How much energy (in kJ) is absorbed/released to cool the gas?

Question

A 130.2 gram sample of an unknown substance (MM = 189.50 g/mol) is cooled from 233.0 °C to 147.6 °C. (heat capacity of liquid = 1.62 J/g･ °C; heat capacity of gas = 1.04 J/g･ °C; ∆Hvap = 78.11 kJ/mol; normal boiling point, Tb = 160.3 °C) How much energy (in kJ) is absorbed/released to cool the gas?

Answer 1

step 1: heat released to cool sample from 233º (a gas) to a gas @160.3º:

q = mC∆T = (130.2 g)(1.04 J/gº)(72.7º) = 11236 J = 11.24 kJ

step 2: heat released for phase change from gas to liquid @ 160.3º:

q = m∆H_vap (note: ∆H_vap is given in kJ/mol so we need to convert 130.2 g to moles as follows:

q = m∆H_vap = (130.2 g)(1 mol/189.50 g)(78.11 kJ/mol) = 53.67 kJ

step 3: heat released to cool liquid from 160.3º to 147.6º:

q = mC∆T = (130.2 g)(1.62 J/gº)(12.7º) = 2679 J = 2.68 kJ

step 4: sum the heat from each step:

11.24 kJ + 53.67 kJ + 2.68 kJ = 67.59 kJ of heat energy released = 67.6 kJ (3 sig.figs.)

(please be sure to check all of the math)