Question

The enthalpy of vaporization of liquid water at 100°C is 2257 kJ/kg. Determine the enthalpy for apordato of iuod eeat capacity of liquid water is 4 154 kJl(kg °C) and for water vapor is 1.859 kJ/(kg °C)

Answer 1

Explanation:

The given data is as follows.

         $T_{1} = 100^{o}C$,       $T_{2} = 10^{o}C$

       $\Delta H_{vap1}$ = 2257 kJ/kg,     $\Delta H_{vap2}$ = ?

For water, $C_{p}$ = 4.184 $kJ/kg ^{o}C$

Formula to calculate heat of vaporization is as follows.

  $\Delta H_{vap1} - \Delta H_{vap2}$ = $C_{p}(T_{1} - T_{2})$

Hence, putting the values into the above formula as follows.

$\Delta H_{vap1} - \Delta H_{vap2}$ = $C_{p}(T_{1} - T_{2})$

$2257 kJ/kg - \Delta H_{vap2}$ = $4.184 kJ/kg ^{o}C (100 - 10)^{o}C$

            $\Delta H_{vap2}$ = 2257 kJ/kg - 376.56 kJ/kg

                                       = 1880.44 kJ/kg

Thus, we can conclude that enthalpy of liquid water at $10^{o}C$ is 1880.44 kJ/kg.