# Homework Help: A demonstration on the necessary positive change in the entropy

1. Nov 23, 2012

### Rulonegger

1. The problem statement, all variables and given/known data
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. $$\bigtriangleup S>0$$
2. Relevant 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 $$N_{1}\ln{\frac{T}{T_{1}}}+N_{2}\ln{\frac{T}{T_{2}}} > 0$$

3. 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.

2. Nov 24, 2012

### haruspex

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

3. Nov 24, 2012

### 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.

4. Nov 24, 2012

### 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.

5. Nov 24, 2012

### 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?

6. Nov 24, 2012

### haruspex

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