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. You may want to reference (Pages 450 - 453) Section 11.4 while completing this problem. Part A Calculate the heat required to convert 75.0 g of C2Cl3F3 from a liquid at 10.60 ∘C to a gas at 78.80 ∘C. Express your answer using two significant figures.

Answer 1

Amount of heat required , Q = heat for the conversion of liquid at 10.60oC to liquid at 47.6 oC + heat for the conversion of liquid at 47.6 oC to gas at 47.6 oC+ heat for the conversion of gas at 47.6 oC to gas at 78.80 oC

Q = mcdt + mL + mc'dt'

Where m = mass of C2Cl3F3 = 75.0 g

c = specific heat capacity of C2Cl3F3(l) = 0.91 J/g.K

dt= change in temperature of liquid = 47.6 - 10.60 = 37.0 oC

c' =specific heat capacity of C2Cl3F3(g) = 0.67 J/g.K

dt' = change in temperature of gas = 78.80-47.60 = 31.2 oC

L = heat of vaporization for the compound = 27.49 kJ/mol.x (1000 J/kJ) x ( 1 mol / 187.4 g) = 146.7 J/g

Plug the values we get Q = 15.095x1000 J

                                     = 15 kJ