A heat transfer question for the chemistry tutors

Question

If I drop a hot solid of mass m₁, specific heat c₁, and temperature T₁ in a cool liquid of mass m₂, specific heat c₂, and temperature T₂, I know they will reach an equilibrium temperature T that may be found from

 

m₁c₁(T₁ - T) = m₂c₂(T - T₂).

 

But what if I mix a hot gas and a cold gas?  Each gas has its own c_p and c_v.  Can I still use the relation above?  Do I use c_p or c_v, or is there another method to find the equilibrium temperature for a mixture of gases?

Answer 1

Very good question, Arturo!  I had not thought about this in a while.  The above expression assumes that, for solids and liquids, with no phase change, the heat transfer is directly related to temperature change.  My first thought is that, for gases, it can be a bit more complicated.  However, at a fundamental level, it might be very similar.  The two specific heat expressions just refer to "at constant pressure" and "at constant volume," so -- once you have determined if either of those cases applies -- you may very well be able to use the same expression, as long as you can assume that all the heat coming out of one gas is going into another.

 

There are plenty more situations gases can be in, but -- as a first cut -- I would say the same expression should work.  But let me think about it more.  Thanks for the question!