Question

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 of this substance must evaporate to freeze 200 g of water initially at 15°C? (The heat of fusion of water is 334 J/g; the specific heat of water is 4.18 J/g-K.)

Answer 1

Step-by-step explanation:

The given data is as follows.

Specific heat of water =
`4.18 J/g^(o)C`

Heat of fusion of water = 334 J/g

Mass of water = 200 g

On bringing water at
`0^(o)C`, heat released will be as follows.

`q_(1) = m * C * \Delta T`

=
`200 g * 4.18 J/g^(o)C * (0 - 15)^(o)C`

= -12540 J

or, = -12.540 kJ (as 1 kJ = 1000 J)

Now, calculate the heat releasedwhen water freezes at
`0^(o)C` as follows.

`q_(2) = mass * \text{-heat of fusion}`

=
`200 g * -334 J/g`

= -66800 J

or, = -66.80 kJ

Therefore, total heat released in freezing water will be as follows.

`q_(total) = q_(1) + q_(2)`

= (-12.540 - 66.80) kJ

= -79.34 kJ

Hence,

amount of heat released in freezing water = heat used to vaporize
`CCl_(2)F_(2)`

Now, heat of vaporization of
`CCl_(2)F_(2)` = 289 J/g

Total heat released in freezing water = -79.34 kJ

Heat consumed to vaporize
`CCl_(2)F_(2)` = 79.34 kJ = -79340 J

Therefore, calculate the mass of
`CCl_(2)F_(2)` vaporized as follows.

Mass of
`CCl_(2)F_(2)` vaporized =
`\frac{\text{heat consumed}}{\text{heat of vaporization}}`

=
`(79340 J)/(289 J/g)`

= 274.53 g

Thus, we can conclude that 274.53 g mass of this substance must evaporate to freeze 200 g of water initially at
`15^(o)C`.