Question

CCl F is considered to be environmentally safe and is used as a refrigerant. The heat of vaporization of CCl F is 289 J/g. What mass of this substance must evaporate in order to freeze 2 moles of water initially at 20°C? The heat of fusion of water is 334 J/g and the specific heat of water is 4.18 J/g-K.

Answer 1

Number of moles of water present = 2

initial temperature is 20⁰C

Inoder to reach the fusion point the temperature of water should be 0⁰C first and then it is fused

For the change of temperature the amount of heat released = m s dt

where m is the mass of the water, s is the specific heat of water and dt is the difference in temperature.

m = 2 X 18 = 36 grams

s = 4.18 J/g-K

dt = 20⁰C - 0⁰C = 20⁰ C

= 293 K - 273 K = 20 K

Amount of Heat released =Q₁ =36 X 4.18 X 20 = 3009.6 Joules

Phase Conversion from water at 0⁰C to ice at 0⁰C

The amount of heat released = Q₂= m L = 36 X 334 = 12024 Joules

Net heat released = Q₁ + Q₂ = 3009.6 + 12024 = 15033.6 Joules

Heat of Vapourization of CCl F is 289 J/g

Let the mass of refrigerant vapourized = m grams

Amount of heat liberated due to the evaporation of the refrigerant = Q₃ = m X 289 Joules

For water to freeze the amount of CCl F vapourized should produce the same amount of heat

i.e., Q₁ + Q₂ = Q₃

15033.6 = m X 289

=> m = 52.019 grams

52.019 grams of CCl F refrigerant is vapourized to freeze 2 moles of water at 20⁰C