Arigid tank initially contains 1.4 kg saturated liquid water at 200∘c. at this state, 25 percent of the volume is occupied by water and the rest by air. now heat is supplied to the water until the tank contains saturatd vapor only. determine (a) the volume of the tank. (b) the final temperature and pressure. (c) the internal energy change of the water.

Volume of tank = 6.4792 × 10 ³ m³

Final temperature = 371°C

Final Pressure = 21.3 kilo pascal

Change in temperature = 1373.54 kilo joule per kilogram.

Explanation:

Saturated liquid water = 1.4 kilogram

T1 = 200°C

At T1, = 25% of liquid  = V1 = water

At T1 =  75% = air

T2 = ? , V2 = ? ΔT = ?

Referencing the question,

to be able to calculate the volume of the tank, we need to know the volume of water in the tank.

Therefore, Volume of water = 1.4 kilogram × 0.001157 m³/kg

= 1.6198 × 10³ m ³

Now going back to calculate the volume of the tank,

We have, Volume of tank = 4 × Volume of water

= 4 × 1.6198 × 10³ m ³

6.4792 × 10 ³ m ³

Now, to calculate the pressure and the temperature, we need to calculate the specific volume of the mix,

So we have, α vapor =  volume of tank ÷ saturated liquid vapor

we have, 6.4792 × 10 ³ m ³ ÷ 1.4

0.004628 m³ / kilogram

T2 now therefor is 371°C and P2 = 21.3 kilo pascal

Change in temperate of water = T2 - T1

= 2224 - 850.46

1373.54 kilojoule/kilogram

answer:

true

explanation: