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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.