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 46.0g of C2Cl3F3 from a liquid at 11.10?C to a gas at 81.20?C.

Answer 1

Acc to the formula

Q = mCT

The liquid CFC from 12.35 degrees celsius to its boiling point (47.6 degrees) is heated i.e.35.25 degree

C = Q / M x dT

where
C is the specific heat of the liquid CFC (.91 J/G/C)
Q is the amount of heat required ( to find out)
M is the mass (46 grams)
and dT is the temperature difference (35.25 degrees)

.91 = Q / 46 x 35.25
.91 = Q / 1621.5
Q = 1475.565 J ( heat up the liquid CFC)

vapourizing the CFC

latent heat of vaporization 27.49 kJ/ mole. now convert this to kJ/g.

the molar mass of CFC is 187.5 .

Dividing the latent heat by 187.5 to convert it into grams.
27.49 / 187.5 = .1466133 kJ/g. or 146.6133 J/g

=46 x 146.6133 = 10629.4666 J

Now heat it from 47.6 degrees to 86.85 degree
C = Q / M x dT
.67 = Q / 46 x 39.25
.67 = Q / 2845.625
Q = 1209.5 J ( to heat up to the gas)
now adding all 3 values

1475.565 + 10629.4666 + 1906.56875 = 14005.599 J

So the total energy we need is 14005.599 J