Work done on condensing steam in a piston

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Homework Help Overview

The problem involves a metal piston in a cylinder containing water and steam, where the piston is observed to fall slowly due to heat loss. The discussion focuses on calculating the rate of heat leaving the chamber and the change in internal energy as steam condenses.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore different methods for calculating work done by the piston, questioning the assumptions about pressure and the forces acting on the piston. There is a discussion about whether the work should be calculated using gravitational force or pressure differences.

Discussion Status

Some participants have provided alternative perspectives on the mechanics involved, leading to a revised understanding of the work done on the system. There is acknowledgment of differing calculations and interpretations of the work involved, but no explicit consensus has been reached.

Contextual Notes

Participants note that the steam is not at atmospheric pressure, which may affect the calculations. There is also mention of the need for a more detailed diagram to clarify the problem setup.

mettw
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Homework Statement



A cylinder has a well fitted 2.0Kg metal piston whose cross-sectional area is 2.0cm^2. The cylinder contains water and steam at constant temperature. The piston is observed to fall slowly at a rate of 0.30cm/s 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 6.0\times 10^{-4} g/cm^3 and the atmospheric pressure is 1.0 atm.

(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 Q = 0.813 J/s (Book says 0.814 J/s)

For (c)

\Delta U = Q - W

The Attempt at a Solution



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

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

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 W = mgh + p\Delta V 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.
 
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The steam is not at atmospheric pressure.
 
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.
 
Thanks!

I see now: mg/A \times \Delta V = mgh. The correct equation is W = (p_{atm} + mg/A)\Delta V = 0.11964 J/s , 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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