Answer:
given,
liquid water in the rigid tank = 1.4 kg
The saturated liquid properties of water at 200◦C are
![v_f = 0.001157 m^3/s\ and \ u_f = 850.46 kJ/kg](https://img.qammunity.org/2020/formulas/physics/college/4v70nsa7k8l62yue7pgprfxljha13btfpe.png)
a) tank initially contains saturated liquid and water so, volume occupied by water
![V_1 = m v_1](https://img.qammunity.org/2020/formulas/physics/college/u8pgzi0nk8etei5b0iqt0jf1fy3n2cwiio.png)
= 1.4 × 0.001157
= 0.001619 m³
total volume =
=
=
![0.006476 m^3](https://img.qammunity.org/2020/formulas/physics/college/w1zc5jip0m7w0n4klxnsmvols1ehp1n26w.png)
b)
![v_2 = (V)/(m) =(0.006476)/(1.4)](https://img.qammunity.org/2020/formulas/physics/college/ioxsmlq8uh8le9pjewrxgl39p8d7j2uvvi.png)
=0.004626 kg/m³
for v₂ = 0.004626 kg/m³
T₂ = 371.3 ⁰C P₂ = 21,367 kPa u₂ = 2201.5 kJ/kg
c) total internal energy
![\Delta U = m(u_2-u_1)](https://img.qammunity.org/2020/formulas/physics/college/s4s2h7k0otwfeomtijyw9nn07e5cz1g4oi.png)
= 1.4 (2201.5 - 850.46) kJ/kg
= 1892 kJ