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Rate of evaporation of a spherical drop

  1. Nov 23, 2008 #1
    1. The problem statement, all variables and given/known data

    The rate of evaporation of a spherical drop of liquid is proportional to its surface area.
    Derive R(t) assuming constant density, where R(t) is the radius of the drop as a function of time.

    2. Relevant equations

    I know that the surface area of the sphere is [tex]A = 4 \pi R(t)^{2}[/tex]

    3. The attempt at a solution

    So would I just take the time derivative of the surface area?
    Which gives:
    [tex]\frac{d A}{dt} = 8 \pi R(t) [/tex]

    Some how this seems wrong.
  2. jcsd
  3. Nov 23, 2008 #2
    For this problem you need the volume of a sphere, and the area of a sphere. Now, what does it mean if the "rate of evaporation is proportional to its surface area". Also, assume that as the drop evaporates, it is the VOLUME that is changing. i.e. dV/dt = ? (look at the statement in quotes). Now do you know what to do? You'll need to use the chain rule too.
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