Does invariant mass depends of temperature? From special relativity theorie we know that $E = mc^2$.
When a system acquires energy, mass becomes greater.
That is clear for kinetic energy, because we have a formula that gives m as a function of $v$: 
$$m = \gamma M$$
with $M$ the invariant mass. 
So when the system has a greater velocity, m is greater and E is greater.
But what when we heat a system?  The energy of the system also becomes greater, so m is greater too. If the system is at rest, v = 0, so m = M and E = m c² = M c².
Is it right to conclude that the invariant mass M becomes greater?
Is it reasonable to think that there is a formule (not already found, but maybe one day??) to calculate m as a function of the temperature to calculate m (or M) and E when heating a system?
Sorry for my bad English. I hope you understand my question. 
 A: The internal energy of a system must be considered as part of the total energy of that system.  This would include all of the parts that are being measured to obtain the total rest mass.
In that case we still have $E=mc^2$, where $E$ is the total energy; thus the increase or decrease in temperature changes the rest mass of that system.
OTOH, the translational (kinetic) energy of the system as a whole does not contribute to total energy in the rest frame.
Albert Einstein discussed this when analyzing his thought experiment about light emitted from a mass; see http://www.fourmilab.ch/etexts/einstein/E_mc2/www/
A: The translational/rotational energy of a system has nothing to do with its thermal energy. As you increase T of the object, you are increasing the internal kinetic energy of the phonons degrees of freedom. But, any system with a high T would emit radiation whose energy depends on T. The higher the T, the more energetic (shorter wavelength or higher frequency) the emitted photons. So, the energy that you are injecting into the system by heating it is actually able to escape through emission depending on the magnitude of its emissivity. However, the system may or may not be in thermal equilibrium with its environment. If it is in thermal equilibrium then for sure there is no net increase of its internal energy and hence no change in its invariant mass. In the second case, when it is not in equilibrium with its environment, then it is losing energy in the form of radiation across an infinite electromagnetic spectrum. The rest mass is fixed and is truly a fundamental invariant even as you heat up your system. All you are doing is to increase the internal energy and hence its total energy with respect to some frame. To back up this argument, let's think of accelerating a single charged fundamental particle such as electron. You are increasing its total energy as you accelerate it. But the increase in its total (kinetic) energy is traded off by losing energy in the form of emitted radiation. The balance between its kinetic and emitted radiation is such that its rest mass is always the same. There is actually a nice proof of this statement in Electrodynamic II in graduate physics textbook where it is shown that regardless of total (kinetic) energy of the charged particle, the emitted radiation is such that the rest mass is always conserved. Now, in an extended object, we have an Avogadro's number of such fundamental particles which could act the same if found to be free. If increase in T is not that severe, then you are triggering only free phonons but you are not changing energy levels of bound quantum states inside the sytem. 
