Rate of evaporation of a spherical drop

castusalbuscor
Messages
15
Reaction score
0

Homework Statement



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.

Homework Equations



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

The Attempt at a Solution



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

Some how this seems wrong.
 
Physics news on Phys.org
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.
 
To solve this, I first used the units to work out that a= m* a/m, i.e. t=z/λ. This would allow you to determine the time duration within an interval section by section and then add this to the previous ones to obtain the age of the respective layer. However, this would require a constant thickness per year for each interval. However, since this is most likely not the case, my next consideration was that the age must be the integral of a 1/λ(z) function, which I cannot model.
Back
Top