Question

Suppose that 1.10 g of rubbing alcohol (C3H8O) evaporates from a 74.0 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. Heat of vaporization of the alcohol at 25 ∘C is 45.4 kJ/mol

Express your answer using two significant figures.

Answer 1

Heat provided by block + Heat required for vaporization of rubbing alcohol = 0

Specific heat capacity of Aluminium = 0.903 J/gC

Molar mass of alcohol = 3 * 12 + 8 * 1 + 16 = 60 gm/mol

Number of moles of alcohol = 1.10/Molar mass of alcohol = 1.10/60 = 0.018333 moles

Heat of Vaporziation required = 45.4 KJ/mol * 0.01833333 moles = 832.33 J

-832.33 = 74 * 0.903 * (change in temperature)

(change in temperature) = -12.45C

Hence the final temp of aluminium = 25 - 12.45 = 12.55C