Suppose that 1.00 g of rubbing alcohol (C3H8O) evaporates from a 57.0g 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.

What values are you using for heat vaporization of rubbing alcohol and heat capacity of Al. My numbers are from an old text and probably won't agree with yours.

To find the final temperature of the aluminum block after the evaporation of the alcohol, we need to calculate the amount of heat transferred from the aluminum block to the rubbing alcohol during the evaporation process.

We can use the equation:

q = m * ΔT * C

where:
q = heat transferred
m = mass
ΔT = change in temperature
C = specific heat capacity

First, we need to find the heat required for the vaporization of the alcohol. The heat of vaporization of rubbing alcohol is the amount of energy required to convert 1 gram of rubbing alcohol from a liquid to a gas at its boiling point.

For rubbing alcohol, the heat of vaporization is typically around 853 J/g. So, for 1.00 g of rubbing alcohol:

q_alcohol = (1.00 g) * (853 J/g)
q_alcohol = 853 J

Now, let's calculate the heat transferred from the aluminum block to the alcohol during evaporation.

q_aluminum = m * ΔT * C

The mass of the aluminum block is given as 57.0 g and its specific heat capacity is approximately 0.897 J/(g°C). We are looking to find the change in temperature (ΔT) of the aluminum block.

Since the heat transferred from the aluminum block is equal to the heat required for vaporization, we can set up the equation:

q_aluminum = q_alcohol

m * ΔT * C = q_alcohol

(57.0 g) * ΔT * (0.897 J/(g°C)) = 853 J

Simplifying the equation:

ΔT = 853 J / ((57.0 g) * (0.897 J/(g°C)))

Calculating ΔT:

ΔT ≈ 16.42°C

Therefore, the final temperature of the aluminum block after the evaporation of the alcohol is approximately 25°C + 16.42°C = 41.42°C.