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 60.5 g of C2Cl3F3 from a liquid at 14.50 ∘C to a gas at 88.85 ∘C .

Express your answer using two significant figures.

Q =

kJ

Answer 1

Molar mass of C2Cl3F3 is 187.3756 g/mol

Q = mc∆T

Q = heat energy (Joules, J) = ?

  m = mass of a substance (g) = 60.5 gm

c = specific heat (units J/kg∙K) = 0.91 J/g⋅K and 0.67 J/g⋅K

∆ is a symbol meaning "the change in"

Until boiling point

∆T = change in temperature = 47.6 -  14.50 = 33.1 ∘C

After Boiling point

∆T = change in temperature = 88.85 ∘C - 47.6 ∘C = 41.25∘C

Hence C2Cl3F3 has undergo following 3 changes

1) C2Cl3F3 will reach it boiling point from 14.50

2) Then liquid converts to gas

3) As a gas temperature will reach 88.85 ∘C

Q = 60.5 x 0.91 x 33.1 + 60.5 x 27490 / 187.37 + 60.5 x 0.67 x 41.25 = 12370 Joules or 12.37 Kilo Joules

Hence 12.37 Kilo Joules of heat required to convert 60.5 g of C2Cl3F3 from a liquid at 14.50 ∘C to a gas at 88.85 ∘C.