What mass of this substance must evaporate to freeze 190 g of water initially at 17 ∘c? (the heat of fusion of water is 334 j/g; the specific heat of water is 4.18 j/(g⋅

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.

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:

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.

yea hi

explanation:

a boiling water reactor does not boil the water. a pressurized water reactor boils the water faster. a pressurized water reactor is heated directly by the core. a pressurized water reactor does not use fission.