________ kJ of heat is required to convert 34.084 g of liquid 2-propanol, C3H7OH, at -0.9 oC to gaseous 2-propanol at 107.3 oC.

Relevance
• In order to answer this question you need to know some physical properties of 2-propanol.  The source below gives them all:

355.5 K  boiling point

39.85 (kJ/mol) heat of vaporization at 355.4 K

Heat capacity (liquid): 165.6 (J/mol*K)

Heat capacity (gas): 103.06 (J/mol*K)

(165.6 J/mol*K) x (34.084 g C3H7OH / (60.095 g C3H7OH/mol)) x

(82.3 - (- 0.9))K = 7814.4 J to heat the liquid to its boiling point

(39.85 kJ/mol) x (34.084 g C3H7OH / (60.095 g C3H7OH/mol)) = 22.60167 kJ =

22601.67 J  to vaporize the liquid

(103.06 J/mol*K) x (34.084 g C3H7OH / (60.095 g C3H7OH/mol)) x ((107.3 + 273.) K - 355.5 K) = 1449.62 J to heat the vapor to 107.3°C

7814.4 J + 22601.67 J + 1449.62 J = 31865.69 J = 31.9 kJ total

(There seems to be considerable variation in the literature about the necessary physical constants, so the final answer here is only a rough estimate.  Whoever wrote this question knew that you would need these four constants so you may be sure that they are given somewhere in your text or web site where this question was asked, and you should use those constants instead of what I looked up here.)