Do this in steps.
1. heat released in lowering steam from 114.5 to steam at 100 c.
q1 = mass steam x specific heat steam x (Tfinal-Tinitial)
q2 = heat released in changing steam at 100 C to liquid H2O at 100 C.
q2 = mass x heat vap
q3 = heat released in moving liquid H2O at 100 C to liquid H2O at 67 C
q3 = mass H2O x specific heat H2O x (Tfinal-Tinitial)
Total q = q1 + q2 + q3
NOTE that mass in in grams and specific heat and heat vap is in kJ/mol or J/mol.
How much energy (in kilojoules) is released when 13.3g of steam at 114.5∘C is condensed to give liquid water at 67.0∘C? The heat of vaporization of liquid water is 40.67 kJ/mol, and the molar heat capacity is 75.3 J/(K⋅mol) for the liquid and 33.6 J/(K⋅mol) for the vapor.
2 answers
By the way you will get faster answers if you don't change screen names.