An
2 answers:
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Answer: Option (a) is the correct answer.
Explanation:
The given data is as follows.
= 4.19
= 1.9
Heat of vaporization () at 1 atm and
is 2259 kJ/kg
= 0
Therefore, calculate the enthalpy of water vapor at 1 atm and as follows.
=
+
= 0 + 2259 kJ/kg
= 2259 kJ/kg
As the desired temperature is given and effect of pressure is not considered. Hence, enthalpy of liquid water at 10 bar and
is calculated as follows.
=
= 334.781 kJ/kg
Hence, enthalpy of water vapor at 10 bar and is calculated as follows.
=
= 2410.81 kJ/kg
Therefore, calculate the latent heat of vaporization at 10 bar and as follows.
=
= 2410.81 kJ/kg - 334.781 kJ/kg
= 2076.029 kJ/kg
or, = 2076 kJ/kg
Thus, we can conclude that at 10 bar and latent heat of vaporization is 2076 kJ/kg.