# Help with this interpolation (change in entropy while heating water)

• Chemistry
• Noob of the Maths
In summary, the conversation discusses a problem with interpolation of state values in a superheated water table. The final state is a combination of saturated liquid and saturated vapor at 240 F and 25 psi, making interpolation unnecessary. The speaker suggests using the saturated water table and the mass fractions of liquid and vapor to calculate the combined entropy in the final state.

#### Noob of the Maths

Homework Statement
A piston-cylinder device initially contain 7 lbm of liquid water at 25 psia and
75 F the water is now heated at constant pressure by the addition of 4520
Btu of heat. Determine the entropy change of the water during this process.
Relevant Equations
Heat and pression
Hello, everyone :).
I try to resolve this common problem. But, when i got in the interpolation of state 2, the values not make the sense.
I have 25 psia and 75 F, but, in the superheated water table, there are not values with 25 psia (only 20 psia and 40 psia). And, the temperature values starting with 1000 F, so, its not possible the interpolation.

Last edited by a moderator:
The final state is not superheated. It is a combination of saturated liquid and saturated vapor at 240 F and 25 psi.

Noob of the Maths
Chestermiller said:
The final state is not superheated. It is a combination of saturated liquid and saturated vapor at 240 F and 25 psi.
So, i can use the saturated water table and use the 240 F with that 24.985 psia, and just use btu/lbm *R of the Sfg? In this case interpolation its not necessary?

Noob of the Maths said:
So, i can use the saturated water table and use the 240 F with that 24.985 psia, and just use btu/lbm *R of the Sfg? In this case interpolation its not necessary?
View attachment 289215
No way. You need to find the mass fraction liquid and the mass fraction vapor in the final state, and then use this to get the combined entropy in the final state.

Noob of the Maths