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People of science - a little help, please. There is a fallacy being propagated in pop culture today and I would like to put an end to it. 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. Not empricaly, like these guys did.

I know the basics here, but I've been out of college far too long to remember the formulas. Yes, I know how to use google but I don't care to. Yes, I've been drinking - shut up.

Assumptions:

Whiskey starts warm, at say 70 deg F.

Ice, either spherical or any other shape with greater surface area, is of identical mass and temperature throughout - say 0 deg F.

Mass of ice is approximately (or exactly, if you prefer) equal to the mass of whiskey.

The drink completely covers the ice in the glass, so you have essentially a cylinder of icy liquid in a glass (both systems have the same boundary conditions.)

The drink is being handled, or swirled Q-family style (back of hand just above the rim in a casual grasp, bottom of glass moved in a circle whilst speaking with a locked gaze on your audience, head tilted slightly downwards.)

OK, do this in the metric system.

First of all, obviously the ice with more surface area will cool faster than a sphere, so don't bother to prove that. The transient response is trivial, so let's move to steady-state. The whiskey-water has cooled to N degrees (frankly, I don't care what that temperature is, but know that it's the same regardless of the ice shape inside the glass, because the thermodynamic boundary conditions are the same, damn it.)

OK, so now that the liquid has cooled to N (32?) degrees, how much ice has melted? I assert that it's the same regardless of the original shape of the ice, but this is where I need your help. In fact, less high surface area ice may have melted than in the sphere given that the center of the sphere may still be <N degrees while the other ice (snowflakes) may be at N.

Second - once cooled to the steady-state temperature, assuming the boundary conditions are the same (air temperature, hand temperature on the glass, etc) can we show that the rate of melting is the same regardless of the shape of the ice? As in, Joules in = grams melted.

This guy kind of gets it, but basically says big cubes are better than small, but I'd like to prove spherical vs non-, with everything else being equal.

Any engineering majors out there taking thermo?

Hi. Mechanical engineering degree here, with post-grad research in fluid mechanics and 10 years in acoustics (fluid mechanics).

I wouldn't model this with a ten foot pole.

Here are some things you're not considering:

1) You give a shit about steady-state. Steady-state is room temperature. The steady-state form of this experiment is only dependent on the mass of ice and the mass of bourbon.

2) The transitional phase is driven by a *shit ton* of variables that will require polynomial math. To whit:

- "swirling speed" will have more effect on a fresh (sharp) ice cube than on a not-fresh (round-edged) ice cube.

- Conductive and convective heat transfer will change non-linearly depending on the size of the cube and the size of the sphere.

- Glass shape and swirl method will put the sphere out of round and the cubes out of square in an unpredictable, nonrepeatable way.

- state change heat transfer will change non-linearly depending on the size of the cubes and size of the spehere.

- thermal conductivity of your drink will change as the ethanol/water blend changes. For that matter, results will be different with Wild turkey 80 and Wild Turkey 101. Not enough to overcome your swirling, but "how fast you swirl" is probably your biggest variable anyway.

- Room temperature matters. Room humidity matters. "big sphere half out of bourbon" is going to absorb more heat from the room; small cubes wholly in bourbon" are going to absorb more heat from the glass. By the way - glass? Ceramic? Crystal? Borosilicate? Cylindrical glass? Prismatic? 4 sides? 6? 8? Yeah, you can control for this, but the more you control for it, the more specific (and less generally applicable) your results will be.

That's off the top of my head. The important factors are this:

- How fast you swirl your drink matters.

- How much contact the cubes have with the drink matters.

- How fast you drink matters.

For the record, I'm a long goddamn ways away from being a purist. I put ice cubes in my bourbon, I drink my scotch straight (although blends with a splash of soda water are tasty). I know whiskey stones are a waste (most of the cooling done by ice is done by transition - IE, melting - not by conduction or convection). The original iteration - all $1500 of it - was much more about table flourish than anything else. BUT: the advantages are this:

1) You can use more ice, because most of it is sitting outside your drink.

2) That additional ice provides some thermal inertia, keeping your drink closer to the beginning of that empirical curve you found.

3) You will finish your drink and have most of your ice left.

Balance that with the fact that HOLY FUCK IT'S JUST ICE and it's been hitting you in the face the entire time you've been drinking. Also, you look like you're guzzling snowball juice, you douche.

This is why I drink Ardbeg, Laphroag and Talisker, and only buy MacAllan when the Chinese are making everything else too goddamn expensive. Although Ralph's 30% off six got me a bottle of Hennessy for $18, a bottle of Woodford for $24 and a bottle of Chivas for $16 so *fuck* pretense.

If you care that much, drink cask strength. A little melted ice brings it back down to earth quite nicely.

* * *

There's a saying: "You know you're an engineer if you've ever modeled a horse as a sphere." The amount of math going on here is formidable, chaotic, difficult to repeat and tedious. Far more efficient - and far more fun - to solve it empirically. If you're serious about it, run a bunch of experiments, graph the results, then find a curve fit. Maybe even write a grant proposal. I'd fuckin' love to see someone using research dollars on booze.

- If you're serious about it, run a bunch of experiments, graph the results, then find a curve fit.

Reminds me of a big science experiment we had to do in the last year of highschool. We were given a week off to work on a large experiment, any experiment we wanted. So the two of us chose the Mpemba-effect, which claims that warmer water freezes *faster* than colder water. Counter-intuitive, so it's fun to test empirically. Your post made me dug it out again.

In retrospect, we were quite thorough about it, testing for lots of possible factors: the position in the freezer, demi vs normal water, hard vs not hard water, shape of the glass and the volume of water. No effect. Only when we added ions (calcium, magnesium) we managed to delay the freezing. The only way we managed to get an effect is when we boiled water with ions (to get rid of the ions) and froze it simultaneously with water that still had a lot of ions in it.

The best part of this all was that it was doable in a day or three, in the freezer in my home, while the other groups had experiments that lasted the full workweek in the chemlab. One group tried to make an e-ink screen with ferromagnetic fluid and magnets but failed when the chemistry teacher accidentaly destroyed their setup.

I'm not really sure why I tell you this. I hope you didn't mind.

- I'm not really sure why I tell you this. I hope you didn't mind.

'cuz it's awesome! Experimentation is how we learn about the world; repeated experimentation is how we learn about precision.

The "round ice melts into the drink more" problem could be solved *really easily* through experimentation:

1) Make 2 kinds of ice of the same mass.

2) Pour two glasses of alcohol of the same volume.

3) Let each sit for 10 minutes and remove ice from both glasses.

4) Measure volume of liquid in glasses; weigh amount of ice from both glasses.

5) Repeat a few times for certainty.

And nobody has to argue about heat sink equations.

The opening post has an empirical experiment, but they chose to measure it with a probe thermometer in the glass itself, which is highly inconsistent at best due to the placement of the ice cubes. Your method sounds much simpler and more accurate.

- And nobody has to argue about heat sink equations.

It's basically why I didn't choose to study civil engineering. I know I could if I found a way to make myself plow through the math. It just never seemed that interesting, especially a future job as a human calculator. Not in the I-hate-math-its-useless-anyway kind of way, just a very low score when you divide how interesting it is by the necessary effort .

What have you done with your mechanical engineering degree? It was one of my options, mostly because you get to make awesome contraptions. But it seems a far stretch from audio engineering afaik.

That experiment was about cooling, not dilution. You could do the cooling measurement here, too; it wouldn't much influence the outcome.

SO THE PATH FROM MECHANICAL ENGINEERING TO MIXING TELEVISION

When I was working on my pre-engineering I took a couple classes in mixing in order to make my keyboard stuff sound better. When I transferred I decided to see if I could get a job mixing bands in clubs to help pay the bills. mission accomplished; in addition to a full engineering load, I mixed a good 50-60 hours a week for my last three years of college.

When you spend that much time with sound, you hear "acoustics" muttered about in much the same way "latin" was likely muttered about in pre-Enlightenment Europe. However, since I was pursuing a mechanical engineering degree, and since acoustics is an offshoot of fluid mechanics, "acoustics" was an undeclared minor I could pursue. It was also handy in bioengineering, the other job I had in addition to mixing in clubs. Bioacoustics is interesting in and of itself; not only that, but the fluid mechanics of blood are so horrifically non-linear that they teach you that life happens in the empirical regime.

"building awesome contraptions" is something that most MEs don't do much of. In my market, my choices were thus:

1) designing logging and paper-processing equipment for Weyerhauser

2) designing construction equipment for Genie

3) designing subsystems for Boeing

4) designing HVAC systems for any number of mechanical contractors

In pursuit of employment, I ended up applying to a firm that did HVAC design, acoustical consulting and audiovisual design. they were looking for an HVAC CAD guy; I mentioned that I had post-grad stuff in acoustics and had paid for college by mixing bands in clubs. Which is how I ended up being the youngest acoustical consultant west of the Mississippi.

Acoustics is actually *all* mechanical engineering. Problem is, fluid mechanics doesn't work unless air is a massless particle, and because "sound" is energy transmitted through a fluid which means the particles *must* have mass. So you're yanked right out of your theoretical ivory tower and cast down into empirical curve-fitting hell. Sometimes you have to look at the problems, run some numbers, re-derive a few equations and curve-fit a whole new equation in order to get your answer. Despite my abiding hatred for math, I used my degree hella more than anyone I knew who had pursued any of steps (1)-(4) above.

And, of course, sound system design is essentially electroacoustics.

That's a really interesting path to have taken. Did you ever plan that ahead or was it just smaller good ideas building upon each other?

I really think that educating yourself on a ton of different skills can be vital to your later career path. But I don't see myself finding a skill that I can make a job out of during my study. Not much GIS or Illustrator jobs without experience I know of.

- "building awesome contraptions" is something that most MEs don't do much of.

Well, that's the impression I got from the propaganda machines they call 'open days', visiting campus to get a rehearsed talk about how great the study is. They showed off how they made a machine students made to play darts. Ensured me how anyone who took the necessary maths in high school will be able to handle matrix equations. *Sure*. What they never tell you is the list you gave, where 80% of the students will land. Hard to sell, I think.

- Which is how I ended up being the youngest acoustical consultant west of the Mississippi.

*those damn acoustic east-coast kids, ruining your consultancy dominance*

I could have spent many more hours modeling this but it was a fun exercise to just go through it, despite everything you brought up (which is all accurate).

Hey now, it's basically the adult alcoholic version of a snow-cone. Oh man, why don't we all just start using shredded ice with alcohol infused to make adult snow-cone (yes, this is already a thing and it's just as fantastic as it sounds.

Going on your research dollars on booze thing. We had an experiment this past semester where we used a Gas Chromatograph solely for the purpose of measuring the alcohol content in samples of Seagram's 7 and Black Velvet...and to set some calibration curves but that's beside the point.

What's that, you say? Post more examples of ridiculous pandering to Americans by Scottish companies?

"Drambuie. Because kilts, parkour and mid '90s garage house."

The ice ball maker is, I'm reasonably certain, Japanese in origin. They've always been weird about ice. Back before their economy crashed they used to import the shit from glaciers in Greenland because it had no bubbles.

I can't speak to Macallan and its cringes; I wonder if they're pandering to the Japanese because they're one of the few distilleries that the Chinese haven't snapped up. I used to drink Laphroag because my dad did and it was cheap - about half as much as MacAllan. Now it's double. That's all Beijing bourgeoisie.

Well, you're point 1) is spot on. Steady state would be back to room temperature. What I meant was the state that the liquid and the solid are the same temperature, while the ice is melting. A graph of the liquid temp would be a slope down, then a flat line, then a gradual return to room temp. By steady state I meant the flat line of constant temperature (0C if at STP).

Points 2) ... all moot. Ice shape A vs ice shape B - **all else being equal**. Remember the ice is submerged in the drink. And if the sphere is not, you pointed out yourself that it would loose heat (melt) even faster. I was only trying to prove that it **doesn't melt less**.

So, two identical glasses, both containing a cylindrical-shaped water-alcohol mix of the same ratio, both cooled to the same temperature, both in the same ambient conditions. 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

Is that formula wrong because it doesn't contain a shape factor? Does the energy to melt one gram of ice differ based on the shape of the ice? (That's rhetorical, I know it does not.) Are the boundary conditions of 0C water/alcohol/ice somehow different? Maybe only in how much solid to solid contact we have between ice and glass..

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?

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.

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.

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.

*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.

The issue of shapes is going to depend on shape factors. There a bunch of different equations that deal with thermal conduction, which is all that I feel like bothering with with how late it is. I'm going to ignore your swirling since that is going to invoke convection and at steady state I'm assuming the liquid and solid are static, therefore conduction will be only the heat transfer that is relevant.

At this point we have the following equation for heat transfer for the system:

q = S*k*dT,

where S is the shape factor, k is thermal conductivity of water (in units of W/(m*K)), and dT is the change in temperature from the temperature of the ice to the temperature of the liquid. For future reference, the thermal conductivity of ice is 2.4 W/(m*K) per this source that I can't get to format properly...

http://www.its.caltech.edu/~atomic/snowcrystals/ice/ice.htm

Everything is similar between the systems except the shapes, as you have specified. For the sake of this exercise I'm just going to go ahead and calculate q1 and q2 with all of the variables. I had to look up the shape factors (you don't have to memorize these things), from this source.

We can then see that the shape factor for a spherical medium is S=(2*pi*D)/(1-D/(4*z)). D is the diameter of the sphere, z in this case is the distance between the center of the sphere to the surface of the glass. As seen in the diagram, the only restriction is that this distance is greater than the radius of the sphere. I will use a .05m diameter tumbler and a .025m diameter ice sphere for this example. Therefore z=.025m, and D=.025m (R=.0125 and our restriction is satisfied). Given this, the S factor is .209440m. This then results in q1=10.611 W for the ice sphere with T1=21.1111°C (70°F) and T2=0°C (32°F).

The cube is a bit more difficult and I had to do some more research on it. All of my findings and assumption are based upon a Heat Transfer book by Anthony Mills, page 143-146 of the version in Google Books. The S value of a cube is dependent upon heat transfer occurring on it's 6 sides, 8 corners, and 12 edges. This results in a couple of equations combining to form the following equation:

S = (6*W^2)/L + (12*.54*W)+(8*.15*L), therefore for a cube where W=L=D=.025m, S = .342m

this is for a rectangle so W=L for a cube. The coefficients should remain the same in this simplified shape. Given this shape factor, q2 = 17.328 W for the cube shaped ice.

This shows that more heat is being transferred to the cube shaped ice as opposed to the sphere shaped ice. This accounts for the faster melting time of the shapes and rate differences from shape to shape.

I'm not entirely confident that I did this problem properly, it's been a while since I've dealt with Heat Transfer, it's late at night, and I'm slightly drunk. If there are any issues please let me know.

This is a good start, but it's going down the path of calculating heat flux, or how fast heat can transfer between two bodies (via convection or whatever mode.) I think it's like trying to solve billiard ball problems using F=Ma, when the easier approach is conservation of energy (1/2 MV^2).

Now that I'm sober with a bit of coffee, maybe we could set up the problem like this to make it easier:

Assume we take two identical glasses chilled at 0 deg C, add one lump of ice with different shapes of ice (sphere and cube) to each, also frozen to exactly 0C. Now add whiskey that is also from the same freezer at 0C. This is all mixed in the walk-in freezer, where the air is also 0C.

If the above two drinks are left sitting there, they would each remain in stasis, with no ice melting, right? Right.

Now we pick each glass up and hold it in our bare hands for a while, swirling it allowing 1 kJ of energy to transfer to the glass/whiskey/ice system. Assuming all of that energy makes it to the ice, how much melts? (If you want to nit-pick, let's say 2 kJ goes into the glass, then 1kJ radiates/convects back out of the glass to the atmosphere, the other 1 is cooled by the ice.)

Here's the formula I was looking for.

q = m·ΔHf

1000 J = m x 334 J/g m =~ 3g

Note the amount of ice melted is independent of the shape.

If we take the drinks out of the freezer and out on the veranda, the heat flux into the glasses could be a difficult calculation, but it is the same for either shaped cube, as they are both submerged within the fluid.

I don't know about all this nonsense but I got my boss spherical ice cubes + glasses they look neat and make you look neat and that's about all that matters to me (someone who likes to look neat). She sent me photos last night and loves them. :)

Personally, I have my square king ice cubes (as well as just about everyone who got an xmas gift from me this year) and I love them. The only problem is that all my regular drinking glasses are too small, save for my coffee mugs. The classiness is infinitely diminished when drinking rye from a coffee mug.

So, cooling or no cooling, math or no math, big square ice cubes and spherical ice cubes just look fucking neat.

I wouldn't have a problem with it if the admen were saying "buy these ice sphere makers because they look slick." It's because they saying something that is factually inaccurate, and consumers at large are being deceived and ultimately dumbed-down as a result.

This is just one example; I see this kind of thing all the time. It usually starts with one bit that is true, establishing credibility, then followed by one or more fallacies: "Spheres have less surface area than cubes..." True. "...and are therefore cools your drink faster with less melting!" False and false.

Bah! excuse me I need to go chase some kids off my lawn..

Due to the Isoperimetric inequality the sphere has the minimal surface area compared to ANY other solid. With a greater surface area, the drink probably will cool faster with ice cubes, thus the drink will reach the "steady state" where ice melts slower faster.

I would guess that once at the steady state, the rate of melting will be the same for any shape ice. But by the time it's the steady state, won't the cube turn into spheres anyways? Maybe I would rephrase the conditions of the question to adding spheres or cubes to an already cooled drink... But then the answer does not have many practical applications. The easy solution is to buy whiskey stones :p

I'm no engineer and i'm pretty curious if anybody can provide a mathematical answer to this question :)

False. Whiskey stones would require more mass just too cool the initial measure. The phase change of ice is a major part of its ability to cool.

Assuming you put enough mass of stone in to actually cool the drink to the desired temperature, the boundary conditions would once again be the same, then the drink with the stones in it would rise in temperature, while the drink with ice in it would remain cold.

Think of it this way. It takes X Joules to cool a drink from T1 to T2 (70 to 32F, let's say.) Depending on the specific heat of the stones (be they granite, stainless steel, or brass) will determine how much mass are needed, or alternatively how cold they themselves need to be from the start. My freezer only sets to one temp, so figure they start off the same.

The same Techies did a a similar study with stones, steel, and an iced glass. Empirical once again.

Add more Joules from your hand and the ambient air. Temperature of the system goes up. Remember we swirl, so the drink/ice or drink/stones keep essentially a homogeneous temperature. If you had ice, the temp would stay the same and you'd get more phase change (melting.)

To your first point, yes, obviously a spherical shape cools the slowest. Your second paragraph (guess) is also correct - but that's the part I want proven with math. It could be a double helix vs a sphere, or a "snowflake" (theoretically infinite surface area) vs sphere. My point is that once we're at steady state the amount of surface area between solid and liquid no longer matters, as energy into the system is the same either way.

I'm drunk now, either way.

I received a set of whiskey stones for xmas. They are probably okay for keeping something chilled, but they don't chill a drink down as fast as ice. I actually prefer my bourbon and water at room temp anyway.

In fairness, it doesn't overlook that. I'm trying to compare spherical ice vs cubes or whatever. The determination that ice *is preferred* has already been made by the person adding the ice. The only thing we're trying to figure now is the best shape for the ice.

- The same Techies did a a similar study with stones, steel, and an iced glass. Empirical once again.

You're missing a big point in all your analyses. The science of convection is almost all empirical. Predicting heat transfer in a fluid over complex shapes is not something that can have a general solution in all but the most simplified and idealized cases. Complex heat transfer problems are pretty much always solved by numerical modeling. In the end, in a perfectly insulated system, only the initial temp and mass of the ice and whiskey matter, but there's no such thing. The swirling, in particular, is going to vastly accelerate heat loss to the surroundings.

A few constants, pulled from the Internet (these are in Joules per gram per degree Kelvin, same magnitude as Celsius):

Specific heat capacity, ice: 2.108 J/g-K

Specific heat capacity, water: 4.20 J/g-K

Specific heat capacity, whiskey: 3.40 J/g-K

Latent heat of melting, ice: 334 J/g

Density of ice: 0.9167 g/cm3

The specific heat of water actually changes a bit with temperature, but not from one glass of water to the next. http://www.engineeringtoolbox.com/water-thermal-properties-d... So If we start with a fluid (whiskey) sitting in a glass (jar?) at room temperature. The whiskey, glass, and air around it are all the same temperature, so there is no net heat energy going into or out of the system. Let’s start with 100 grams of whiskey at 21C. If we then add a lump of ice to it and swirl it around, at some point in time the temperature of that whiskey, the original 100grams of whiskey, will be at an even 15 deg C. At that moment in time, a fixed amount of heat energy must have been removed from the fluid. That amount of energy is calculated with this formula:

Q = cp m dT

Q = (3.4J/g-K) (100g) (21C – 15C)

Q = 2040 J

It doesn’t matter how you cool it, the answer is always the same. You could blow cold air over it, put it in a plastic bag and drop it in the snow, or whatever. In all cases, if you want to get those molecules of whiskey cooled down by 6 degrees, you need to remove exactly 2040 J of energy from them. Any less, and it’s warmer, and more and it’s colder.

In our ice lump case, there is only one way we are comparing removal of that heat energy – by the ice we’ll put it in contact with. Some energy may come from the warming of the ice, some may come from the melting of the ice, and some fraction may come from the melted ice (water) warming up to the same temperature as the whiskey.

If the goal is minimal ice meltage, we would want to design our ice lump to warm up evenly. With 100 grams of ice to work with, and a starting temperature of -10C, it would be possible to cool this measure of whiskey by 6 degrees without even melting any:

Q = (2.108) (100g) (10C) = 2108 J

More than what is needed to cool the whiskey by six degrees, so it’s possible to design an ice lump to cool it without melting. Basically we’d want a lot of surface area and no thick parts to insulate bits of ice – basically every frozen molecule will need to pull their weight in order to cool the whiskey without any of their frozen colleagues melting.

The amount of energy that has to be removed from the whiskey in order to cool to a lower (drinking) temperature of, say, 10C, is:

Q = (3.4J/g-K) (100g) (21C – 10C) = 3740 J

So, now it’s not possible for 100g of ice to cool without melting at least a little. We can calculate the bare minimum of melting (let w = amount of ice melted, in grams):

3740 = (2.108) (100g) (10C) + (w) (344 J/g) + (w) (4.2) (10 - 0)

W = 4.4 grams

This is the minimum amount of ice that has to be melted in order to cool 100g of whiskey from 21C to 10C (using the starting ice lump of 100g at -10C.) If any of the ice in our lump doesn’t warm up to 0C, then some other portion of our lump will have to melt in order to remove that additional energy from the whiskey.

So, design-wise, if any ice is insulated from the whiskey (by being surrounded by more ice) it will not be effective at cooling, and the ice which is at the boundary condition will have to melt.

100g of ice will have a volume of 109.087 cm3. As a sphere, it will have a radius of 29.64mm, giving it a surface area of 110.4 cm2. Compare this to 4 cubes of 250g each – 30.1mm to each side, 217.4 cm2 of surface area – nearly double. This means that the 4 cubes of ice will be able to cool the whiskey nearly twice as fast as the sphere.

Furthermore, the ice at the center of the sphere has nearly twice as much insulation as the ice at the center of the cubes. Because the ice cubes warm more evenly than the sphere, there will be less melting as the fluid passes temperature T (10C).

If we then also consider the time factor, we see the case for the sphere getting even worse. Because it will take longer for the sphere to cool the whiskey, there will be more time for heat to transfer into the whiskey via the warm glass, the warm air, and your warm hand (on the glass). The rate of heat transfer is a function of delta T for each of these boundary conditions. The total heat energy transferred is directly proportional to the time. The longer the spherical ice takes to cool the liquid, the more the liquid will heat up from outside conditions, the more it will have to melt in order to cool to the given temperature.

Spherical ice is the *worst* possible shape in terms of its ability to cool a drink and not melt into it.

- That amount of energy is calculated with this formula:
Q = cp m dT

Still wrong. You're oversimplifying, as discussed. This is not a first law of thermodynamics problem it is a:

- convection heat loss problem (ice to whiskey)

- conduction heat loss problem (ice to whiskey)

- laminar flow problem (convection caused by thermoclines within whiskey)

Do you understand the difference between "accuracy" and "precision?" You have some very *precise* calculations going on there - numbers and equations and logic oh my - but you do not have any *accuracy* as to what you're calculating. Yes, in a completely magical closed system something would have to lose so many joules in order to cool so many kelvin. And then the real world stepped in and smashed your dreams. You keep holding on to this as if wishing would make it true:

- It doesn’t matter how you cool it, the answer is always the same.

It does. It *really* does. All the stuff you're discounting is the difference between "real world performance" and "ideal performance" and even "ideal performance" doesn't get anywhere without conduction and convection at a bare minimum.

I know that you only want to use the math that you understand, but the math you *don't* understand is why I'm going to keep telling you you're wrong. And once more with feeling: I wouldn't model this. Period. I'd test it, fit a curve to it and use the data empirically in the future. When you model a complex system with a simplification bad things happen.

It's not wrong. There is a change in energy taking place; the corresponding change in temperature for each body is governed by that formula. Your inability to understand that is why you can't solve this problem.

The amount of energy it takes to raise or lower a gram of water by 1C *doesn't* change depending on how you do it. It takes 1 calorie, period. It doesn't matter whether you are using natural gas, electricity, cold air in the freezer, or contact with another body. You saying that it matters doesn't make it true. Find a credible source that says otherwise and show me.

- The amount of energy it takes to raise or lower a gram of water by 1C doesn't change depending on how you do it.

No. But the **SPEED** does. The **EFFICIENCY** does. And your entire line of questioning is about "how long."

- Find a credible source that says otherwise and show me.

- No.

Ah, so you agree that the energy transfer doesn't differ. Good. You said earlier that it *did*, which had me confused.

So, yeah, at point in time t, when the fluid being cooled has experienced delta T, the amount of energy pulled from it is the same, regardless of how long it took to get to delta T. Good, that's what I was saying. The heat energy went into the ice, either warming it, melting it, or warming the melted water. We know it didn't go into the glass or atmosphere, that would violate one of the LAWS of thermodynamics.

So, knowing that all of that heat energy went into the ice, all we have to do is decide how it was distributed. If any of the solid ice remain un-heated (from -10 to 0C), then the balance of the energy must have been removed by phase change or heating the water.

So, does one large sphere warm more evenly than 4 small cubes? I don't need a precise answer, just >, <, or =. Or is it sometimes yes, sometimes no, depending???

- You said earlier that is did, which had me confused.

Show me where.

- So, yeah, at point in time t, when the fluid being cooled has experienced delta T, the amount of energy pulled from it is the same, regardless of how long it took to get to delta T

But point in time t is variable and indeterminate, delta T is variable and indeterminate. *THAT'S WHAT YOU'RE SOLVING FOR.*

- We know it didn't go into the glass or atmosphere, that would violate one of the LAWS of thermodynamics.

It would do no such thing. You're presuming the glass and atmosphere are perfect insulators, which *would* violate the laws of thermodynamics.

- So, knowing that all of that heat energy went into the ice,

You know no such thing.

- all we have to do is decide how it was distributed.

…and how it got there.

…and how long it took.

…and what the efficiency was.

…all of which are variable based on the geometry of your ice and the convection caused by it, which is - again - a variable you're attempting to solve for.

- If any of the solid ice remain un-heated (from -10 to 0C), then the balance of the energy must have been removed by phase change or heating the water.

wat

- So, does one large sphere warm more evenly than 4 small cubes?

Remember when I said I wouldn't model this with a ten foot pole? That statement has not changed veracity lo these many weeks.

- You keep holding on to this as if wishing would make it true:

```
It doesn’t matter how you cool it, the answer is always the same.
```

It does. It really does. All the stuff you're discounting is the difference between "real world performance" and "ideal performance" and even "ideal performance" doesn't get anywhere without conduction and convection at a bare minimum.Right there. You said it *really does* matter how you cool it. "The answer" being Q in the equation right before the quote of mine you used. "Q" stands for heat energy removed, which you said can be different depending on *how* you cool the 100g of whiskey from 21C to 15C.

- It would do no such thing.

That's exactly what I said.

- You're presuming the glass and atmosphere are perfect insulators

No, I simply know that heat flux doesn't go from cool bodies to warmer ones. Heat goes from warm to cool (glass to liquid.) That's one of the laws (2nd, I believe.)

- So, knowing that all of that heat energy went into the ice,

You know no such thing.

Yes, I do. If the ice didn't make the whiskey colder, what did? The warm glass, the warm air, or the warm hand holding the glass? 2nd law, bro.

It's clear to me that you're not going to wrap your head around this problem or how it's solved. I was hoping to get the light bulb to come on for you. You still want to calculate how long the cooling will take. (You and I seem to be in agreement that the spherical ice will cool the whiskey more slowly, so I'm confounded that you think you need to calculate exactly how much slower it is. For me, "well, it's ain't gonna cool faster, that's for damn sure!" is good enough.) I'm calculating **how much ice will melt** once the whiskey has reached the desired temperature. And as I've said before, the longer it takes, the more energy will have to be removed from the whiskey by the ice due to the heat transfer from the surrounding environment. Again, I don't even care for the precise answer, only "it sure as hell ain't gonna melt less" will do.

- Right there. You said it really does matter how you cool it.

What I said:

- It does. It really does. All the stuff you're discounting is the difference between "real world performance" and "ideal performance" and even "ideal performance" doesn't get anywhere without conduction and convection at a bare minimum.

What you said:

- The amount of energy it takes to raise or lower a gram of water by 1C doesn't change depending on how you do it.

What I said:

- No. But the SPEED does. The EFFICIENCY does. And your entire line of questioning is about "how long."

So what you're saying is that I agreed with you when I quoted you to say you were incorrect.

Just so we're clear:

Your standpoint is that the *amount* of energy is the same, therefore nothing else needs to be calculated.

My standpoint is that the *speed* of energy transfer is what we're focused on, therefore we have to calculate a whole bunch of ugly shit.

Worse, you're no longer arguing fact, you're attempting to depose me - you're trying to argue my statements hold no value because you think you've tricked me into contradicting myself. This isn't court, this is physics, and no matter how much you wish to dance around the issue, I have a lot more training in it than you do - which means, I suspect, that you can't even follow my arguments closely enough to understand them. Further, it doesn't matter how much smarter you think you are or how incorrect you presume me to be, the fact remains: I spent several thousands of dollars being tested on this stuff in an ABET-certified 4-year institution (one of the top ten in the country at the time, in fact) and I feel like I'm beating a dead horse: *This isn't a Physics 101 problem.* Which only makes the following that much more insulting:

- It's clear to me that you're not going to wrap your head around this problem or how it's solved. I was hoping to get the light bulb to come on for you.

So. You and your high school physics can be as self-assured as you wish about this particular problem. I said before and I'll say again: it's a lot more complex than you think. I'll add this, though: I no longer have any interest in helping to illuminate the problem to you. I've been purposefully ignoring your condescending tone in the face of an exceptional amount of patience, but I can't do it any more.

Good luck with this and your future endeavors.

So, what is that, the fourth time you've removed yourself from this conversation?

- My standpoint is that the speed of energy transfer is what we're focused on

No, it's what you're focused on.

- ignoring your condescending tone

Wow. Go back and read your own posts in this thread. You have been extremely condescending towards me.

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

Would you speak to your work peers or boss the way you have to me?

Here's a rundown of your total contribution here:

1. This problem is too difficult for me; I have a masters degree in engineering. (Where'd you get it, Ohio State?)

2. Ambient conditions matter. Like, on a humid day, spherical ice melts less, but on a dry day, square cubes will melt less. Or is it the other way around? You never enlightened us on that, you only said that it mattered...

3. Spherical ice cools faster because it's bigger. (Highlighting to me that you don't understand the difference between shape and size.)

4. The laws of thermodynamics are over-simplifications of the real world and do not apply to something as complex as melting ice. (Even though those laws were conceived and proven by observations from the real world.)

Thanks for all your thoughtful contributions!

OK, after doing a bit more web surfing I basically found somebody that's done this problem already, calculating for ice vs stones.

http://scottf.wordpress.com/2011/12/20/whiskey-stones-coolin.../

Again, note there is no place in the formula to consider shape. He does not calculate the rate of change, either, but it should be clear that spherical shape is the slowest possible (least surface area for heat exchange.)