A rigid tank contains 5 lb of a two-phase, liquid-vapor mixture of H2O, initially at T = 260 F with a quality of 0.6. Heat transfer to the contents of the tank occurs until the temperature is T = 320 F. Show the process on a p-v diagram. Determine the mass of vapor, in lb, initially in the tank in the final pressure, in lbf/in^2.
V = Vf + X ( VG - VF
Various thermodynamic charts found in any thermodynamics book.
The Attempt at a Solution
I can get the first part of the question fine. I took a look at a solution for this problem, and they randomly make the statement that while in phase 1 the mixture is a saturated vapor, in phase 2 the substance is a superheated vapor. This is what I need help with. I am not exactly sure why the substance is a superheated vapor in phase 2.
I understand that since the the system is within a "rigid tank" that the specific volume must remain constant between the two phases. I've calculated the specific volume of phase one (using the quality, temperature, and the specific volume of both fluid and gas) and I've come up with V = 7.0688 ft3/lb. I know that the specific volume must be the same in phase two, but I do not have a Quality in which to calculate it.
I used the same quality from phase 1 (although I know it would not be the same) and I've also used an assumptive quality of 1 (MAX). In both cases the specific volume of phase two is less than that of phase 1. Is this why I can justify in calling the substance a "superheated vapor in phase 2?
This is my first time posting here, I hope I am not breaking any of your rules. I'm not looking for the answer to the question, I just need help understanding how I can determine why the substance became a "superheated vapor."