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Specific Gravity of Alcohol.

  1. Nov 21, 2008 #1
    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. jcsd
  3. Nov 21, 2008 #2

    Doc Al

    User Avatar

    Staff: Mentor

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