CFCs, or chlorofluorocarbons,were once used as refrigerants, but were banned due to their role in the depletion of ozone in our upper atmosphere. The evaporation of CFCs can be used to drive the freezing of liquid water. Given the heat of vaporization for one such CFC is 289 J/g, how many grams of liquid water at 25.0 degrees Celsius (oC) could be frozen at 0.0 oC when 52.0 g of the CFC evaporates at its boiling point?
Useful constants
specific heat of ice: 2.09 J/goC
specific heat of water: 4.18 J/goC
specific heat of steam: 1.84 J/goC
water's heat of fusion: 334 J/g
water's heat of vaporization: 2260 J/g
How much heat can be absorbed by the CFC? That's 289 J/g x 52.0g = ?J.
How much heat must be extracted from water to move from 25 C to 0C. That will be mass H2O x specific heat H2O x (25) = ?
How much to freeze H2O. That's mass H2O x heat fusion = ?
heat to freeze H2O + heat to move H2O from 25 to zero = heat absorbed by CFC. I obtained something close to 34 g b ut that's not an exact answer.
For the heat absorbed by the CFC, I got 15028 J. Then for heat extracted from H2O to move from 25 C to 0 C i got 1883.09 J. And for the heat to freeze H2O, 6018.68 J was what I got. When I added the two energy values of H2O, it did not amount to the heat absorbed by CFC. And how did you go from the heat to freeze H2O + heat to move H2O from 25 to 0 = heat absorbed CFC to 34 g of liquid water?
How in the world did you come up with 1883.09J for moving water from 25 to zero? You don't have a mass of water.
mass H2O x specific heat H2O x (Tf-Ti) = mass H2O x 4.184 x (25) = 104.6*mass H2O
How much to freeze H2O. That's mass H2O x 334 = 334*mass H2O
Let x = mass H2O and substitute into
heat to freeze H2O + heat to move H2O from 25 to zero = heat absorbed by CFC.
334x + 104.6x = 15028
Solve for x = mass H2O. I obtained about 34.3 g H2O. We can check the number.
34.3 x 4.184 x 25 = 3587.8 J
34.3 x 334 = 11,456
11,456 + 3587 = 15,043 which doesn't quite add up to the 15028 BECAUSE the 34.3 is rounded from 34.26357.
To solve this problem, we need to consider the amount of heat required to freeze the water, which is equal to the heat released during the evaporation of the CFCs.
First, let's calculate the heat released by the evaporation of the CFCs:
Heat released = mass of CFCs evaporated × heat of vaporization for CFCs
Given:
Mass of CFCs evaporated = 52.0 g
Heat of vaporization for CFCs = 289 J/g
Heat released = 52.0 g × 289 J/g
Heat released = 15028 J
This amount of heat is released when the CFCs evaporate, and it will be used to freeze the water.
Next, let's calculate the heat required to freeze the water:
Heat required = mass of liquid water × specific heat of water × temperature change to freeze
Given:
Mass of liquid water = ?
Specific heat of water = 4.18 J/goC
Temperature change to freeze = 25.0 oC (initial temperature of water) - 0.0 oC (final temperature of ice) = 25.0 oC
Heat required = mass of liquid water × 4.18 J/goC × 25.0 oC
To solve for the mass of liquid water, we need to equate the heat released by the CFCs to the heat required to freeze the water:
Heat released = Heat required
15028 J = mass of liquid water × 4.18 J/goC × 25.0 oC
Now, let's isolate the mass of liquid water:
mass of liquid water = 15028 J / (4.18 J/goC × 25.0 oC)
mass of liquid water = 143.427 g
Therefore, approximately 143.43 grams of liquid water at 25.0 oC can be frozen at 0.0 oC when 52.0 g of the CFC evaporates at its boiling point.