Answer:
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 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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