Okay, so check this out.

What I meant was the state that the liquid and the solid are the same temperature, while the ice is melting.

That state is totally not flat, though. You've got the enthalpy of fusion to mess with. All throughout that, you've got the ice giving up its specific heat (and staying ice) and the water absorbing specific heat (and staying liquid). You've got the convection induced by the temperature gradient across the fluid, not to mention the convection and thermoclines induced by differing solvent chemistry. It looks simple from a mile up but it really ain't. You'll get a different graph from a rocks glass vs. a highball, even if they've got the same ice of the same mass and physical geometry. Stack a couple ice cubes and you have the added chaos of convection modeling around those two surfaces.

Ice shape A vs ice shape B - all else being equal.

You can't just wish it so, though. If I have two identical glasses, I sit them in a vacuum chamber on a laser table, use the exact same pour and the exact same ice cubes, convectional cooling is going to be different depending on how the ice cubes rearrange themselves as they melt. This isn't to say the problem is unsolvable. It's not to say that there isn't an answer. This is to prove that - wait for it:

you're modeling the wrong thing.

See, I get this:

I was only trying to prove that it doesn't melt slower.

And what I'm saying is that you aren't proving that. You're doing math on the stuff that, in the larger system, has much less influence on the problem than all the stuff you're saying is "moot."

Ice cools drinks through convection of specific heat potential. Heat transfer through convection is complicated. Like, a lot complicated. The whole point of those stupid ice balls is they aren't submerged in your drink. "for equal amounts of ice" doesn't model well because when we drink something on the rocks, our rocks are mostly submerged… while when we drink something over an ice ball, the ice ball is largely surfaced. That alone changes the conversation, as I discussed above.

| However complex the system is, I think it's fair to say that the two cylinders of fluid at the boundary conditions will absorb heat at the same rate. q = m·ΔHf|

Right. it's a cute equation. Positively adorable.

Is that formula wrong because it doesn't contain a shape factor?

No, it's a gross oversimplification of the problem at hand.

Li'l story. I have a mechanical engineering degree. And it's all mohr's circle and stress equations and tensor analysis and shit and then when you finally get to capstone engineering, they show you something that looks like this:

(sigma)= n* k(a)k(b)k©k(d)k(e)k(f)k(g)k(h)k(i)k(j)

That n is your answer, derived so slavishly over the course of lo these many years. Call it whatever you want. The cross-sectional area of the chain on a playground swing.

That (sigma) is your real answer, that number you make your drawings match, the value that you're going to give to manufacturing.

All those k-sub-whatevers - and the equation had eleven of them - are your bullshit factors. All of them are under 1. All of them are "real world" empirical curve fits to account for the unknowns you have no idea about but here, in this pathetic, miserable equation, you can solve for. Each and every one of them has a look-up table. As in, "don't bother running the numbers, it's better to plug and chug."

What I'm saying here is that all those little k-sub-whatevers dominate this particular experiment while you're solving for N.

Does the energy to melt one gram of ice differ based on the shape of the ice?

Here's our misunderstanding. you're not fully melting the ice. That's the whole point of the big stupid sphere. Any understanding you come across from fully melted ice is irrelevant because the boundaries of the experiment do not include steady-state.

Do you understand?