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.
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.If you're serious about it, run a bunch of experiments, graph the results, then find a curve fit.
'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.I'm not really sure why I tell you this. I hope you didn't mind.
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. 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.And nobody has to argue about heat sink equations.
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. 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. those damn acoustic east-coast kids, ruining your consultancy dominance"building awesome contraptions" is something that most MEs don't do much of.
Which is how I ended up being the youngest acoustical consultant west of the Mississippi.
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? Delighted you asked! "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. 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. 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: 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. 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. 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?What I meant was the state that the liquid and the solid are the same temperature, while the ice is melting.
Ice shape A vs ice shape B - all else being equal.
I was only trying to prove that it doesn't melt slower.
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?
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: 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: 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. 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. 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. Well, step back for a minute: Your initial argument was: 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 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: 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: 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.I believe that you are looking to model the entire response of the melting ice.
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.
And as far as me saying "all else being equal" that's how you address a given 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.
The transient response is trivial, so let's move to steady-state.
What I meant was the state that the liquid and the solid are the same temperature, while the ice is melting.
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.
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. 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 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.T1 to T2 will be faster for the ball, not slower. It will be faster through sheer volume.
obviously the ice with more surface area will cool faster than a sphere
sigh. More ice, more cooling potential. That's one factor that you can model pretty safely. Double sigh. Remember last year when I said: ??? 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. 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. Here's your key question: 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.Are you saying it's cooling faster because the ball is bigger?
If so, how would a bigger cube cool in comparison?
I wouldn't model this with a ten foot pole
2. I never said that the rate of cooling for spheres vs cubes should be the same.
That chart does not show how much of the ice has melted, which is the key question I had posed.
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.