Re: A "heat" question

[John Mallinckrodt <ajm@CSUPOMONA.EDU>]

Bob LaMontagne wrote:

> May I give an example to see if I follow you correctly? Two
> containers of fluids at different temperatures are put in thermal
> contact with each other. At the interface molecules collide. Because
> of the finite collision time, there are equal and opposite forces on
> each molecule along with corresponding displacements during the
> collision time - work is done. The work results in one molecule
> ending up with a higher KE value than it had and the other ends up
> with a lower one. On average, the higher temperature molecules will
> tend to lower values of KE, and the lower temp ones will tend to
> higher values. Eventually, both containers come to the same
> temperature and the average work done tends to zero - positive and
> negative canceling over arbitrary time intervals.  Therefore this
> "Q" is simply work done at the molecular level.

and Jim Green wrote:

> Bob gets an "A".

I'd like to offer a more qualified endorsement.  The picture of heat
being nothing more than microscopic work is highly appealing--I use
it myself--but it fails to answer the critically important question:
"Why does reversible heat change the entropy of a system while
reversible work does not?"  Entropy is "fundamental" so reversible
heat is *fundamentally* different from reversible work.

As has been discussed here before, quantum mechanics offers
significant insight into this difference and suggests that the "heat
as microscopic level work" point of view is at least somewhat naive.

See, for instance

 http://lists.nau.edu/cgi-bin/wa?A2=ind0110&L=phys-l&P=R42727&D=0

and

 http://lists.nau.edu/cgi-bin/wa?A2=ind0111&L=phys-l&D=0&P=25612

--
John Mallinckrodt        mailto:ajm@csupomona.edu
Cal Poly Pomona          http://www.csupomona.edu/~ajm

This posting is the position of the writer, not that of SUNY-BSC, NAU or the AAPT.