The question is incomplete, here is a complete question.
of 
is 289 J/g. What mass of this substance must evaporate in order to freeze 190 g of water initially at 17 degrees C? (heat of fusion of water is 334 J/g; specific heat of water is 4.18 J/g.K).
Answer : The mass of this substance evaporate to freeze must be, 258.2 grams.
Solution :
First we have to calculate the total heat absorbed.
![\text{Total heat absorbed}=[m\times c_{p,l}\times (T_{final}-T_{initial})]+m\times \Delta H_{fusion}](https://tex.z-dn.net/?f=%5Ctext%7BTotal%20heat%20absorbed%7D%3D%5Bm%5Ctimes%20c_%7Bp%2Cl%7D%5Ctimes%20%28T_%7Bfinal%7D-T_%7Binitial%7D%29%5D%2Bm%5Ctimes%20%5CDelta%20H_%7Bfusion%7D)
where,
m = mass of water = 190 g
= specific heat of liquid water = 
= enthalpy change for fusion of water = 
= final temperature = 
= initial temperature = 
Now put all the given values in the above expression, we get:
![\text{Total heat absorbed}=[190g\times 4.18J/g.K\times (290-273)K]+190g\times 334J/g](https://tex.z-dn.net/?f=%5Ctext%7BTotal%20heat%20absorbed%7D%3D%5B190g%5Ctimes%204.18J%2Fg.K%5Ctimes%20%28290-273%29K%5D%2B190g%5Ctimes%20334J%2Fg)
Now we have to calculate the mass of substance.
As, 298 J of heat are absorbed by 1 g of
So, 76961.4 J of heat are absorbed by 
of
Therefore, the mass of this substance evaporate to freeze must be, 258.2 grams.
