I was wondering how to model the rate of mass loss of liquid as it evaporates from a hemispherical bowl. Assuming that the rate of change of volume is proportional to the exposed surface area, I know that as the liquid evaporates, the radius of exposure decreases (the top circle gets smaller).
I thought that the radius would change with this:$$\frac{dV}{dt} = -kA $$ $$\frac{d(\frac{2}{3}\pi r^3)}{dt} = -k\pi r^2$$ $$r = - \frac{1}{2}kt + c$$
However I am not sure how to continue and arrive at a final equation for $ \frac{dV}{dt}$ to plug in to: $$\frac{dm}{dt} = \rho \frac{dV}{dt}$$ to find the final relationship for the rate of mass loss.
Thanks