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.

[edit] : 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..!
 A: 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.
A: 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?
