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⋅Kand 0.67 J/g⋅K, respectively. The heat of vaporization for the compound is 27.49 kJ/mol.

Calculate the heat required to convert 60.0 g of C2Cl3F3 from a liquid at 14.70 ∘C to a gas at 83.10 ∘C

Answer 1

Ti = 14.7 oC

Tf = 83.1 oC

here

Cl = 0.91 J/g.oC

Heat required to convert liquid from 14.7 oC to 47.6 oC

Q1 = m*Cl*(Tf-Ti)

= 60 g * 0.91 J/g.oC *(47.6-14.7) oC

= 1796.34 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

= 60.0/187.37

= 0.3202 mol

Heat required to convert liquid to gas at 47.6 oC

Q2 = n*Lv

= 0.3202 mol *27490 J/mol

= 8802.9033 J

Cg = 0.67 J/g.oC

Heat required to convert vapour from 47.6 oC to 83.1 oC

Q3 = m*Cg*(Tf-Ti)

= 60 g * 0.67 J/g.oC *(83.1-47.6) oC

= 1427.1 J

Total heat required = Q1 + Q2 + Q3

= 1796.34 J + 8802.9033 J + 1427.1 J

= 12026 J

= 12.0 KJ

Answer: 12.0 KJ