Question

Compounds like CCl₂F₂ are known as chlorofluorocarbons, or CFCs. These compounds were once widely used as refrigerants but are now being replaced by compounds that are believed to be less harmful to the environment. The heat of vaporization of CCl₂F₂ is 289 J/g. How many grams of this substance must evaporate to freeze 201 g of water initially at 33°C. (The heat of fusion of water is 334 J/g; the specific heat of water is 4.18 J/g-K.)Compounds like CCl₂F₂ are known as chlorofluorocarbons, or CFCs. These compounds were once widely used as refrigerants but are now being replaced by compounds that are believed to be less harmful to the environment. The heat of vaporization of CCl₂F₂ is 289 J/g. How many grams of this substance must evaporate to freeze 201 g of water initially at 33°C. (The heat of fusion of water is 334 J/g; the specific heat of water is 4.18 J/g-K.)

Answer 1

Answer: Mass Of CFC that needs to evaporate for the freezing of water = 328.24 g

Explanation: Heat gained by the CFC = Heat lost by water

Heat lost by water = Heat required to take water's temperature to 0°c + Heat required to freeze water at 0°c

Heat required to take water's temperature from 33°c to 0°c = mCΔT

m = 201g, C = 4.18 J/(gK), ΔT = 33

mCΔT = 201 × 4.18 × 33 = 27725.94 J

Heat required to freeze water at 0°c = mL

m = 201g, L = 334 J/g

mL = 201 × 334 = 67134 J

Heat gained by CFC to vaporize = mH = 27725.94 + 67134 = 94859.94 J

H = 289 J/g, m = ?

m × 289 = 94859.9

m = 328.24 g

QED!!