Temperature in the Maxwell-Boltzmann Distribution

I am attempting to generate the probability for the specific speed of a molecule using the Maxwell-Boltzmann distribution, but I cannot decide on which temperature I should use in the equation.

For example, I have a very, very cold liquid (let's say 70 K) at room temperature (we'll call that 296.15 K), but would I use the temperature of the liquid in the equation, or should I use the temperature of the air around the liquid?

• from wiki (MB distribution) "Particle in this context refers to either gaseous atoms or molecules, and the system of particles is assumed to have reached thermodynamic equilibrium." Therefore, liquid with different temperature than environment is not very suitable for Maxwell-Boltzmann distribution, though it might be possible for a very crude estimates (temperature gradually varing from 70K to 296.15K) Mar 1 '14 at 9:35
• I thought that Maxwell Boltzmann distributions were applicable in systems in equilibrium. In this case, you certainly are not in equilibrium. The calculation would bequite complex. I dont think you can calculate that with just temperature. I second what user26143 said. However, please let me know if you find something that counters this notion. I am interested in it. Mar 2 '14 at 5:06
• I was indeed working on a crude estimation, and the liquid is certainly not going to make it all the way up to room temperature under the circumstances that I'm in Mar 3 '14 at 3:42
• The problem is what is your system? If you are considering the liquid itself, then you direct apply 70K to the equation. But if you include the room, then you have an inequilibrium system and you cannot do it. May 30 '14 at 3:43