Answer : The heat required is, 10.3178 KJ
Solution :
The conversions involved in this process are :

Now we have to calculate the enthalpy change.
![\Delta H=n\times \Delta H_{fusion}+[m\times c_{p,l}\times (T_{final}-T_{initial})]](https://tex.z-dn.net/?f=%5CDelta%20H%3Dn%5Ctimes%20%5CDelta%20H_%7Bfusion%7D%2B%5Bm%5Ctimes%20c_%7Bp%2Cl%7D%5Ctimes%20%28T_%7Bfinal%7D-T_%7Binitial%7D%29%5D)
where,
= enthalpy change or heat required = ?
m = mass of water = 15 g
= specific heat of liquid water = 
n = number of moles of water = 
= enthalpy change for fusion = 6.01 KJ/mole = 6010 J/mole
Now put all the given values in the above expression, we get:
![\Delta H=0.83mole\times 6010J/mole+[15g\times 4.18J/g^oC\times (85-0)^oC]](https://tex.z-dn.net/?f=%5CDelta%20H%3D0.83mole%5Ctimes%206010J%2Fmole%2B%5B15g%5Ctimes%204.18J%2Fg%5EoC%5Ctimes%20%2885-0%29%5EoC%5D)
(1 KJ = 1000 J)
Therefore, the enthalpy change is, 10.3178 KJ