From: jmgreen@pilot.njin.net (Jim Green)
Newsgroups: sci.physics
Subject: Question 5 Reprive
Date: 30 Dec 92 23:58:26 GMT
Organization: Rutgers Univ., New Brunswick, N.J.



Thank you all for your responses to my questions.  Especially to
those of you who have diverted your attention from the likes of
temporal inertia and other imponderables, which I am sure are
more fun.  I am one of those who, long after student days and
now back in the classroom in another guise, realizes that there
is still a lot to learn.  I therefore enjoy these little chats
and, eventhough they extend over several weeks, there is time to
meditate about the issue and perhaps finally get it right.  I am
still meditating about the following one:
 
First, I'll repeate the question; then I'll ruminate a bit on
what the answers might be (about which I continue to solicit
comments), and then I will add a rambling commentary on the
responses received to date which only the truly interested will
tolerate, others having n'd long since.
 
===============================================================
 
A system consists of a compressible fluid which is contained
within a vertical cylinder. The cylinder has adiabatic walls
OBand bottom and is fitted with the following: a heavy adiabatic
piston, a small diathermal window in the side, an internal
paddle wheel mounted on an horizontal shaft which protudes
through the side wall and is not thermally conducting, and an
electrical resistor which may be connected to an external
battery.
 OB
Consider further the following independent processes:
 
1)  Additional weight is added to the piston slowly thereby
    compressing the internal fluid further.
 
2)  The shaft is turned thereby agitating the internal fluid.
 
3)  A hot acetylene torch flame is applied to the window.
 
4)  An electrical current is applied to the resistor.
 
State which of these processes does work and which does heat
and substantially defend your answer. If you know of a
reference where this kind of issue is discussed, please give it.
 
The important point here is to explain the rules _you_ use to
decide which is work and which is heat. The best answers will
either use the phrase "do heat" or not use the word heat at all.
Don't merely state your opinion which is which, state _why_ you
come to your conclusions.
 
===============================================================
 
There are several reasonable definitions for "work" and "heat":
 
For all of the following, heat is ABSOLUTELY NOT something
which "flows".
 
A)  Heat is microscopic work done by molecules very much the
same as molecules of the piston do work, except that micro-
scopic work is random while macroscopic work is concerted.
Thus, the torch, the wheel, and the resistor do heat and only
the piston does work.
 
B)  Heat is an energy exchange induced by a temperature
difference.  Certainly this conformes to the views of
Clausius and Carnot except of course they thought that heat
"flows".  Thus, the resistor (with the assumption that there
is a dT between the resistor and the fluid) and the torch do
heat and the wheel and the piston do work.
 
C)  Work is any process which can be written in the form
W = |F.ds.  The resistor does W = |EdZ = |Pdt = |EIdt; the
wheel does W = |Tdtheta: and the piston does W = - |PdV.
So only the torch does heat.
 
D)  Doing heat leads to an increase in entropy and work not so.
Certainly this is true for reversible processes.
Reversible W = - |PdV does not change the entropy, while
reversible DQ = |TdS does.  However, ALL irreversible processes
change the entropy.  Hence, only the piston does work and only
to the extent that the process is reversible.
 
E)  It is obvious what is heat and all else is work!
 
Comments are encouraged re the above.
 
================================================================
 
Unless you are very interested, delete now.
The rest is really babbling.
 
Next, may I give some incidental reflections on an amalgam of
some points which were posted by several respondents to this
question.  Quoted material is preceded by >> with only slight
editing.  The authors are now lost to me.
 
>>...I think it is a sloppy and perhaps misleading term that
>> you use, "do heat."  I think one can add heat, or remove heat,
>> but doing heat is something you must define, since it is not
>> a widely used term.
 
This response was of course predictable -- the question did
intentionally bait a bit -- but, folks, you would not say
"add work" or "remove work" or "work is energy transferred",
would you?  Then why say that about heat?  Haven't we gotten
past thinking that heat is _inside_ the object?  It was my
belief that Rumford and, later, Joule laid to rest the concept
of caloric.  Why do we continue to nourish this concept?
If you wish to do so, tell me just what is meant when it is
said --
 
>> HEAT:   A measure of the energy transferred by thermal
>> process.
 and
>> ...flame will heat the window....temperature difference
>> between...therefore, heat will flow.
and
>> ...temperature is defined....the idea that heat flows from
>> hot to cold),
and
>>  The paddle, torch, and resistor give heat to the fluid.
 
Just what is it that is "flowing"?  What is it that is being
"given"? This simply looks like bad physics to me!
 
>> ...have never heard of the term "diathermal"........
 
Really?!!  From greek: dia is to pass through thus diathermal
is to "pass heat through" ie conducting.  The term was of
course coined before the idea of caloric was discredited
(at least in some circles :-)); it is also an example of gross
misuse and misleading use of language (along with "heat
capacity" ugh!).  Most standard introductory texts explain
the word "diathermal" -- Halliday & Resnick, Sears & Zemansky,
Serway, etc etc etc
 
>> ...weight bounces because the liquid is compressible so the
>> liquid is doing work
 
Oh my, let's not complicate the issue with irreversible work
done by the added weight.  Let's say that the weight which is
added is a grain of sand at a time -- so the work done is
reversible
 
>> ...When the torch heats the fluid it expands and does work
>> against gravity to lift the weight.
 
Well, yes, in this case it does, but only because the piston
is free to move.  It is NOT the case that the heating process
will always cause work to be done by the piston -- I guess
that is obvious to all.
 
>> ...since the fluid is compressible that it will also expand
>> when heated....
 
That is an interesting conclusion.  It's probably true but how
does one show a meaningful functional relationship between
compressibility and thermal expansion -- even for a perfect
gas -- Help me out here!
 
>> As justification for my answers, I will use the following
>>definitions of work and heat.
>>
>> WORK:  Quantitatively work is expressed as  W = |F.ds
>> Where F is some [generalized] force, which may be a function
>> of s, and ds is the incremental [generalized] displacement
>> in the vector direction of F.
 
I trust that there is no question in anyone's mind that the
slowly-added sand does work on the system.  The important
question here is whether the paddle wheel does work or heat.
Is W = |T.dtheta in this case?  Well then is W = |E.dZ for the
battery and W = |H.dB for a magnetic field, etc.?
 
>> ...all of the above processes change the internal energy of
>> the system, but work is only done when an external source
>> can be shown to be applied over path.
 
I am listening intently, but I am very worried about this "path".
ShoOBuld one expect that this "path" have the units of length?
 
>> The transfer of heat [yuch] alters the internal state of a
>> system, but through the contact of surfaces with different
>> temperatures.  No external force is needed.
 
This seems very important!  If we are to use the concepts of
Carnot and Clausius, then we must assume that doing heat is
associated with a temperature difference.  But, if this is the
case then the resistor does work not heat, because it is held
at the same temperature as the fluid. Again the assumption is
that the process is "slow" and the system remains in equilibrium.
But note that the process is NOT reversible; electricity cannot
be made to flow in the reverse direction with an incremental
change in temperature.  So the entropy must increase.
 
>> 1)  W  = - |PdV.
 
>> 1) This is work, done reversibly.  PdV = FdL = work done.
 
No one disagrees here.  (Except that the process need not be
reversible)  If it is reversible, then there is NO CHANGE in
the entropy.
 
>> 2)  W = |T.do .....the work done will be converted to heat
>> by means of viscous dissipation of the resultant fluid
>> motions, this heat will raise the temperature of the fluid,
>> and thus the pressure, which in turn will DO work ON the
>> piston, which the piston will store in potential energy,
>> manifested as a displacement of the piston.
 
>> 2)  The shaft is turned thereby agitating the internal fluid.
>> This is work, done irreversibly.
>> \vec T \cdot \vec omega dt = work done.
 
First, any work done is NOT "converted to heat"!  Yes, indeed,
the temperature rises, but that is due to the increase in
internal energy!  May I again ascend my soapbox:  Look folks,
dU = W + Q.  Either W or Q can change the internal energy.
Have we misunderstood the relation Q = mcdT? This is a very
misleading equation.  It should be Q = dU = mcdT. It is dU that
is mcdT not Q!  U(T) is a function of temperature, not Q!
 
In any case the process is irreversible and entropy increases.
 
>> 3)  The torch will transfer heat to the fluid by means of
>> radiative heat transfer, driven by the temperature difference
>> between the inside of the system and torch flame ....
>> (Assuming that the temperature of the flame is higher than
>> the fluid.)
 
Certainly at these temperatures there is some radiation absorbed
by the fluid, but there is much more microscopic work done on
the window and thence upon the fluid by the very high speed
collisions of the gas molecules with the window and thence by
the window molecules upon the fluid.
 
>> 3)  A hot acetylene torch flame is applied to the window.
>> This is heat transfer.  You have a thermal energy source
>> (energy)  (From the reaction of burning) raising the
>> temperature of the gas which thermally heats one side of the
>> window, which propagates through the window to the other side,
>> which 'heats' the gas inside.  Almost no energy is
>> transferred due to the expansion of the window.
 
There is likely no disagreement that the torch does heat, but
I have no idea of a precise meaning of "heat transfer"!
 
>> 4)  An electrical current is applied to the resistor.
>> Since the 'system' is the gas inside the cylinder, the
>> resistor gets 'hot,' and the gas inside the cylinder is
>> 'heated.'  Almost no energy is transferred due to the
>> expansion of the resistor.
 
No one said that the fluid is a gas, even so, if the resistor
and the fluid remain in equilibrium, does the resistor then do
work or heat?  I have no meaning for "energy transferred" in
this case, and certainly not for "energy transferred due to the
expansion of the resistor".
 
>> 4) An electric current applied to a resistor of finite
>> resistance will dissipate heat in the resistor.  This will be
>> transferred as heat to the fluid, and as in 2) above,.......
and 
>> 4) If the resistor is contained in the fluid, the battery
>> will do work against the resistor to force the flow of current
>> (the resistor is grounded). Also, as the resistor heats the
>> fluid ....
and
>> 4) External work will be done by the battery:
>> Work=Integral(vi)dt;where v=potential and i=current gas by
>> being 'hotter' than the gas and being in 'thermal contact'
>> with the gas will fill in the blanks.
 
By the way both Sears and Salinger and Zemansky and Ditman  say
that the resistor does work on the system!!!
 
>>......This is work.  If any I*V loss is called work, even if
>> you're just looking at the wires entering the box without
>> looking at the generator, then it is still work.  The resistor
>> dissipates the energy of the electrons via scattering
>> processes....... The energy being dissipated heats the
>> resistor, and the only mechanism for transfer of energy to
>> the gas seems to be heat.
 
Does this mean that the resistor does work or does it do heat???
 
===================================================================
 
If you have gotten this far, you are very interested indeed,
you are on vacation, or you need to be doing something
productive and don't want to. I know; me too!
 
Best ......  Jim Green
-- 
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