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 42.5 g of C2Cl3F3 from a liquid at 11.00 ∘C to a gas at 99.05 ∘C.

Answer 1

specific heat of C2Cl3F3(l) = 0.91 J/g⋅K

specific heat C2Cl3F3(g) = 0.67 J/g⋅K

heat of vaporization for C2Cl3F3 = 27.49 kJ/mol

Mass of C2Cl3F3 = 42.5 g

Molar mass of C2Cl3F3 = 187.38 g/mol

Moles of C2Cl3F3 = 42.5 / 187.38

= 0.227

Initial temperature = 11.0 oC

Final temperature = 99.05 oC

Heat required to raise the temperature of C2Cl3F3(l) from 11.0 oC to 47.6 oC:

q1 = m*c*T

= 42.5*0.91*(47.6 - 11.0)

= 1415.5 J

= 1.415 kJ

Heat required to vaporize C2Cl3F3:

q2 = H_vap * moles of C2Cl3F3

= 27.49 * 0.227

= 6.24 kJ

Heat required to raise the temperature of C2Cl3F3(g) from 47.6 oC to 99.05 oC:

q3 = m*c*T

= 42.5 * 0.67 * (99.05 - 47.6)

= 1465.04 J

= 1.465 kJ

So, total heat required  to convert 42.5 g of C2Cl3F3 from a liquid at 11.00 ∘C to a gas at 99.05 is:

Q = q1 + q2+ q3

= 1.415 kJ + 6.24 kJ + 1.465 kJ

Q = 9.12 kJ