Answer : The heat energy required in total for this process was 273529.7 joules.
Solution :
The conversions involved in this process are :

Now we have to calculate the enthalpy change.
![\Delta H=[m\times c_{p,l}\times (T_{final}-T_{initial})]+n\times \Delta H_{vap}+[m\times c_{p,g}\times (T_{final}-T_{initial})]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5Bm%5Ctimes%20c_%7Bp%2Cl%7D%5Ctimes%20%28T_%7Bfinal%7D-T_%7Binitial%7D%29%5D%2Bn%5Ctimes%20%5CDelta%20H_%7Bvap%7D%2B%5Bm%5Ctimes%20c_%7Bp%2Cg%7D%5Ctimes%20%28T_%7Bfinal%7D-T_%7Binitial%7D%29%5D)
where,
= enthalpy change or heat required = ?
m = mass of water = 110.0 g
= specific heat of liquid water = 
= specific heat of water vapor = 
n = number of moles of water = 
= enthalpy change for vaporization = 40.67 KJ/mole = 40670 J/mole
Now put all the given values in the above expression, we get
![\Delta H=[110.0g\times 4.184J/gK\times (100-50)^oC]+6.11mole\times 40670J/mole+[110.0g\times 1.84J/gK\times (110-100)^oC]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B110.0g%5Ctimes%204.184J%2FgK%5Ctimes%20%28100-50%29%5EoC%5D%2B6.11mole%5Ctimes%2040670J%2Fmole%2B%5B110.0g%5Ctimes%201.84J%2FgK%5Ctimes%20%28110-100%29%5EoC%5D)

Therefore, the heat energy required in total for this process was 273529.7 joules.