Question

Problem 11.46 -

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.

Part A

Calculate the heat required to convert 59.0 g of C2Cl3F3 from a liquid at 12.30 ∘C to a gas at 79.80 ∘C. Express your answer using two significant figures.

q = kJ

Answer 1

Ti = 12.3

Tf = 79.8

here

Cl = 0.91 J/g.oC

Heat required to convert liquid from 12.3 oC to 47.6 oC

Q1 = m*Cl*(Tf-Ti)

= 59 g * 0.91 J/g.oC *(47.6-12.3) oC

= 1895.257 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

= 59.0/187.37

= 0.314885 mol

Heat required to convert liquid to gas at 47.6 oC

Q2 = n*Lv

= 0.314885 mol *27490 J/mol

= 8656.188291 J

Cg = 0.67 J/goC

Heat required to convert vapour from 47.6 oC to 79.8 oC

Q3 = m*Cg*(Tf-Ti)

= 59 g * 0.67 J/goC *(79.8-47.6) oC

= 1272.866 J

Total heat required = Q1 + Q2 + Q3

= 1895.257 J + 8656.188291 J + 1272.866 J

= 11824 J

= 11.8 KJ

Answer: 12 KJ