Respuesta :
Answer: Option (a) is the correct answer.
Explanation:
The given data is as follows.
[tex]C_{p}_{liquid}[/tex] = 4.19 [tex]kJ/kg ^{o}C[/tex]
[tex]C_{p}_{vaporization}[/tex] = 1.9 [tex]kJ/kg ^{o}C[/tex]
Heat of vaporization ([tex]\DeltaH^{o}_{vap}[/tex]) at 1 atm and [tex]100^{o}C[/tex] is 2259 kJ/kg
[tex]H^{o}_{liquid}[/tex] = 0
Therefore, calculate the enthalpy of water vapor at 1 atm and [tex]100^{o}C[/tex] as follows.
[tex]H^{o}_{vap}[/tex] = [tex]H^{o}_{liquid}[/tex] + [tex]\DeltaH^{o}_{vap}[/tex]
= 0 + 2259 kJ/kg
= 2259 kJ/kg
As the desired temperature is given [tex]179.9^{o}C[/tex] and effect of pressure is not considered. Hence, enthalpy of liquid water at 10 bar and [tex]179.9^{o}C[/tex] is calculated as follows.
[tex]H^{D}_{liq} = H^{o}_{liquid} + C_{p}_{liquid}(T_{D} - T_{o})[/tex]
= [tex]0 + 4.19 kJ/kg ^{o}C \times (179.9^{o}C - 100^{o}C)[/tex]
= 334.781 kJ/kg
Hence, enthalpy of water vapor at 10 bar and [tex]179.9^{o}C[/tex] is calculated as follows.
[tex]H^{D}_{vap} = H^{o}_{vap} + C_{p}_{vap} \times (T_{D} - T_{o})[/tex]
[tex]H^{D}_{vap}[/tex] = [tex]2259 kJ/kg + 1.9 \times (179.9 - 100)[/tex]
= 2410.81 kJ/kg
Therefore, calculate the latent heat of vaporization at 10 bar and [tex]179.9^{o}C[/tex] as follows.
[tex]\Delta H^{D}_{vap}[/tex] = [tex]H^{D}_{vap} - H^{D}_{liq}[/tex]
= 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 [tex]179.9^{o}C[/tex] latent heat of vaporization is 2076 kJ/kg.
