Question

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⋅K).)

Answer 1

The question is incomplete, here is a complete question.

$\Delta H_{vap}$ of $CCl_2F_2$ 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.

$\text{Total heat absorbed}=[m\times c_{p,l}\times (T_{final}-T_{initial})]+m\times \Delta H_{fusion}$

where,

m = mass of water = 190 g

$c_{p,l}$ = specific heat of liquid water = $4.18J/g.K$

$\Delta H_{fusion}$ = enthalpy change for fusion of water = $334J/g$

$T_{final}$ = final temperature = $17^oC=273+17=290K$

$T_{initial}$ = initial temperature = $0^oC=273+0=273K$

Now put all the given values in the above expression, we get:

$\text{Total heat absorbed}=[190g\times 4.18J/g.K\times (290-273)K]+190g\times 334J/g$

$\text{Total heat absorbed}=76961.4J$

Now we have to calculate the mass of substance.

As, 298 J of heat are absorbed by 1 g of $CCl_2F_2$

So, 76961.4 J of heat are absorbed by $\frac{76961.4J}{298J}\times 1g=258.2g$ of $CCl_2F_2$

Therefore, the mass of this substance evaporate to freeze must be, 258.2 grams.