Question 2.6: Calculate ΔU and ΔH for 1 kg of water when it is vaporized a...
Calculate ΔU and ΔH for 1 kg of water when it is vaporized at a constant temperature of 100°C and a constant pressure of 101.33 kPa. The specific volumes of liquid and vapor water at these conditions are 0.00104 and 1673 m^3·kg^{−1}, respectively. For this change, heat in the atexmount of 2256.9 kJ is added to the water.
We take the 1 kg of water as the system because it alone is of interest, and we imagine it contained in a cylinder by a frictionless piston that exerts a constant pressure of 101.33 kPa. As heat is added, the water evaporates, expanding from its initial to its final olume. Equation (2.12) as written for the 1 kg system is:
ΔH = Q = 2256.9 kJ
By Eq. (2.14),
ΔU = ΔH − Δ(PV ) = ΔH − P ΔV
For the final term:
P ΔV = 101.33 kPa \times (1.673 − 0.001) m^{3}
= 169.4 kPa⋅ m^{3} = 169.4 kN⋅ m^{-2}⋅m^{3} = 169.4 kJ
Then
ΔU = 2256.9 − 169.4 = 2087.5 kJ