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

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.

2b2t

explanation:

2b2t

if you're asking what comes out, it's steam.