Question

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/gK and 0.67 J/gK, 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 5.0°C to the gas at 80.0°C is __________ kJ.

Answer 1

The fluorocarbon compound 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 for the compound is 27.49 kJ/mol.
Molar mass of C2Cl3F3 = (24 + 106.5 + 57) = 187.5grams
H vap = 27.49 kJ/mol. * 1000 J / kJ * 1 mole/ 187.5grams = 146.6 J/g

Calculate the heat required to convert 50.0 g of C2Cl3F3 from a liquid at 5.0°C to a gas at 80.0°C.

Δ Heat = mass * Sp.Ht. * Δ T

Δ Heat = mass * H vap

There are 3 steps
1) Increase temperature of liquid from 5.0°C to 47.6°C
2) Boil the liquid
3) Increase temperature of gas from 47.6°C to 80.0°C.

1) Δ Heat = 50 * 0.91 * (47.6 - 5.0) = 1938.3
2) Δ Heat = 50 * 146.6 = 7330
3) Δ Heat = 50 * 0.67 * (80 – 47.6) = 1085.4

Total heat = 10353.7 Joules = 10.353 kJ