let's say at some point the surface area is one square cm and the depth is one cm, so in cm units they're numerically the same - of course, they can never be literally the same, since they're quantities with different dimension and therefore incommensurate.
That's right, they are not the same, just numerically the same at that point, which doesn't tell you much. What matters is the ratio, which of course depends on the angle of the cone
What I mean by "the surface area approaches zero faster than the depth" is that for all depths smaller than 1cm, the area in cm^2 will be numerically smaller than the depth in cm.
No it's dependant on the angle, that is to say on the ratio between height and surface area, or just width if you like. There may be a point where the area and height are numerically equivalent, but the two values will continue to converge towards zero at the same rate beyond that point.
Your chosen cone has an angle such that when the height is 1cm the surface area is 1cm^2. That's just a particular case.
If you square a number smaller than 1, the answer will be smaller still.
Well don't do it then!
It's not really smaller than 1 is it? It's only said to be so in relation to some point you have arbitrarily defined as being '1' because it happens to coincide with a particular unit of measurement. Again, the rates remain the same.
I don't see how this is a 'problem', since the depth, area and volume are all dependent on each other. Each quantity could be used to uniquely define the amount of water left, and it's all gone when any (i.e. all) of these quantities reaches zero.
Erm, what you really need to know is the height to width ratio, i.e. the angle of the cone.
I agree about the angle of the cone, but the only difference that makes is the point at which the rate of decrease of area 'overtakes' the rate of decrease of depth, I think. So for a shallow cone the area will initially shrink must faster than the depth and for a very narrow cone it's the reverse, but for all cones the rate of area shrinkage will eventually be faster once the remaining volume goes lower than some threshold value.
The ratio between the height and area doesn't actually change though does it? So the ratio between the rates of change stays constant as well.
I concede that from a physics point of view, this is where the 'paradox' breaks down; namely, a smaller volume of water will be able to absorb heat from its surroundings faster than a larger volume. Then again, the precise way the rate of heat absorption behaves might be non-trivial to work out, depending possibly on some combination of depth, area and volume.
I would say that the above argues that the ratio between the rates of change of height and area remains constant, and such that they will reach zero at the same time. Of course.
However the additional factor that less water will evaporate faster given the same amount of energy might mean that the rates themselves also increase as the volume of water decreases. I think the ratio would still stay the same though.