# Second principle of thermodynamic and heat flow

My question is about the rigorous proof of the fact that the heat goes from hot body to the cold one. There is a part of the proof I don't understand.

We consider 2 systems : one is at temperature $T_1$ and the other at $T_2$. The ensemble of those two systems is isolated. I assume they only exchange heat (no work).

I apply the second principle of thermodynamic on the ensemble :

$$dS=dS_1+dS_2=\delta S_1^e+\delta S_2^e + \delta S_1^c + \delta S_2^c=\delta S^c \geq 0$$

Where : $\delta S^e$ is the entropy exchanged and $\delta S^c$ is the entropy created.

Now, I know : $\delta S_1^e = \frac{\delta Q_1}{T_1}$, $\delta S_2^e = \frac{\delta Q_2}{T_2}$

Applying the first principle on $1+2$ I find that $\delta Q_1 = - \delta Q_2$, I end up with :

$$dS=\delta Q_1(\frac{1}{T_1}-\frac{1}{T_2})+\delta S_1^c + \delta S_2^c = \delta S^c \geq 0$$

To prove the direction of the heat transfer, I need to have : $\delta S^c -(\delta S_1^c + \delta S_2^c) \geq 0$

But using only the classical thermodynamic (first and second principle) I don't know why it would be true ?

Do we need extra postulate to prove it ? I thought that the heat transfer direction can directly be shown using classical thermodynamics.

 : What I tried in link with the answer.

Ok, let's assume I have two systems $1$ and $2$ with temperatures $T_1 \neq T_2$.

I take in consideration an interface system $I$ between those two systems. I'm forced to take it because else I couldn't have thermodynamic equilibrium and two different temperatures for my systems.

I write the variation in internal energy : $dU=dU_1+dU_2+dU_I$.

I assume the system $I$ is very small, so it's internal energy variation is negligible (conceptually I can take it as small as I want, thus in the limit its variation of energy can be considered as $0$).

$$dU_I=0$$

My whole system $\{1+2+I\}$ is isolated, so $dU=dU_1+dU_2=0$. What's more, $1$ only exchange heat (such as $2$), so I have :

$$dU_1=C_1(T_1) dT_1=\delta Q_1=T_1(dS_1-\delta S^c_1)$$

$$dU_2=C_2(T_2) dT_2=\delta Q_2=T_2(dS_2-\delta S^c_2)$$

$$dS=dS_1+dS_2+dS_I=C_1 \frac{dT_1}{T_1}+C_2 \frac{dT_2}{T_2}+(\delta S^c_1+\delta S^c_2 + \delta S^c_I)=\delta S^c \geq 0$$

Where the last equality use the fact the entropy of the whole system must increase. And I don't necesseraly have $\delta S^c=\delta S_1^c+\delta S_I^c+\delta S_2^c$ : the created entropy is not additive.

In the end, I have :

$$C_1 \frac{dT_1}{T_1}+C_2 \frac{dT_2}{T_2}=(\delta S^c-(\delta S^c_1+\delta S^c_2 + \delta S^c_I))$$

which is neither positive or negative, so I don't really see how to conclude. And I don't find the same entropy variation that you have. How did you end up with such a result ?

If I assume $C_1=C_2=C$ independant of temperature (I would like to avoid any such assumptions but let's assume it just to see some of my problems), I would have something like :

$$C ln(\frac{T_1^f}{T_1^i}\frac{T_2^f}{T_2^i})=S^c-(S^c_1+S^c_2 +S^c_I)$$

And I don't see how to conclude anything from here... :S

[edit 2] :

As you suggested for now, I don't take in account the entropy creation terms.

I assume : $T_1^i \leq T_2^i$. I thus need to prove $T_2^f-T_2^i \leq 0$ (the hot system gets cold and reciprocally).

I assume that my $\Delta S$ is only due to the log (i forget about the creations as suggested).

Thus I have the following inequality :

$$\Delta S \geq 0 \Leftrightarrow 1-(\frac{T_2^f-T_1^f}{T_2^i+T_1^i})^2 \geq 1-(\frac{T_2^i-T_1^i}{T_2^i+T_1^i})^2 \Leftrightarrow (T_2^f-T_1^f)^2 \leq (T_2^i-T_1^i)^2$$

Thus :

$$-(T_2^i-T_1^i) \leq T_2^f-T_1^f \leq T_2^i-T_1^i$$

So, I find :

$$T_2^f-T_2^i = \Delta T_2 \leq T_1^f-T_1^i = \Delta T_1$$

$$\Delta T_2 \leq - \Delta T_2 \Leftrightarrow \Delta T_2 \leq 0$$

We find the good result.

But now, why could I "forget" about those creation terms ?

Do you use an argument like the entropy is a state function and thus its variation only depends on the initial and final states.

So we choose a reversible transformation in all the reservoirs and in the global system that has the same final and initial temperatures ?

Using this we find a positive variation of entropy.

Is it the final idea ?

The little thing that confuse me is that either the transformation is reversible or not we would have the same heat exchanged (because same starting and ending temperature in both systems). So it is like "nothing change" physically if the transformation is reversible or not here.

But maybe it is not the idea..!

• I don't understand what you mean by entropy "exchanged" and "created". $dS_{1,2}$ is the entropy variation and it's equal to $\delta Q_{1,2}/T_{1,2}$... – valerio Mar 8 '18 at 13:15
• @valerio In general we have $dS \geq \frac{\delta Q}{T}$. So we define the entropy exchanged by $\delta S^e=\frac{\delta Q}{T}$ and the entropy created by $\delta S^c \geq 0$ such that $dS=\delta S^e + \delta S^c$. – StarBucK Mar 8 '18 at 13:21
• I agree with @valerio. There is no conversation of entropy or any way to keep track of a flow of it. How can you distinguish between those two types of entropy? – Aaron Stevens Mar 8 '18 at 13:21
• Ok, then said differently, I don't see why the variation of entropy in $1$ and $2$ would be equal to $\frac{\delta Q}{T}$ and not be greater than this according to second principle of thermodynamics. $1$ and $2$ can follow irreversible transformations for example (in addition to the fact that $\{1+2\}$ eventually follows irreversible transformation). – StarBucK Mar 8 '18 at 13:22

The reason that this is not working out correctly is that the entropy exchange for each of the bodies is not being expressed correctly. The entropy exchange takes place at the interface between the two bodies, and the temperature at the interface $T_I$ is neither T1 nor T2. It is somewhere in-between the two starting temperatures (The temperature at the interface cannot be both T1 and T2 at the same time). So the exchanged entropy between the bodies should, in this example, be $$\delta S_1^e=\frac{Q_1}{T_I}$$and $$\delta S_2^e=-\frac{Q_1}{T_I}$$Accordingly, the two exchanged entropies cancel out, and you are left with:$$\delta S=\delta S^c\geq0$$
Now, if you really want to prove that the entropy can only increase if heat flows from the hot body to the cold body, you should derive the equation for the change in entropy of each body to a final state in which the hot body is hotter and the cold body is colder (for a specified amount of heat transferred). You will find that this will come out negative. Do you know how to do this derivation?

• Thank you. I understand my mistake but now I tried to prove what you said. I introduced another system that I called "I" between $1$ and $2$ such that is has the temperature $T1$ when in contact with $1$ and temperature $T2$ but I still have a problem (I would like to check with you it is the good procedure to do before editing with calculations). – StarBucK Mar 8 '18 at 15:39
• I don't think that this approach is going to give you what you want. Suppose that the mass of each body is M and the heat capacity of each body is C. If the initial temperatures of the bodies are $T_{1i}$ and $T_{2i}$ and their final temperatures are $T_{1f}$ and $T_{2f}$. In terms of these parameters, what is the heat balance on the overall system? In terms of these parameters, what is the change in entropy of body 1? Of body 2? – Chet Miller Mar 8 '18 at 16:46
• The overall entropy change for the problem I partially posed in my previous post is expressible in the form: $$\Delta S=MC\ln{\left[\frac{1-\left(\frac{T_{2f}-T_{1f}}{T_{2i}+T_{1i}}\right)^2}{1-\left(\frac{T_{2i}-T_{1i}}{T_{2i}+T_{1i}}\right)^2}\right]}$$What is the sign of the entropy change when the difference between the body temperatures finally is greater than initially, and what is the sign when the difference between the body temperatures finally is leas than initially? – Chet Miller Mar 8 '18 at 17:52
• I edited my message with some questions ! I don't understand how you end up with this result. – StarBucK Mar 10 '18 at 17:16
• This result is mathematically exactly the same as the left hand side of your equation. To get this result, I wrote the following: $$T_{2f}=\frac{(T_{2f}+T_{1f})}{2}+\frac{(T_{2f}-T_{1f})}{2}$$$$T_{1f}=\frac{(T_{2f}+T_{1f})}{2}-\frac{(T_{2f}-T_{1f})}{2}$$and the same for the initial temperatures. – Chet Miller Mar 10 '18 at 19:46

I thought about the new approach you were suggesting in your most recent comment on my previous answer, and, after further consideration, agree that this approach will be fruitful.

So you are assuming that you have two ideal infinite reservoirs, one at $T_H$ and the other at $T_C$, and you insert a medium I between the two reservoirs that has very low mass and heat capacity (so that its change in entropy is negligible), but that is able to conduct heat between the two reservoirs via a finite thermal conductivity.

Since the reservoirs are ideal, the changes in their entropies are due only to exchange with the medium I: $$\Delta S_H=-Q/T_H$$ $$\Delta S_C=Q/T_C$$where Q is the heat leaving the hot reservoir and entering the cold reservoir. The change in entropy of the intervening medium I involves both entropy exchange (with the two reservoirs at its two boundary temperatures) and entropy creation (as a result of heat conduction within the medium):$$\Delta S_I=\frac{Q}{T_H}-\frac{Q}{T_C}+\delta^I$$The overall change in entropy of the system is equal to the sum of the three entropy changes: $$\Delta S=\Delta S_H+\Delta S_C+\Delta S_I=\delta^I\geq 0$$ But, since the change in entropy of the intervening medium is negligible, we have: $$\delta^I=\frac{Q}{T_C}-\frac{Q}{T_H}=Q\frac{(T_H-T_C)}{T_HT_C}$$Combining the previous two equations, this means that $$\Delta S=Q\frac{(T_H-T_C)}{T_HT_C}\geq 0$$ This is only possible if Q, the heat flow from $T_H$ to $T_C$ is positive. Otherwise, the 2nd law is violated.

• Thank you for your answer and sorry for my late response. I have some questions about it but I will firstly talk about the thing that cause the more problem for me. Why do you assume that the hold and cold reservoir are ideal ? Actually what I want to see here is that it is never possible to transfer heat from cold to hot without providing work. So I would like to avoid to use hypothesis at the maximum. Indeed, from what I understood the second principle was created to always avoid transferring heat from cold to hot without providing work, so it should be possible to have a very general – StarBucK Mar 10 '18 at 15:30
• proof of it ? Because here it looks like that if my reservoir creates entropy then we could imagine transferring heat from cold to hot without providing work which would be surprising ? I will take a look about your other proof now ! – StarBucK Mar 10 '18 at 15:31
• The comments in my other answer solves the problem for two bodies with finite heat capacities (not ideal infinite reservoirs). But I thought you wanted to solve the problem using something similar to your initial approach, so I introduced this answer, using infinite reservoirs to simplify the analysis. The analysis definitely shows that, without doing work, the entropy of the system increases only if heat flows from the hot to the cold. – Chet Miller Mar 10 '18 at 15:41
• I read your edited addition. For one thing, it doesn't make sense to me to include entropy generation (i.e., always having to be positive) if we are going to examine cases in which the temperature difference can increase. So your analysis to just see what the entropy change would be for both the temperature difference decreasing and the temperature difference increasing is adequate to prove what you intended to prove. Secondly, there is a difference between the heat being transferred spontaneously in an isolated system and the heat being exchanged reversibly with external sources. – Chet Miller Mar 11 '18 at 2:24
• Why doesn't it make sense to include entropy generation if the temperature difference can increase ? Btw you are talking about temperature difference between 1 and 2 at a given time, or about $T_f-T_i$ ? Bc for me we have to include entropy generation as soon as the transformation is not reversible (so no link with the difference of temperatures). I agree with your last point but I don't see where you want to go. In my problem it is indeed an internal heat transfer, but what I found strange is that either the whole thing is reversible or not nothing changes (excepted the fact the transfo – StarBucK Mar 11 '18 at 11:56