Heat of vaporization/final temperature

[luffyiskind]

Homework Statement

Suppose that 2.05g of rubbing alcohol (60.90g/mol) evaporates from the surface of a 40.0g aluminum block. If the aluminum block is initially at 25 C, what is the final temp of the block after the vaporization of the alcohol?

Heat of vaporization for alcohol: 45.4 kJ/mol
Heat capacity of Al : 0.903 j/g*C

Homework Equations

q=m cs T

The Attempt at a Solution

I finished the problem but I got it wrong. Here are my steps.

Step 1: 2.05g alcohol x 1 mol / 60.09g x 45.4 kJ / 1 mol = 1.55 kJ

I did this step to find out how much energy was needed to vaporize the alcohol.

Step 2: 1550 J = 40 g x 0.903 j/g*c x delta T

solved for delta T and got 42.9 C

since 1.55 kJ was needed, this energy was taken from Al

Step 3: I added 42.9 C to the original 25 C to get 67.9 C

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What did I do wrong?

 

[Rebooter]

since 1.55 kJ was needed, this energy was taken from Al

Step 3: I added 42.9 C to the original 25 C to get 67.9 C

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What did I do wrong?

Reread these two steps over and over and over and over until you realize your mistake.

Hint:
" taken from Al"
taken

 

[luffyiskind]

So I should have subtracted right? If so, I knew it! Dx

 

[Rebooter]

So I should have subtracted right? If so, I knew it! Dx

Correct,
Remember any temperature above 0 K for an object technically has energy.

 

[DeeNos]

Your calculations and approach seem correct. However, in step 2, you have used 1.55 kJ instead of 1.55 kJ/mol. This may have affected your final answer. Also, make sure to use the correct units throughout your calculations (kJ instead of J).