We can use steam tables to solve this problem.

a) When half of the mass has condensed, the system contains saturated vapor-water mix. Thus, the final temperature is the same as the saturation pressure at the final pressure. This, we can find from the steam table A-5, for a pressure of 2.5 MPa (= 2500 kPa).

The temperature at this pressure = **223.95 degree
Celsius**.

b) Specific volume at the initial stage, T = 300 degree Celsius and P = 2.5 Mpa = 2500 kPa can be obtained from table A-6.

Sp vol. = 0.09894 m^3/kg

At the final stage, P = 2.5 MPa = 2500 kPa, x = 0.5

v2 = vf + x2 vg = 0.001197 + 0.5 (0.079952 - 0.001197)

= 0.0405745 m^3/kg

The change in volume = m (v2- v1) = 0.46 (0.0405745 - 0.09894)

= **-0.02685 m^3**.

Hope this helps.

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