# Specific Gravity of Alcohol.

1. Nov 21, 2008

### Soojin

1. The problem statement, all variables and given/known data

An aluminum cylinder weighs 1.03 N. When this same cylinder is completely submerged in alcohol, the volume of the displaced alcohol is 3.89 multiplied by 10-5 m3. If the cylinder is suspended from a scale while submerged in the alcohol, the scale reading is 0.732 N. What is the specific gravity of the alcohol?

2. Relevant equations

Fscale + FBuoyancy = mg

FBouyancy = DVg

Specific gravity = Dobject / Dmedium

3. The attempt at a solution

First I used Fscale + FBuoyancy = mg
(.732)(FB) = 1.03
FB = .298 N

Then I used FB = DVg
.298 = Dalcohol(3.89e-5)(9.8)
Dalcohol = .298 / [(3.89e-5)(9.8)]
Dalcohol = 781.7009 kg.m3

Then I used Specific gravity = Dobject / Dmedium, where D = mv.

Specific gravity = (m/v) / Dalcohol
Specific gravity = [(1.02/9.8)/3.85e-5] / (781.7009)
Specific gravity = 3.458

I tried this answer and it was wrong. Can anyone help out? Thank you!

2. Nov 21, 2008

### Staff: Mentor

The specific gravity of a substance is the ratio of the density of that substance to the density of water.