A piston-cylinder device initially contains 0.6 kg of water with a volume of 0.1 m3 . The mass of the piston is such that it maintains a constant pressure of 1000 kPa. The cylinder is connected through a valve to a supply line that carries steam at 5 MPa and 500 o C. Now the valve is opened and steam is allowed to flow slowly into the cylinder until the volume of the cylinder doubles and the temperature in the cylinder reaches 280 o C, at which point the valve is closed. The pressure remains constant during the process. Determine:

Hello the missing piece of your question is attached

question : Determine mass of steam that has entered ( in kg )

answer : 0.206 kg

Explanation:

V1 = 0.1 m^3 ,

v' = V1 / m1 = 0.1 / 0.6 = 0.167 m^3/kg

V2 = 0.2 m^3

using the steam tables

at ; P = 1000 kPa, v' = 0.167 m^3/kg

U1 = 2321  KJ/kg

at ; P = 1000 kPa , T2 = 280°C

v'2= 0.2481 m^3kg

U2 = 2760.6

at ; P = 5MPa ,  T = 500°C

h1 = 3434.7 KJ/Kg

calculate final mass ( m2 )

M2 = V2 / v'2

      = 0.2 / 0.2481 =  0.806 kg

therefore the mass added =  m2 - m1

                                            = 0.806 - 0.6 =  0.206 kg

i don't understand is there any other way you can rephrase it plz let me know in the comments and i will do it.

explanation: