2 kg of saturated liquid water is to be boiled at a constant-pressure of 100 kPa in a piston-cylinder device until the water is all saturated vapor. Determine the work and heat transfer for the process.
Given: Water; m = 2 kg; P = 100 kPa; x1 = 0; x2 = 1.0
Assume: ?KE = ?PE = 0.
What will be an ideal response?
Solution: The First Law for closed systems reduces to
Q – W = m (u2 – u1)
For a constant pressure process, the moving boundary work is W = P (V2 – V1) = mP(v2 – v1)
For water at 100 kPa:
v1 = 0.0010431 m3/kg
u1 = 417.33 kJ/kg
v2 = 1.694 m3/kg
u2 = 2505.55 kJ/kg
So, W = 339 kJ
Q = 4515 kJ
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