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Re: CONSERVATION OF ENERGY



On Mon, 14 Jul 1997 Leigh Palmer <palmer@sfu.ca> wrote:

1)
Ludwik has now qualified c (as I suggested) and a further answer is
appropriate.

It was qualified implicitly from the beginning by saying "all in a vacuum".

2)
I wrote that c is a constant, that no mass is lost, and that T is a state
variable. Thus the product m*c*dT is path independent. Leigh called this
quantity heat. But thermodynamics textbooks say that heat is not a state
function; it is a path-dependent quantity. That is why I think that the
phrase "a change in thermal energy" is more appropriate for c*m*dT. I am
still not convinced that this triple product, called dH (ethalpy) by
chemists, should be called heat. Enthalpy is a state function. Are we to
tell students, who studied chemistry before physics, that "ethalpy is heat"?
Enthalpy and energy are identical in our specific (vacuum) situation. Nobody
responded to that line of reasonning. Is it correct or not?

3)
Another line of reasoning was that "heat is that part of internal
energy which is transfered through a system boundary due to a difference
of temperatures". Leigh says "that definition makes the first law of
thermodynamics redundant. There should be no mention of internal energy
in the definition of heat." Please explain the redundancy pronouncement.

4)
In the example used, dQ is not a difference of temperature responsible for
the transfer of energy into the system. In fact dT=T2-T1 is a well defined
quantity while the "energy-driving" dT, if I am allowed to use this term,
is not a strictly defined quantity; the system is not in equilibrium during,
and shortly after, sliding. But, as I wrote before, I do not feel competent
to discuss the numerical nuances of the first law. I am primerilly interested
in qualitative statements which are either correct or not (in an itroductory
physics course). How can you say, that "there is no heat involved in the
problem you stated"? How can I describe the so-called "work-heat equivalency"
experiment of Joule without saying that heat is involved?

5) Leigh wrote:
The transfer of energy that occurs is, in the equilibrium thermodynamic
limit (where the system evolves sufficienly slowly that it is never far
from a state of thermodynamic equilibrium), classified as work.

I noticed this observation; it triggered my thinking about how slow is
slow enough to be acceptable? This, however, is another quantitative
consideration which would only distract us from the main topic. Work is
now described as an energy transfer process. Is this the same as to say
that "work is force times distance"? Is it correct to say that work is
"done, accomplished, performed", etc.?

6) Leigh:
You have now qualified c to be the heat capacity per unit mass at
zero constant pressure. I take it m is the mass of your sliding cube,
and dT is the temperature increase resulting from the process. Then
m*c*dT is equal to the change in the internal energy of the block in
this process. Note that I say it is equal to the change, not that it
*is* the change. The equality arises because of a concept called the
mechanical equivalent of heat. This change in internal energy is the
same as would have been achieved through heating the block through
the came temperature difference at zero pressure *or* doing an equal
amount of work W = m*c*dT on it. In the latter case the change in
the internal energy would be equal to W.

In the original formulation the cube and the block were considered as
one system whose mass is M+m and where M>>m (to make sure the plate is
always at rest). The whole thing is one system. In a later formulation
m really ment to be M+m.

The difference between saying "equal to the change" and "the change is
equal" is beyond my grasp; please elaborate, if you think it is an
important part of your argument.

A change of x on the left is "equal to the change" of x on the right.
A change of x on the left and "the change on the righ are equal". Aren't
these two ways of saying the same thing? Yes, I mean absolute values of
all changes. Another distraction from the main topic. The original
situation was simple. Why do we have so many distractions to clarify
proper thermodynamic terminology for it? Perhaps it is my fault. I will
wait to see what other people have to say about proper thermodynamic
terms for the situation.
Ludwik Kowalski