Ccl2f2 is considered to be environmentally safe and is used as a refrigerant. The heat of vaporization of ccl2f2 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.

The mass of  CCl2F2 is  52 g.

From the question, we known that the heat lost during vaporization is used to cool water. Hence;

Heat lost by vaporization = Heat gained by water

We have the following information;

Heat of vaporization of CCl2F2(L) = 289 J/g

Mass of CCl2F2(ma) = ?

Heat capacity of water(c) = 4.18 J/g/K

Mass of water(mb) = Number of moles × molar mass = 2moles × 18 g/mol = 36 g

Freezing point of water = 0°C

Initial temperature of water = 20 °C

Heat of fusion of water (Lb) = 334J/g

We know that;

maLa = mbcθ + mbLb = mb(cθ + Lb)

Where θ = T2 - T1 = Temperature change of water

So;

-maL = mbcθ

Since heat is lost during vaporization

-ma(289) = 36 × 4.18 × (0 - 20)

ma = -[36 (4.18 × (0 - 20) + 334)/289)

ma = 52 g

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52.1 gram of CCl2F2 must evaporate

Explanation:

Step 1: Data given

The heat of vaporization of CCl2F2 is 289 j/g

Number of moles water = 2.0 moles

Initial temperature of water = 20.0 °C

The heat of fusion of water is 334 j/g

The specific heat of water is 4.18 j/g *k

Step 2: Calculate mass H2O

Mass = 2.0 moles * 18.02 g/mol

Mass H2O = 36.04 grams H2O

Step 3: Cooling of liquid H2O from 20.0 °C to 0°C

Q = m*c* ΔT

⇒Q = heat lost when cooling = TO BE DETERMINED

⇒ m = the mass of water = 36.04 grams

⇒c = the specific heat of water = 4.18 J/g*K = 4.18 J/g°C

⇒ΔT = the change in temperature =  20.0 °C

Q = 36.04 * 4.18 *20.0

Q = 3012.9 J

Step 4: Converting liquid H2O to solid H2O (ice) = freezing

Q = m*Hfus

⇒with Q = the heat lost when cooling = TO BE DETERMINED

⇒with m = the mass of water = 36.04 grams

⇒with Hfus = The heat of fusion of water = 334 j/g

Q = 36.04 * 334

Q = 12037 .4 J

Step 5: The total heat lost

Q = 3012.9 + 12037.4

Q = 15050.3 J

Step 6: Calculate mass CCl2F2 needed

Mass CCl2F2 = Q / hvap

Mass CCl2F2 = 15050.3 J / 289 J/g

Mass CCl2F2 = 52.1 grams

52.1 gram of CCl2F2 must evaporate

the answer is 0.8m.

by: nicholas