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Aluminum cylinder in alcohol

  • Thread starter mikefitz
  • Start date
  • #1
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An aluminum cylinder weighs 1.01 N. When this same cylinder is completely submerged in alcohol, the volume of the displaced alcohol is 3.88 × 10-5 m3. The apparent weight of the cylinder when completely submerged is 0.778 N. What is the specific gravity of the alcohol?

m=W/g
m(cylinder)=.10296 kg
p=m/V
.10296kg/3.88*10^5 => 2653.608kg/m^3 (avg density)
SG= 2653.608/1000=> 2.65

What did I do wrong? Thanks
 

Answers and Replies

  • #2
Doc Al
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You found the specific gravity of the cylinder, not the alcohol. Hint: Buoyant force is involved.
 
  • #3
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Doc Al,

So I can find the mass of the alcohol, then the avg density, divide by 1000 and get the alcohol's SG?

Also, how do I relate Buoyant force to this problem? Thanks again
 
  • #4
Office_Shredder
Staff Emeritus
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The force applied upwards by the alcohol is equal to the weight of the displaced alcohol
 
Last edited:
  • #5
Doc Al
Mentor
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mikefitz said:
So I can find the mass of the alcohol, then the avg density, divide by 1000 and get the alcohol's SG?
That'll work.

Also, how do I relate Buoyant force to this problem?
The buoyant force equals the weight of the displaced fluid (alcohol). You can find the buoyant force by comparing the actual weight of the cylinder to its apparent weight when submerged in the alcohol.
 

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