Respuesta :
Explanation:
The given data is as follows.
Specific heat of water = [tex]4.18 J/g^{o}C[/tex]
Heat of fusion of water = 334 J/g
Mass of water = 200 g
On bringing water at [tex]0^{o}C[/tex], heat released will be as follows.
[tex]q_{1} = m \times C \times \Delta T[/tex]
= [tex]200 g \times 4.18 J/g^{o}C \times (0 - 15)^{o}C[/tex]
= -12540 J
or, = -12.540 kJ (as 1 kJ = 1000 J)
Now, calculate the heat releasedwhen water freezes at [tex]0^{o}C[/tex] as follows.
[tex]q_{2} = mass \times \text{-heat of fusion}[/tex]
= [tex]200 g \times -334 J/g[/tex]
= -66800 J
or, = -66.80 kJ
Therefore, total heat released in freezing water will be as follows.
[tex]q_{total} = q_{1} + q_{2}[/tex]
= (-12.540 - 66.80) kJ
= -79.34 kJ
Hence,
amount of heat released in freezing water = heat used to vaporize [tex]CCl_{2}F_{2}[/tex]
Now, heat of vaporization of [tex]CCl_{2}F_{2}[/tex] = 289 J/g
Total heat released in freezing water = -79.34 kJ
Heat consumed to vaporize [tex]CCl_{2}F_{2}[/tex] = 79.34 kJ = -79340 J
Therefore, calculate the mass of [tex]CCl_{2}F_{2}[/tex] vaporized as follows.
Mass of [tex]CCl_{2}F_{2}[/tex] vaporized = [tex]\frac{\text{heat consumed}}{\text{heat of vaporization}}[/tex]
= [tex]\frac{79340 J}{289 J/g}[/tex]
= 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 [tex]15^{o}C[/tex].
