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Compounds like CCl2F2 are known as chlorofluorocarbons, or CFCs. These compounds were once widely used as refrigerants but are now being replaced by compounds that are believed to be less harmful to the environment. The heat of vaporization of CCl2F2 is 289 J/g.

What mass (in grams) of this substance must evaporate to freeze 200g of water initially at 15 degrees C?

(The heat of fusion of water is 334 J/g; the specific heat of water is 4.18 J/g-K.)

Respuesta :

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

1. Given Data:

  • 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 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

2.The heat release to cool the water from (15.0 °C → 0°C):  

  • Q = m*c*ΔT
  • Q = 200 * 4.18 * (15)
  • Q = 12540 J

3.The energy needed to freeze the water:

  • Q = m*ΔHfus  
  • Q = 200 grams * 334J/g  
  • Q = 66800 J

4.The total heat released

  • Q= Heat release+Energy needed to be freeze
  • Q = 12540 + 66800
  • Q = 79340 J

5.The mass of this substance must evaporate

The total heat released = 79340 J

 CCl2F2 = 289 J/g.  

  • Q = m*ΔH(vaporization)
  • 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 water of 15 °C.

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