I was thinking that the primary purpose of
this thread was to question if (I^2)R(int)
is the proper way to characterize these
internal losses. Therefore, I was trying
to avoid using the R(internal) idea yet
come up with a statement as to why
the terminal potential difference
is less than the emf.
As I expressed it, the primary idea is
that when a thermodynamic process
is carried out in a nonreversible
manner, we necessarily cannot obtain
all the reaction energy in the
form of work. I was not specifying
the mechanism by which the non-work
energy shows up as heat.
Michael D. Edmiston, Ph.D.
Professor of Chemistry and Physics
Bluffton College
Bluffton, OH 45817
(419)-358-3270
edmiston@bluffton.edu