Answer : The correct option is, (a) 
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 = 5.00 g
= specific heat of liquid water = 
= specific heat of liquid water = 
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=[5.00g\times 4.18J/g^oC\times (100-14)^oC]+0.278mole\times 40670J/mole+[5.00g\times 1.84J/g^oC\times (115-100)^oC]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B5.00g%5Ctimes%204.18J%2Fg%5EoC%5Ctimes%20%28100-14%29%5EoC%5D%2B0.278mole%5Ctimes%2040670J%2Fmole%2B%5B5.00g%5Ctimes%201.84J%2Fg%5EoC%5Ctimes%20%28115-100%29%5EoC%5D)

Therefore, the enthalpy change is, 