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