3) The fluorocarbon C2Cl3F3 has a normal boiling point of 47.6°C. The specific heats of C2Cl3F3 (l) and C2Cl3F3(g) are 0.91 J/g-K and 0.67 J?
/g-K, respectively. The heat of vaporization of the
compound is 27.49 kJ/mol. The heat required to convert 50.0 g of the compound from the liquid at
10.0°C to the gas at 65.0°C is __________ kJ.
A) 2301 B) 29.78 C) 9.63 D) 1454 E) 8.46
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(0.91 J/g-K) x (50.0 g) x (47.6 - 10.0) K = 1710.8 J to warm the liquid to its boiling point
(27.49 kJ/mol) x (50.0 g C2Cl3F3 / (187.3756 g C2Cl3F3/mol)) = 7.3355 kJ = 7335.5 J to vaporize the liquid
(0.67 J/g-K) x (50.0 g) x (65.0 - 47.6) K = 582.9 J to heat the vapor to 65.0°C
1710.8 J + 7335.5 J + 582.9J = 9629.2 J = 9.63 kJ total
So answer C).