Answer:
274.5 grams of this substance must evaporate to freeze 200 grams of waterof 15 °C
Explanation:
Step 1: data given
The heat of vaporization of CCl2F2 is 289 J/g.
Mass of water (H2O) = 200 grams
Temperature of water = 15.0 °C = 15 - 273 = 288 K
The final temperature of water = the freezing temperature = 0°C = 273 K
The heat of fusion of water is 334 J/g
the specific heat of water is 4.18 J/g*K
Step 2: Calculate the heat release to cool the water from (15.0 °C → 0°C)
Q = m*c*ΔT
⇒with Q = the heat released = TO BE DETERMINED
⇒with m = the mass of water = 200 grams
⇒with c = the specific heat of water is 4.18 J/g*K
⇒with ΔT = the change of temperature of water = 15 °C
Q = 200 * 4.18 * (15)
Q = 12540 J
Step 3: Calculate the energy needed to freeze the water
Q = m*ΔHfus
Q = 200 grams * 334J/g
Q = 66800 J
Step 4: Calculate the total heat released
Q = 12540 + 66800
Q = 79340 J
Step 5: Calculate the mass of this substance must evaporate
Q = m*ΔH(vaporization)
⇒with Q = the total heat released = 79340 J
⇒with m = the mass of CCl2F2 = TO BE DETERMINED
⇒with ΔH(vaporization) = the heat of vaporization of CCl2F2 = 289 J/g.
79340 J = m * 289 J/g
m = 79340 J /289 J/g
m = 274.5 grams
274.5 grams of this substance must evaporate to freeze 200 grams of waterof 15 °C