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The heat of vaporization of CCl2F2 is 289J/g
The heat of vaporization of CCl2F2 is 289J/g. What mass of this substance must evaporate in order to freeze 100g of water initially at 18
C? ( The heat of fusion of water is 334 J/g; the specific heat of water is 4.18 J/(gk) Hint: Start by determining how much energy is needed to cool the water fom 18
C to 0
C, and then to freeze it at 0
C.
Expert Solution
Answer:
The heat released by the fusion of water from 18 oC is
Q = mcdt + mL
Where m = mass of water = 100 g
c = specific heat of water = 4.18 J/(gk)
dt = change in temperature = 18 - 0 = 18 oC
L = heat of fusion of water = 334 J/g
Plug the values we get Q = 40924 J / mol
The mass of CCl2F2 required to freeze the water ,
m' = Q / heat of vaporization of CCl2F2
= 40924 J / 289 J/g
= 141.6 g
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