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Pie
3 years ago
12

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).)
Chemistry
1 answer:
Nikolay [14]3 years ago
7 0

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.

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