Question

Suppose that 1.13 g of rubbing alcohol (C₃H₈O) evaporates from a 66.0 g aluminum block.

Part A

If the aluminum block is initially at 25℃, 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℃. The heat of vaporization of the alcohol at 25℃ is 45.4 kJ / mol, the specific heat of aluminum is 0.903 J / g ·℃

Answer 1

Amount of rubbing alcohol = 1.13 g

Molecular weight of rubbing alcohol = 60.1 g

Therefore number of moles of rubbing alcohol = 1.13/60.1 = 0.018802

Heat of vaporization of alcohol = 45.4 kJ/mole

Therefore heat of vaporization for 1.13 g of rubbing alcohol = 45.4 x 10³ x 0.018802

= 854 J

Specific heat capacity of aluminium block, C = 0.903 J/g ^oC

Weight of aluminium block = 66 g

Initial temperature, T1 = 25 ^oC

Let final temperature of block be x^oC

Heat absorbed by aluminium block = mc(T1-T2)

= 66*0.903(25-x)

Heat of vaporization = heat absorbed by aluminium block

=> 854 = 66*0.903(25-x)

=> 25 - x = 854/59.598

=> 25-x = 14.3

=> x = 25-14.3

=> x = 10.7^oC

Therefore the final temperature of the block = 10.7^oC