The fluorocarbon compound C2Cl3F3 has a normal boiling point of 47.6 °C. The specific heat of this compound in the liquid state is 0.91 J/g-K and in the gas state is 0.67 J/g-K. The heat of vaporization is 27.5 kJ/mol. What is the amount of heat required to convert 5.6 g of the compound from a liquid at 30.0 °C to a gas at 60.5 °C?

1. First, use Q=mCdeltaH for liquid C2Cl3F3
   5.6g x .91 J/g-K x (47.6-30) = 89.69

   b. Then, use the same equation to find the heat for it to vaporize.
   (5.6g/187.5 g/mol) x 27.5 kJ/mol x 1000J = 821.33

   c. Last, find what the energy would be to warm it to 81 from its boiling point.
   5.6g x .67J/g-k x (60.5-47.6) = 48.4

   Adding a+b+c to get the total energy (J) and convert to KJ

89.69 + 821.33 + 48.4 = 959.42 KJ