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comment by kleinbl00
kleinbl00  ·  1849 days ago  ·  link  ·    ·  parent  ·  post: Spherical Ice Fallacy

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?




q-  ·  1847 days ago  ·  link  ·  

Yes, I understand what you're saying, and I think I see where our understanding of the problem posed is different. I do appreciate everything you're saying, as you are clearly the right person to consult on this.

I believe that you are looking to model the entire response of the melting ice. As in, at time t1, the state of melt is X, at time t2 the state of melt is Y, etc. A full set of curves for temperature and amount of ice melted at any given time. A difficult problem to solve, for sure - I would use a computer model to solve this one.

My approach is different. I'm just trying to show that the amount of ice melted taking fluid from T1 down to T2 will be the same regardless of the shape of the ice. The time that it takes to get from T1 to T2 will be longer for the ball than the cube, so one could claim that "the ball of ice melts slower" - which is true, but the amount of ice that has melted by the time the temperature reaches T2 (the drinking temperature) will be the same. Furthermore, the rate of melting once the liquid has reached the melting temperature of ice (~0C) will be the same for either shape of ice.

And as far as me saying "all else being equal" that's how you address a given claim. I'm not wishing it so, I'm setting it as the parameters of the problem so I can address the claim of "it melts less because it has less surface area." It's not my claim. And there's no point in comparing A to B if you're going to have a bunch of other factors that are different. If the admen said "it melts less because it's only half in the drink," I'd have set up a problem around that claim. (But since I'm challenging shape, I might want to compare a sphere half in compared to a cube half in.)

As for the formula being the correct one, it is. I learned it in my high school Chem 1 class my freshman year, and it is still used today. In fact, you'll notice the same formula in the Enthalpy of Fusion wiki link.

In fairness, yeah, one could argue that all those Newtonian physics formulas are dead wrong over-simplifications now that one has taken a relativity course, but for validating a claim that there is a difference in A to B it should be a perceivable difference. The small stone and large stone dropped off a tower may not accelerate at exactly 9.8m/s/s and may not hit at exactly the same time, but most people would agree it's close enough; the small stone does not fall faster.

kleinbl00  ·  1847 days ago  ·  link  ·  

Awright. We're closer to sympatico on this. One thing at a time:

    I believe that you are looking to model the entire response of the melting ice.

More specifically put: My experience with thermodynamics and heat transfer has me convinced that the "entire response" as you put it is a lot more dynamic and a lot more relevant to the discussion than:

    I'm just trying to show that the amount of ice melted taking fluid from T1 down to T2 will be the same regardless of the shape of the ice.

And I get that. What I'm saying is that you can't "just" do that the way you want because the variables you're choosing to model are a lot less important than the variables you're choosing to ignore.

    The time that it takes to get from T1 to T2 will be longer for the ball than the cube, so one could claim that "the ball of ice melts slower" - which is true, but the amount of ice that has melted by the time the temperature reaches T2 (the drinking temperature) will be the same.

T1 to T2 will be faster for the ball, not slower. It will be faster through sheer volume. That's the argument for "ice balls." Simply put, "more ice, less of it in your drink." The argument for an ice ball is similar to the argument for an ice luge. Note that these are not arguments I put much faith in - not to say they aren't going to do anything, just that they're not worth the hassle.

You wanted to eliminate the "ice ball advantage" to make for a simpler (but much less practical) experiment. Okay, but in doing that you're tilting the dominant effects of convection and specific heat change further into the fore, and they're chaotic. That cue 1st law of thermo - all things being equal - drops right out of the problem and gets replaced with something that's like, but isn't quite, heat sink equations. Note that in addition to all the heinous math here, you're also dealing with variable geometry, variable hf, variable k and the specific heat equations.

The discussion really comes down to this: You think in terms of T1 and T2. You're presuming the rest of that stuff doesn't matter. The part you're missing is that when you've got ice in a drink, it's the specific heat part of the equation that dominates: T3, that of the ice, will never be reached by the drink and will never be left by the ice throughout the period of the experiment. Thus your entropy equation is inapplicable for the experiment you wish to run on it.

    And as far as me saying "all else being equal" that's how you address a given claim.

No, that's how you pick an inappropriate model. For the third time, at the scale you're dealing with all the stuff you're ignoring matters more than the stuff you're focusing on. And I'm going to have to appeal to authority here: I spent three years of my life being tested on thermodynamics, and I spent ten years of my life applying fluid mechanics. You looked up an equation on Wikipedia. I'd say "take my word for it" but I've done a lot more than that. Let me restate it in nice, scientific terms:

IF: the system under investigation is characterized by

A) initial liquid volume within an order of magnitude of initial solid volume

B) investigated thermal behavior within the transition temperature of the system

C) a primary characteristic of study is the effect of solid geometry on melt speed

THEN: Effects of convection and transition temperature will dominate the system while fundamental equations of entropy will not appreciably impact the model.

You linked to experimental investigation of this. They did one run, but at least they did a run. They found a difference. Based on your equation, they shouldn't have. That in and of itself should demonstrate that your approach is flawed: theoretically, it shouldn't matter; experimentally, it did: QED, the theory is wrong.

    And there's no point in comparing A to B if you're going to have a bunch of other factors that are different.

Well, step back for a minute: Your initial argument was:

    The transient response is trivial, so let's move to steady-state.

My initial argument was that steady-state is irrelevant (you agreed with this, and restated your argument) and that the transient response is not only what you care about, it's messy as fuck. You circled back to

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

And again - that's not an "entropy" equation. That's a convection and specific energy equation.

So we're back to where we started: you want this to be simple, but it just isn't. That's why I'm saying "you're wishing it so." I've described a bunch of considerations based on an engineering background. You've described a bunch of reasons why they don't matter because you want things to be simple:

    As for the formula being the correct one, it is. I learned it in my high school Chem 1 class my freshman year, and it is still used today. In fact, you'll notice the same formula in the Enthalpy of Fusion wiki link.

You want to use what you learned in high school chemistry but refuse to acknowledge that I might have learned something in 500-level fluid mechanics.

And that's where the problem isn't that your understanding is "different" it's that you won't accord me the respect of acknowledging that I might just know more about this than you do. You aren't trying to understand my arguments, you're trying to nullify them. You aren't trying to apply my insights to your problem, you're arguing they're invalid. See, check this out:

    In fairness, yeah, one could argue that all those Newtonian physics formulas are dead wrong over-simplifications now that one has taken a relativity course, but for validating a claim that there is a difference in A to B it should be a perceivable difference.

You linked to this yourself:

I've now spent a thousand words explaining why those curves don't line up, and explaining the dismal science in putting math to that gap.

Your response is "but they have to line up, I took chemistry in high school." So:

| The small stone and large stone dropped off a tower may not accelerate at exactly 9.8m/s/s and may not hit at exactly the same time, but most people would agree it's close enough; the small stone does not fall faster.|

Drop a pound of water and a pound of snowflakes. Which one hits the ground faster?

I had to do the math to explain why. You didn't. So when I say "they're different, here's why" I can offer an explanation as to why the observed experiment does not align with vf=vi+at. Meanwhile, you've got "snow isn't slower, you're observing it wrong."

There are only so many ways I can point this out. I've hit my limit. Frankly, the fact that you're rubbing high school chemistry in my face three layers deep just shows how little respect you accord me, and makes me regret trying in the first place.

Happy new year.

q-  ·  1847 days ago  ·  link  ·  

Well, first of all, I do respect your authority on this subject, and I do appreciate you taking the time to teach me/us something about it. That's why I wanted to engage on this subject in the first place.

At the risk of infuriating you, let me just seek/make clarification on these points:

1.

    T1 to T2 will be faster for the ball, not slower. It will be faster through sheer volume.

Are you saying it's cooling faster because the ball is bigger? If so, how would a bigger cube cool in comparison? Or, more precisely, a cube of the same size (mass) as the sphere? Or do you mean the distance of the surface of the sphere to the center where T3 is? If so, isn't the distance from the surface to the center of the cube sometimes further (depending on the surface point)?

2. I never said that the rate of cooling for spheres vs cubes should be the same. In fact, from the very beginning I've asserted that

    obviously the ice with more surface area will cool faster than a sphere

The graph in your response shows that, yet you are insisting that I'm wrong. That chart does not show how much of the ice has melted, which is the key question I had posed.

kleinbl00  ·  1847 days ago  ·  link  ·  

sigh.

    Are you saying it's cooling faster because the ball is bigger?

More ice, more cooling potential. That's one factor that you can model pretty safely.

    If so, how would a bigger cube cool in comparison?

Double sigh.

Remember last year when I said:

    I wouldn't model this with a ten foot pole

???

That was me saying "I don't fucking know." More than that, that's me saying "I don't fucking want to find out mathematically." Even more, that's me saying "Anyone with any experience in modeling and curve fits would run this three times, Look at the line, pick linear or quadratic and throw a coefficient on it and call it a day."

What I mean is "I wouldn't model this with a ten foot pole." I've pointed out all the factors I can think of that make it complicated - I do this not to say "I have the answer" but to say "all these factors preclude an easy answer." The fact that you keep asking questions about "the center where T3 is" indicates that you're just not listening. Look:

Ice cools water through phase change. While it's changing phase, it can release energy into the surrounding fluid without changing temperature. "the center where T3 is" is pretty much everywhere where ice is during a transition. That omission alone should indicate that you're underthinking this.

    2. I never said that the rate of cooling for spheres vs cubes should be the same.

And I never said otherwise. You threw some math up there to indicate your thinking. I threw up some caveats to show that your thinking was over-simplified.

To be clear:

I wouldn't model this. It's heinous. I've been forced to model enough in my life to know that it's heinous.

You're trying to model this. You refuse to acknowledge the heinousness of it because you've never been forced to confront the heinousness of heat transfer in an empirical environment.

    That chart does not show how much of the ice has melted, which is the key question I had posed.

Here's your key question:

    It is being said that spherical ice cubes cool your drink quickly and minimize the dilution. (Humbug, I say!) I would like a proof in the purest form: by math.

Humbug all you want, yo. Pure math has no business here. I've said that five different ways. Each time it comes down to you doubting my expertise.

I'm over it.