A demonstration on the necessary positive change in the entropy

[Rulonegger]

Homework Statement

Hello everyone. My problem is as follows: In a spontaneous process where two bodies at different temperatures T_{1} and T_{2}, where T_{1}>T_{2}, are put together until they reach thermal equilibrium. The number of atoms or molecules of the first is N_{1} and N_{2} for the second one, with N_{1} \neq N_{2}, and they have heat capacities equal to C_{V_{1}}=aN_{1}k and C_{V_{2}}=aN_{2}k, with a given with the appropriate units. Past some sufficiently large time, the system reaches a temperature T, provided that T_{1}>T>T_{2}, 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. \bigtriangleup S>0

Homework Equations

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

The Attempt at a Solution

I think that i have to use the two cases (N_{1}>N_{2} and N_{1}<N_{2}), and using the fact that T_{1}>T>T_{2}, 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.

 

[haruspex]

Don't you need conservation of energy? Otherwise there's nothing to stop both ending arbitrarily close to T₂.

 

[Rulonegger]

Actually I've used conservation of energy, and i can determine the temperature T of equilibrium, which is T=\frac{N_{1}T_{1}+N_{2}T_{2}}{N_{1}+N_{2}}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 T in the inequality, but i get nothing again.

 

[haruspex]

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.

 

[Rulonegger]

Thanks haruspex. If i understand, you say that dS=dS_{1}+dS_{2}=\frac{dQ}{T_{1}}+\frac{dQ}{T_{2}}With the fact that T_{1}>T_{2}, 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 C_{V_{1}} nor C_{V_{2}}? But maybe i should take into account a small change in the temperature T_{1} and in T_{2} just to get a formal demonstration, don't you think?

 

[haruspex]

dS=dS_{1}+dS_{2}=\frac{dQ}{T_{1}}+\frac{dQ}{T_{2}}

dS=dS_{1}+dS_{2}=\frac{-dQ}{T_{1}}+\frac{dQ}{T_{2}}

But maybe i should take into account a small change in the temperature T_{1} and in T_{2} 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.