Given it takes 970 btu's to vaporize 1 lb. water at 212 degrees at sea
level. If you condense 1 lb. steam at 212 degrees at sea level do you
get back 970 btu's or is there entropy? If so how much?

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Request for Question Clarification by mvguy-ga on 26 Nov 2002 07:49 PST

What assumptions are you making in terms of losses due to inefficiency?

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Clarification of Question by 3rrotec-ga on 27 Nov 2002 05:18 PST

This is just a hypothetical question. I know there is no such thing as
100% eff. in any energy transfer but if there were and we input 970
btu's boiling water, then we condense that water, will mother nature
give us back 970 btu's or will she keep some entropy? If she does, do
we know what she keeps? Thanks! 3rrotec

Hello again 3rrotec

         I will answer this question in order to prevent accumulation 
 of confusion by letting  more comments to be posted.

 I will use SI units, since it makes reasoning about these
 matters more simple. If you need translation, you may consult pages such as
 http://www.unc.edu/~rowlett/units/dictW.html
 but I recommend you do the calculations in SI (that is in metric units).

 If you do any closed cycle slow enough, (technical and also search term
 is quasistatic) your are aproaching a reversible process. That means, no
 entropy is generated and you are  aproaching 100% efficiency.

 Since you specified 100% efficiency, the answer is yes. You get back all
 energy and entropy will not increase.



 Simple example is a cylinder with well lubricated piston.  You fill it with
 water vapor and insert piston. 

This web page shows the cycle we do talk about:
http://www.mech.utah.edu/~isaacson/tutorial/applet_work.html


 When you push on the piston you deliver energy. Piston goes up. 
 You release piston SLOWLY and piston goes back down
 and you recover all the energy. System acts like a (ideal) spring.  

We made a silent assumption that no heat escaped - your cylinder was a thermos.

  NOW - you push even harder. You may start  at the lower temperature,
but it is not necessary. Eventually part of the vapor will condense into
water.  For each gram converted you recover 'latent heat of evaporation'
which is  2500 J per gram.

  You release the piston SLOWLY and the water will evaporate again, consuming
the same amount of heat. You come back to the original state.

 The reasoning is general, valid for gas  and valid during a
phase transition as well. Any material, any shape of the cylinder....
 (That is the beuty of thermodynamics).


This can be done in real life with much less then 20% losses mentioned 
in one of the comments. 
I would guess, less then 5% if you insulate well against the heat loss.

 Key word is SLOWLY: if your water boils (with visible bubbles), if your vapor
 gets turbulent (you can see motion) then you are not slow wnough and you will
 dissipate more energy.

 I hope that answers your other question on evaporation as well.
 It is useful to separate engineering details (a given machine) from
 basic principles.

 I would recommend a textbook, such as
http://filebox.vt.edu/eng/mech/scott/steam.html

hedgie

Good answer

I wonder how DO you condense 1 lb. of steam at 212 degrees F. at sea
level?

   Would you force it back into it's original container thus making it
return to a liquid state because the molecules have no choice but to
be back to their original distance from each other, incidentally
raising the temperature - not to mention the pressure - to a
phenomenal degree in the fashion of a diesel cylinder?

   Or would you rather leave it at 'Standard Pressure', and allow the
molecules to radiate (or conduct) their energy to another medium, then
re-adhere to one another, and finally drip back into the original
container?  [possible funnel required] :)

   regards, RR

I am assuming that the 970 btu's you specify is the theoretical amount
of heat needed to be absorbed by the water to vaporize it. Based on my
understanding of physics, the actual energy expended to perform this
task will be more than 970. The difference in amount is dependant on
the quality of the method used, and can vary greatly. I have heard of
no methods of energy conversion designed that perform with greater
than 80% efficiency, and 50% or less is more common.

The same can be said of the reverse process. Though I can not think of
an efficient method to campture the heat given from condensing steam
into water, it should theoretically give off as much as was needed to
do the reverse reaction. Again, the efficiency of a system in real
life is dependant on the quality of its design, and is unlikely to
exceed 50% efficiency.

This is just a hypothetical question. I know there is no possibility
of 100% efficiency in any energy transfer but if there were can we
assume: I put in 970 btu's boiling the water.- I then condense that
steam- Does mother nature give me back 970 btu's or does she keep some
entropy? I guess what my question is: In the vaporization and
condensation process are the energies used and given back equal?