Question

How much heat is needed to vaporize 33.3 grams of ethyl alcohol at its boiling point of 78.0°C? The latent heat of vaporization of ethyl alcohol is 857 J/g. Round your answer to three significant figures.

Answer 1

28,500 Joules

Step-by-step explanation:

Q=mL

m = 33.3 g (mass of the ethyl alcohol)

L = 857 J/g (latent heat of vaporization of ethyl acohol)

Q = 33.3(857)

Q = 28538.1 Joules

Round it to 3 significant figures.

Answer 2

28,538 J

Step-by-step explanation:

The heat needed to vaporize the ethyl alcohol is given by:

`Q=m\lambda_v`

where

m = 33.3 g is the mass of the ethyl alcohol

`\lambda_v=857 J/g` is the latent heat of vaporization of alcohol

Since the alcohol is already at its boiling point (78 degrees), we don't need any other heat to bring it at this temperature. Therefore, we can simply calculate the heat needed by putting the numbers inside the formula:

`Q=(33.3 g)(857 J/g)=28,538 J`