Re: Entropy and dQ vs dW

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"Jacques Mallah" <Jacques.Mallah.4b508d5@xxxxxxxxxxxxxxxxx> wrote in message
news:Jacques.Mallah.4b508d5@xxxxxxxxxxxxxxxxxxxx

Uncle Ben;1404059 Wrote:
Just for fun I have posted a question in Thermodynamics that bothered me
in the past. I think I have the answer, but if you want to try your
hand at solving it, see

Uncle Ben's Pages
http://greenba.com
Directory: Other Physics
File: Entropy.rtf


Ah, nice to see an actual physics question here!

"When my turn came as a faculty member, I too felt that my knowledge
of thermodynamics needed brushing up quite a bit. In partricular I was
a puzzled by the definition of entropy, S. The increase in S during a
process is defined as the integral of dQ/T, where Q is the heat that is
added to the body in question during the process and T is the absolute
temperature of the body when the increment of heat is added to it."

Hi. As you've probably figured out by now, Uncle Ben, the above formula
is not quite right. The real formula is

dS = dQ_rev / T

where Q_rev is the amount of heat that would be added to the body by a
_reversible process_, which is a process in which the entropy of some
other body changes by the opposite amount. In practice there are no
fully reversible processes because the temperature of the two bodies
would have to be the same, so the heat flow rate would be zero.
Because of the requirement of reversibility, entropy is a function of
the state of the system, _not_ of the way you got there.

For example, in the free expansion of a gas (suddenly pulling away a
partition between the gas and a vacuum chamber), there is entropy
increase despite no heat being added, and the gas ends up at the
original temperature. The amount of entropy increase is equal to the
heat required to be added for the gas in order for its temperature to
remain constant while pushing a piston with the force of its pressure,
letting the volume expand by the same amount, divided by the
temperature.

In the case of heating something by friction, the entropy increase is
equal to the heat required to heat it to the same warmer state by
reversible means divided by the temperature, integrating as needed over
the temperature range. Assuming the friction work W all goes into
heating the block, and T remains close to constant, this would indeed
equal W/T.

(T of course being the absolute temperature, such as Kelvins measure,
not on a relative scale such as Celsius.)




--
Jacques Mallah


.

Relevant Pages

  • Re: Entropy and dQ vs dW
    ... Directory: Other Physics ... process is defined as the integral of dQ/T, where Q is the heat that is ... which is a process in which the entropy of some ... fully reversible processes because the temperature of the two bodies ...
    (sci.physics)
  • Re: Entropy and dQ vs dW
    ... Uncle Ben;1404059 Wrote: ... process is defined as the integral of dQ/T, where Q is the heat that is ... which is a process in which the entropy of some ... fully reversible processes because the temperature of the two bodies ...
    (sci.physics)
  • Re: Entropy and dQ vs dW
    ... process is defined as the integral of dQ/T, where Q is the heat that is ... which is a process in which the entropy of some ... fully reversible processes because the temperature of the two bodies ... partition between the gas and a vacuum chamber), ...
    (sci.physics)
  • Re: Slow Stealth
    ... that entropy at a higher temperature. ... You have to radiate more heat to ... Free energy has some technical meanings in thermodynamics, ... expensive to get rid of the entropy by increasing the temperature. ...
    (rec.arts.sf.science)
  • Re: Lizard engines and rat engines
    ... >>> situations where entropy production is high. ... >>> Pump heat into the bottom plate, and extract heat from the upper plate. ... > that the investigator could control the heat flow exogenously. ... More heat is flowing than was possible before convection cells ...
    (sci.bio.evolution)
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