Vaporization and Vapor Pressure

Hi, can someone please help me with this problem. Thanks!

Suppose that 1.01 g of rubbing alcohol (C3H8O) evaporates from a 55.0 g aluminium block. If the aluminum block is initially at 25 degrees cel., what is the final temp. of the block after the evaporation of the alcohol? Assume the heat required for the vaporization of alcohol comes from the alum. block and that the alcohol vaporizes at 25 degrees cel.

What I did but got incorrect answer:

1.01 g x 664 J/g = 670.64 J

55g x 0.9J/g = 49.5J

670.64J/49.5J = 13.5

25-13.5 = 11.5 but this is not the correct answer. Can you tell me what I did wrong. Thanks again. :)

Can somebody clear up my thinking, please?

On the face of it, this seems to be a simple problem where you calculate the heat used up by vaporization (presumably your first line)

1.01 g x 664 J/g = 670.64 J

Then you say this heat came from the aluminum, so you use it's specific heat to figure out from that heat loss and that mass, what the temperature difference must be. Although I don't follow your calculation, I assume that's what you're trying to do.

55g x 0.9J/g = 49.5J

670.64J/49.5J = 13.5

25-13.5 = 11.5

The problem I have is that the rubbing alcohol vaporizes at 25C. Following the above thinking, after a little alcohol has vaporized and the aluminum has cooled off a little, the temperature is no longer 25C. The alcohol is no longer at the temperature at which it will vaporize, so the above argument can't be used.

Can somebody clear up my confusion, please?