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Work done on condensing steam in a piston

  • Thread starter mettw
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  • #1
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Homework Statement



A cylinder has a well fitted [itex]2.0Kg[/itex] metal piston whose cross-sectional area is [itex]2.0cm^2[/itex]. The cylinder contains water and steam at constant temperature. The piston is observed to fall slowly at a rate of [itex]0.30cm/s[/itex] because heat flows out of the cylinder walls. As this happens, some steam condenses in the chamber. The density of the steam inside the chamber is [itex]6.0\times 10^{-4} g/cm^3[/itex] and the atmospheric pressure is [itex]1.0 atm[/itex].

(b) At what rate is heat leaving the chamber?
(c) What is the rate of change of internal energy?

Homework Equations



(b) I correctly found that [itex]Q = 0.813 J/s[/itex] (Book says [itex]0.814 J/s[/itex])

For (c)

[itex]\Delta U = Q - W [/itex]

The Attempt at a Solution



[tex] W = mgh = 2.0\times 9.81\times 0.003 = 0.05886 J/s[/tex]
or, since the pressure is constant,
[tex] W = p \Delta V = 1.013\times 10^5\times 2.0\times 10^{-4}\times 0.003 = 0.06078 J/s[/tex]
Which gives, since heat is flowing out and work is being done on the system, [tex] \Delta U = -0.813 + 0.05886 = -0.75414 J/s[/tex]

But the textbook says: [tex] \Delta U = -0.694 J/s[/tex]

Which implies that the value of the work should be twice as large as the value I calculated.

If they take the work to be [tex]W = mgh + p\Delta V[/tex] wouldn't that be counting the same work twice? After all, the piston isn't doing any work, rather gravity is doing work on the piston.
 

Answers and Replies

  • #2
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The steam is not at atmospheric pressure.
 
  • #3
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Here's another way of looking at it:

The piston is essentially in mechanical equilibrium (not accelerating), so the force on the top face of the piston differs from the force on the bottom face of the piston by the weight of the piston. So the pressure exerted by the piston on the steam is atmospheric pressure plus the weight of the piston divided by its cross sectional area.
 
  • #4
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Thanks!

I see now: [itex]mg/A \times \Delta V = mgh[/itex]. The correct equation is [tex]W = (p_{atm} + mg/A)\Delta V = 0.11964 J/s ,[/tex] which gives the correct answer.

I suppose I should have done a more detailed diagram of the problem rather than just trying to work it out in my head.

Thanks a heap.
 
Last edited:

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