Question

The halpy of vaporization of H2O at 1 atm and 100 C is 2259 kJ/kg. The heat capacity of liquid water is 4.19 kJ/kg.C, and the heat capacity of water vapor is 1.9 kJ/kg-C. H20 at 10 bar boils at 179.9 C. What is the enthalpy of vaporization of H20 at 10 bar? You can neglect the effect of pressure. e 2076 kJ/kg e 1924 kJ/kg e 2259 kJ/kg 2442 kJ/kg 2594 kJ/kg None of the above

Answer 1

Answer: Option (a) is the correct answer.

Explanation:

The given data is as follows.

        $C_{p}_{liquid}$ = 4.19 $kJ/kg ^{o}C$

        $C_{p}_{vaporization}$ = 1.9 $kJ/kg ^{o}C$

Heat of vaporization ($\DeltaH^{o}_{vap}$) at 1 atm and $100^{o}C$ is 2259 kJ/kg

        $H^{o}_{liquid}$ = 0

Therefore, calculate the enthalpy of water vapor at 1 atm and $100^{o}C$ as follows.

            $H^{o}_{vap}$ = $H^{o}_{liquid}$ + $\DeltaH^{o}_{vap}$        

                                   = 0 + 2259 kJ/kg

                                   = 2259 kJ/kg

As the desired temperature is given $179.9^{o}C$ and effect of pressure is not considered. Hence, enthalpy of liquid water at 10 bar and $179.9^{o}C$ is calculated as follows.

             $H^{D}_{liq} = H^{o}_{liquid} + C_{p}_{liquid}(T_{D} - T_{o})$

                             = $0 + 4.19 kJ/kg ^{o}C \times (179.9^{o}C - 100^{o}C)$

                              = 334.781 kJ/kg

Hence, enthalpy of water vapor at 10 bar and $179.9^{o}C$ is calculated as follows.

               $H^{D}_{vap} = H^{o}_{vap} + C_{p}_{vap} \times (T_{D} - T_{o})$

             $H^{D}_{vap}$ = $2259 kJ/kg + 1.9 \times (179.9 - 100)$            

                              = 2410.81 kJ/kg

Therefore, calculate the latent heat of vaporization at 10 bar and $179.9^{o}C$ as follows.

       $\Delta H^{D}_{vap}$ = $H^{D}_{vap} - H^{D}_{liq}$              

                         = 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 $179.9^{o}C$ latent heat of vaporization is 2076 kJ/kg.