Question

Suppose that 1.00 g of rubbing alcohol (C3H8O) evaporates from a 68.0 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

Answer 1

Sol.

As mass of alcohol , C3H8O = 1 g

Molar Mass of alcohol = 60.095 g / mol

So , Moles of alcohol = n = 1 / 60.095 = 0.0166 mol

Also , Heat of vapourization of alcohol at 25 °C = deltaHvap = 45.4 KJ / mol = 45400 J / mol

Therefore ,

Heat gained by alcohol = n × deltaHvap

= 0.0166 × 45400 = 753.64 J

Now , Specific heat capacity of Aluminium , Al = C = 0.902 J / g °C  

Initial temperature = 25 °C

Let final temperature =   T

Change in temperature = 25 - T

Mass of Al = m' = 68 g

So , Heat lost by Al = m' × C × ( 25 - T )

= 68 × 0.902 × ( 25 - T )

Now , Heat gained by alcohol = Heat lost by Al

753.64 = 68 × 0.902 × ( 25 - T )

25 - T = 12.29

T = 25 - 12.29 =    12.71 °C