Work done on condensing steam in a piston

[mettw]

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.

 

[nasu]

The steam is not at atmospheric pressure.

 

[Chestermiller]

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.

 

[mettw]

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.

 

[Nickie]

I would like to clarify a few points about the work and energy involved in this scenario. First, the work being done on the piston is not the same as the work being done on the system. The work done on the piston is due to the force of gravity acting on it, while the work done on the system is due to the change in volume caused by the falling piston. These are two separate forms of work and should not be added together.

In this situation, the work being done on the system is given by the change in volume, which is equal to the change in internal energy. This is because the system is at constant temperature and pressure, so any change in volume must correspond to a change in internal energy. Therefore, the correct equation to use is \Delta U = p \Delta V.

As for the discrepancy between your calculated value and the value given in the textbook, it is possible that there is a rounding error or a difference in assumptions made in the calculations. It is also important to note that the density of the steam given in the problem may not be accurate, as it is highly dependent on temperature and pressure. Overall, the important thing is to understand the concepts and equations involved, rather than focusing on exact numerical values.