3.7 kg of a saturated water vapor at 0.4 MPa is isothermally cooled until it is a saturated liquid. Calculate the amount of heat rejected during this process in MJ. (Report your answer in 3 decimal places.)

7.894 MJ

Explanation:

Given that:

The mass of the saturated water vapor = 3.7 kg

The pressure of the saturated water vapor = 0.4 MPa

From saturated properties of steam tables when the pressure is at 0.4 Mpa

The enthalpy in (kJ/kg);

hf = 604.7 kJ/kg

hg = 2738.1 kJ/kg

The rejected heat during this process is:

Q = m(hg - hf)

Q = 3.7(2738.1 - 604.7) kJ/kg

Q = 3.7(2133.4) kJ/kg

Q = 7893.58 kJ/kg

Q = 7.894 MJ

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