In a closed system 9 kg of water is brought to a boil. if the pressure is 1 bar, what will the specific internal energy be in the system when the quality of the water is 0.77. answer in units of kj/kg.

2025KJ/kg

Explanation:

Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)  

through prior knowledge of two other properties such as pressure and temperature.  

in this case we can found the  specific internal energy  using the quality (0.77) and the pressure, with this formula

U=Uf+(Ufg)X

Uf=innternal energy at x=0 p=1bar=417.4KJ/kg

Ufg= internal energy of evaporation at a pressure of 1 bar=2088KJ/kg

x=quality=0.77

Solving

U=417.4+(2088)0.77=2025KJ/kg

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