Question

Suppose that 1.05 g of rubbing alcohol (C3H8O) evaporates from a 75.0 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

Answer – We are given, mass of Al block = 75.0 g ,

mass of rubbing alcohol C3H8O = 1.05 g , ti = 25^oC , Heat of vaporization of the alcohol at 25 ∘C = 45.4 kJ/mol

moles of C3H8O = 1.05 g / 60.1 g.mol^-1

                        = 0.0175 moles

Heat of the vaporization of the alcohol at 25 ∘C

q = m* ∆H vap

= 0.0175 moles * 45400 J/mol

= 793.2 J

Now we know the

Heat lost form the Al block = heat gain by the alcohol

So, q for Al block = -793.2 J

So, the specific heat of Al = 0.900 J/g^oC

q = m*C*∆t

-793.2J = 75.0 g * 0.900 J/g^oC * (tf-25^oC)

-793.2 J = 67.5 tf – 1687.5

-793.2 +1687.5 = 67.5 tf

894.32 = 67.5 tf

So, tf = 13.25^oC.

So the final temperature of the block after the evaporation of the alcohol is 13.25^oC