Let's establish a frame of reference. First invert the cone and use the centre of the base as the origin (0,0) in a traditional x-y plane, so that y is the height of the volume of water evaporated and x is the radius. The volume of a cone radius r and height h is ⅓πr2h. The equation y=3(1-x) represents the slant of the cone so that, as x decreases from 1m y increases towards 3m and the volume, E, of evaporating water increases. The evaporating volume is given by:
E=E0-⅓πx2(3-y) where 3-y is the height of the liquid water remaining in the cone. E0, the initial volume=⅓π(1)2(3)=π m3. So substituting for x=(3-y)/3 and E0:
E=π-π(3-y)3/27. (When y=0, E=0; when y=3m, E=π m3. Also when y=3/2 m, E=⅞π m3, which is the amount of water which has evaporated, leaving only ⅛π m3 of liquid water.) The rate of evaporation is dE/dt=π(3-y)2(dy/dt)/9. But dE/dt=kπx2, where k is a constant of proportionality, and πx2 is the area of the surface of the water. Therefore:
dE/dt=kπ(3-y)2/9=π(3-y)2(dy/dt)/9, and so k=dy/dt. Integrating we get y=kt+c. When t=0, y=0, so c=0 and y=kt. When y=3/2 (half the maximum height), t=1 (hour), so k=3/2 m/hr, and y=3t/2. 3-y is the height of the water in the cone so:
3-y=3-3t/2. When y=3 all the water in the cone will have evaporated, so 0=3-3t/2 and t=2 hours.
(At the halfway mark (1.5m), the evaporation rate is only a quarter of what it was at the start. When the height of the remaining water is 0.75m, the evaporation rate is only a sixteenth of what it was at the start. Hence it takes an hour for the remaining ⅛π m3 of water to evaporate.)
