A saturated mixture is heated, volume is doubled and pressure remains the same.

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A saturated water mixture is heated from an initial state with a pressure of 200 kPa, a mass of 0.5 kg, and an initial volume of 0.3 m^3, resulting in a doubled volume of 0.6 m^3. The temperature at state 2 is calculated to be 120.12°C using pressure tables and interpolation. To find the specific volume at state 2, the formula v2=V2/m is applied. The specific internal energy can be determined from the table and multiplied by the mass to calculate the internal energy. The discussion focuses on using thermodynamic tables to solve for temperature and internal energy in a saturated mixture.
allly
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Homework Statement



A saturated water mixture is heated.

P1=200kPa
m=0.5kg
V1=0.3m^3
V2=0.6m^3

find
T1
T2
Internal Energy

2. The attempt at a solution

Using the pressure tables and interpolation I found T2=120.12

stuck on T2 and internal energy.
 
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allly said:
2. The attempt at a solution

Using the pressure tables and interpolation I found T2=120.12

stuck on T2 and internal energy.

Since you have the mass, for state 2, you can get the specific volume v2 by using v2=V2/m and read that off from the table.

Your table will also have values for specific internal energy, which when multiplied by mass, gives the internal energy.
 
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