Re: Thermo 101 - 2nd Law Clarification

Thanks Man (no pun).

I read yours and Greg's responses and with the help of both I think I
finally understand things clearly.
Cheers.

The_Man wrote:
renai wrote:
I've got what is most likely a very simple thermo 101 question or
misunderstanding that I'm hoping someone can clear up regarding the
second law. I do understand the idea that when anything happens a
portion of that activity goes to heat and that it is irreversible. I
understand the idea of heat flowing from a hot reservoir to a system
which can only convert part of that heat into work it will do on its
surroundings the rest of which will be lost as heat to a colder
reservoir, i.e. no engine is 100% efficient. Then I see the equation
that says that change in entropy is equal to the heat content of a
system divided by the temperature of the system at equilibrium. But I
don't completely understand how it corresponds to the example that I
did understand above. Is this for an isolated system (i.e. no heat
transfer). If so, does that mean that the delta Q that I see in the
equation is simply the internal kinetic energy of the system.
Furthermore, if temperature is basically average internal energy, then
isn't the entropy equation effectively dividing internal kinetic energy
by internal kinetic energy? Lastly I keep getting confused about the
symbol Q. I though Q was the heat flowing in or out of a system. But
if a system is isolated there is no heat flow so why do they use Q in
the entropy equation

For the engine, there are two different q's. One is the heat flow from
the "hot" reservoir to the engine - call it q(h). Then there is the
heat flow from the engine to the "Cold" reservoir - call it q(c). The
work done w=q(h) - q(c).
The entropy change from a heat flow is q/T. Note that the hot and cold
reservoirs are at DIFFERENT temperatures (Th for the hot, and Tc for
the cold) - this is one way of thinking about the 2nd law, that you
can't have work done between two reservoirs at the same temperature.

So the total entropy is q(h)/Th -q(c)/Tc = 0 (if fully reversible)


Any help or guidance would be much appreciated.

Sincerely,

Renai

.

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