helps us understand the direction of energy flow and the efficiency of processes.

2. How does entropy relate to positive change?

Entropy can only increase or remain constant, never decrease. This means that in any process, the total entropy of the universe will always increase. In order for a change to be positive, there must be an increase in order or decrease in disorder, which goes against the natural direction of entropy. Therefore, achieving positive change requires an input of energy to decrease entropy in a specific area.

3. Can entropy be reversed?

No, entropy can never be reversed. Entropy is a natural and irreversible process. However, it can be decreased in a localized area with the input of energy. For example, refrigerators use energy to decrease the entropy inside to keep food cold, but the overall entropy of the universe still increases.

4. How can positive change be achieved in the face of increasing entropy?

Positive change can be achieved by using energy to decrease entropy in a specific area. This can be done through processes such as organization, purification, and maintenance. However, it is important to note that this requires a constant input of energy, as entropy will continue to increase in the universe.

5. What are some examples of positive change in the face of increasing entropy?

Some examples of positive change in the face of increasing entropy include the growth and development of living organisms, the formation of complex structures like crystals, and the organization of systems such as a city or a company. These processes require an input of energy to decrease entropy and maintain order.

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