Suppose that 1.02 g of rubbing alcohol (C3H8O) evaporates from a 72.0 g aluminum block.?

If the aluminum block is initially at 25 ∘C, what is the final temperature of the block after the evaporation of the alcohol? Assume that the heat required for the vaporization of the alcohol comes only from the aluminum block and that the alcohol vaporizes at 25 ∘C. The heat of vaporization of the alcohol at 25 ∘C is 45.4 kJ/mol, the specific heat of aluminum is 0.903 J/g⋅∘C.

Roger the Mole

(45.4 kJ/mol) x ((1.02 g C3H8O) / (60.095 g C3H8O/mol)) = 0.77058 kJ = 770.58 J gained by the alcohol (770.58 J) / (0.903 J/g·°C) / (72.0 g) = 11.85°C change 25°C initially - 11.85°C change = 13°C