Answer : The amount of heat changes is, 56.463 KJ
Solution :
The conversions involved in this process are :

Now we have to calculate the enthalpy change.
![\Delta H=[m\times c_{p,s}\times (T_{final}-T_{initial})]+n\times \Delta H_{fusion}+[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%2Cs%7D%5Ctimes%20%28T_%7Bfinal%7D-T_%7Binitial%7D%29%5D%2Bn%5Ctimes%20%5CDelta%20H_%7Bfusion%7D%2B%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 changes = ?
n = number of moles of water = 1 mole
= specific heat of solid water = 
= specific heat of liquid water = 
= specific heat of liquid water = 
m = mass of water

= enthalpy change for fusion = 6.01 KJ/mole = 6010 J/mole
= 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=[18g\times 4.18J/gK\times (0-(-25))^oC]+1mole\times 6010J/mole+[18g\times 2.09J/gK\times (100-0)^oC]+1mole\times 40670J/mole+[18g\times 1.84J/gK\times (125-100)^oC]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B18g%5Ctimes%204.18J%2FgK%5Ctimes%20%280-%28-25%29%29%5EoC%5D%2B1mole%5Ctimes%206010J%2Fmole%2B%5B18g%5Ctimes%202.09J%2FgK%5Ctimes%20%28100-0%29%5EoC%5D%2B1mole%5Ctimes%2040670J%2Fmole%2B%5B18g%5Ctimes%201.84J%2FgK%5Ctimes%20%28125-100%29%5EoC%5D)
(1 KJ = 1000 J)
Therefore, the amount of heat changes is, 56.463 KJ