Question

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.

Calculate the heat required to convert 40.0 g of C2Cl3F3 from a liquid at 13.20 ∘C to a gas at 83.45 ∘C.

Answer 1

Ti = 13.2 oC

Tf = 83.45 oC

here

Cl = 0.91 J/g.oC

Heat required to convert liquid from 13.2 oC to 47.6 oC

Q1 = m*Cl*(Tf-Ti)

= 40 g * 0.91 J/g.oC *(47.6-13.2) oC

= 1252.16 J

Lv = 27.49KJ/mol =

27490J/mol

Lets convert mass to mol

Molar mass of C2Cl3F3 = 187.37 g/mol

number of mol

n= mass/molar mass

= 40.0/187.37

= 0.2135 mol

Heat required to convert liquid to gas at 47.6 oC

Q2 = n*Lv

= 0.2135 mol *27490 J/mol

= 5868.6022 J

Cg = 0.67 J/g.oC

Heat required to convert vapour from 47.6 oC to 83.45 oC

Q3 = m*Cg*(Tf-Ti)

= 40 g * 0.67 J/g.oC *(83.45-47.6) oC

= 960.78 J

Total heat required = Q1 + Q2 + Q3

= 1252.16 J + 5868.6022 J + 960.78 J

= 8082 J

Answer: 8.08*10^3 J