Thermodynamics-heating water in piston help

[JGreen48]

Homework Statement

A cylinder/piston arrangement contains water at 105 C, 85% quality with a volume of 1L.
The system is heated, causing the piston to rise and encounter a linear spring. At this
point, the volume is 1.5 L, piston diameter is 150 mm, and spring constant is 100 N/mm.
The heating continues, so the piston compresses the spring. What is the cylinder
temperature when the pressure reaches 200 kPa.

Homework Equations

F/A=pressure
F=K*d

The Attempt at a Solution

From what I stated with. I solved for the Force on the spring.
F/(pi/4)(.150^2)mm=200,000 Pa
F=3534.29N

Then I found out the displacement of the spring.
3534.39=100*d
d=35.3439mm

That is where I am stuck at. What I tough might work is to find the volume after the spring is compressed. That is just over 1.5L and comparing that to the saturation tables it is above the saturated vapor. So it would be a gas. But I don't think that is correct. If anyone can help straighten me out and point me in the right direction. Thanks.

 

[Q_Goest]

The expansion from 1 L to 1.5 L is done at constant pressure with heat input. So the final state (ie: quality) of the water/water vapor, can be found from the tables. Do you know how to do that?

Once the water/vapor mix begins pushing against the piston, it is doing work and follows a line of constant entropy. If you can find the state at 1.5L above, then you can find the entropy at that state from the tables. Follow the line of constant entropy up to the pressure specified to find the final state.

 

[piercas]

Based on the given information, it seems that the system is undergoing a phase change from liquid to gas as the volume increases and the quality decreases. In order to determine the temperature at which the pressure reaches 200 kPa, we need to use the ideal gas law, which relates pressure, volume, temperature, and number of moles of gas.

First, we need to calculate the number of moles of water present in the system at 105 C and 85% quality. Using the ideal gas law, we can find the number of moles of water in 1L of the system:

P1V1 = nRT
n = (P1V1)/(RT)
n = (200,000 Pa)(0.001 m^3)/(8.314 J/molK)(378.15 K)
n = 0.0635 moles

Since the quality is given as 85%, we can calculate the amount of liquid and vapor present in the system:

n_liquid = (0.0635)(0.85) = 0.054 moles
n_vapor = (0.0635)(0.15) = 0.0095 moles

Next, we need to determine the volume at which the pressure reaches 200 kPa. Using the ideal gas law again, we can solve for the volume at this pressure:

P2V2 = nRT
V2 = (nRT)/P2
V2 = [(0.054 moles)(8.314 J/molK)(378.15 K)]/(200,000 Pa)
V2 = 0.000999 m^3

Thus, we can see that the volume at 200 kPa is very close to the initial volume of 1L. This indicates that the system is still mostly in liquid form, but with a small amount of vapor present. To find the temperature at which this occurs, we can use the saturation tables to find the temperature corresponding to a pressure of 200 kPa and a quality of 5%. This can be done by interpolation or by using a table of values.

Using this method, we can find that the temperature at which the pressure reaches 200 kPa is approximately 111.5 C. This makes sense, as the system is still mostly liquid but with some vapor present, resulting in a slightly higher temperature compared to the initial temperature of 105 C.

In conclusion