Explanation:
The given data is as follows.
[tex]T_{1} = 100^{o}C[/tex], [tex]T_{2} = 10^{o}C[/tex]
[tex]\Delta H_{vap1}[/tex] = 2257 kJ/kg, [tex]\Delta H_{vap2}[/tex] = ?
For water, [tex]C_{p}[/tex] = 4.184 [tex]kJ/kg ^{o}C[/tex]
Formula to calculate heat of vaporization is as follows.
[tex]\Delta H_{vap1} - \Delta H_{vap2}[/tex] = [tex]C_{p}(T_{1} - T_{2})[/tex]
Hence, putting the values into the above formula as follows.
[tex]\Delta H_{vap1} - \Delta H_{vap2}[/tex] = [tex]C_{p}(T_{1} - T_{2})[/tex]
[tex]2257 kJ/kg - \Delta H_{vap2}[/tex] = [tex]4.184 kJ/kg ^{o}C (100 - 10)^{o}C[/tex]
[tex]\Delta H_{vap2}[/tex] = 2257 kJ/kg - 376.56 kJ/kg
= 1880.44 kJ/kg
Thus, we can conclude that enthalpy of liquid water at [tex]10^{o}C[/tex] is 1880.44 kJ/kg.
