Posted by Kyle on .
Suppose that 1.00 g of rubbing alcohol (C3H8O) evaporates from a 54.0 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.
45.4 kJ/mole for the heat of vaporization of rubbing alcohol
The specific heat of Al is 0.903 J/g*C
heat of vaporization -
1) Convert the 1.00 g of C3H8O to moles. (60 g per mole)
2) Compute the energy required to vaporize 1/60 mole (45.4/50 = 0.757 kJ). Call it Q
3) Compute the drop in temperature of the Al block when that amount of heat is removed. (delta T = Q/(M C) )