When water is boilded under a pressure of 2 atm, the heat of vaporization is 2.2 x 10^6 J/kg and the boiling point is 120 degree C. At this pressure, 1 kg of water has a volume of 10^-3 cubic meter and 1 kg of steam has a volume of 0.824 cubic meter.
a.) Compute the work done when 1 kg of steam is formed at this temperature.
b.) Compute the increase in the internal energy of the water.
boiling point = 393 K
number of moles = 1000 / 18.0
first law of thermo: change in internal energy = energy input - work done by system
The Attempt at a Solution
a.) W = p(V2-V1) since it is isobaric
= 2x10^5 (0.824 - 10^-3)
= 1.646 x 10^5 J (work done by water)
(Is this correct?)
b.) pV = nRT
2x10^5x10^-3 = (1000/18.0) x 8.31 x T
T = 0.43 K
therefore: change in T = 393 - 0.43
= 392.57 K
(Don't I need to know the heat capacitance to work out the heat used to increase the temperature of the water to the boiling point? If I find this, then I can simply add this to the heat of vaporization to get the total heat input. Then, from first law of thermodynamics, I can find the change in internal energy.)