Problem 11.46 The fluorocarbon compound C2Cl3F3 has a normal boiling point of 47
ID: 691881 • Letter: P
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/gK and 0.67 J/gK , 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.
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/gK and 0.67 J/gK , 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 = kJExplanation / Answer
Molar mass of C2Cl3F3 is 187.3756 g/mol
Q = mcT
Q = heat energy (Joules, J) = ?
m = mass of a substance (g) = 60.5 gm
c = specific heat (units J/kgK) = 0.91 J/gK and 0.67 J/gK
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 = 41.25C
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.