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.