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.