Small Ice vs. Big Ice
December 31, 2008 12:04 PM Subscribe
Craft bartenders say that big ice melts more slowly than small ice. How is this possible? Treating the cocktail as a closed system, if you include an equal mass of ice, and the starting and ending temperatures are the same, wouldn't the same amount of heat be transferred into the ice, resulting in an equal quantity of melt?
Is it possible within the constraints of the average kitchen to have a large enough mass of ice with a cold enough starting temperature, such that the ratio of surface area to cocktail affects the total ice melt in the drink? If you take that sufficiently large mass of ice and break it into smaller pieces, will more of it melt before the drink reaches its final temperature? Wouldn't the increased surface area actually reduce total melt, despite what the cocktailians say?
For your math, you can start with several assumptions:
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I have an understanding from a high school science class that ice doesn't go through its phase change until its entire mass has reached the melting point. There are three possibilities here:
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Is it possible within the constraints of the average kitchen to have a large enough mass of ice with a cold enough starting temperature, such that the ratio of surface area to cocktail affects the total ice melt in the drink? If you take that sufficiently large mass of ice and break it into smaller pieces, will more of it melt before the drink reaches its final temperature? Wouldn't the increased surface area actually reduce total melt, despite what the cocktailians say?
For your math, you can start with several assumptions:
- Starting temperature of the liquid ingredients is room temperature of 20° C (68° F). Starting temperature of the ice is -15° C (5° F), presumed to be within range for a home freezer.
- Call it a closed system, assume a perfect vessel with no specific heat of its own, and ignore (for now) the transfer of heat from the air, the drinker's hands, etc.
- For the purpose of this discussion, leave aside the aesthetics of shaking and stirring, and just assume that the cocktail and the ice are well enough exposed to each other to cool the drink to the freezing point.
- Again for the purpose of this discussion, use an 80mm sphere as the largest practical mass of ice. (It's expensive, but it's attainable at home so we'll go with it).
- For the cocktail, use:
50ml Tequila 25ml Cointreau 25ml lime juice
I wouldn't know how to calculate the specific heat of this, but I'm sure someone here will. - For extra credit, address the issue of outside heat from the environment, the hands, etc. Once the drink has been properly cooled, does large ice melt more slowly than an equal mass of small ice over the 30 minutes to one hour a drink might be nursed?
[sidebar]
I have an understanding from a high school science class that ice doesn't go through its phase change until its entire mass has reached the melting point. There are three possibilities here:
- This is completely true, and there is no large enough mass of ice such that the core temperature is lower than the temperature at the surface when the surface has reached its melting point.
- This was believed to be true when I took high school science, but has since been disproved (see also).
- This was known even then to be untrue, but was used as a convenience for the sorts of calculations that happen in a high school science class.
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You don't take into account the increased surface area of the smaller ice pieces.
posted by Ookseer at 12:11 PM on December 31, 2008 [3 favorites]
posted by Ookseer at 12:11 PM on December 31, 2008 [3 favorites]
For an equal mass of ice, breaking it into small bits increases the surface area, thus increasing the heat transfer and the rate of melting. This is why, say, heat sinks on computer chips have lots of fins - to increase surface area and heat transfer.
Once all the ice melts, assuming no heat transfer to the container or atmosphere, the drink will be at the same temperature whether or not the ice was broken up.
posted by driveler at 12:17 PM on December 31, 2008
Once all the ice melts, assuming no heat transfer to the container or atmosphere, the drink will be at the same temperature whether or not the ice was broken up.
posted by driveler at 12:17 PM on December 31, 2008
It's an increasing effect, the surface area difference of the large vs small pieces increases as melt time does. Initially and over time the smaller pieces of ice have a greater surface area exposed to the liquid compared to the large pieces, smaller pieces have a larger surface area exposed as a ratio to their overall mass as time goes on. The more time both sizes are exposed to the environment the more dramatic the melt in smaller pieces will be. There are factors involved relating to the medium the ice sits in, it's volume and it's ability to absorb temperature, but for a simple discussion without considering other external factors it's probably irrelevant.
posted by iamabot at 12:20 PM on December 31, 2008
posted by iamabot at 12:20 PM on December 31, 2008
Your understanding from high school science class isn't right. The ice doesnt' collectively make a decision to no longer be ice once it reaches transition temperature. This is a convenience for calculations, as you mentioned.
Instead individual molecules either remain attached to the ice matrix, or release and become part of the liquid mass around it. In fact this happens even if both the ice and liquid are exactly at freezing temperature (which as you know varies depending on the pressure, and yes will also be lower because of all that alcohol), it's just that if no more ice is melting, the molecules move off the ice and back onto the ice at the same rate, so we macroscopic beings think nothing is happening.
If the ice is still busy melting (and it will be, because the glass isn't going to be at room temperature, much less hand temperature, until the ice is melted, and the liquid is brought up to room temp--- the drink is much closer to being in a heat bath (the room) than being a closed system)...
So the when the ice is still busy melting, the number of molecules falling off the solid and into the liquid is greater than the number reattaching. (If it's freezing, we're going the other way). As you can imagine, only molecules at the surface can do this, and the process takes time (as well as needing heat input, from your hands or the room temp. environment, and as always, increasing the entropy of the system).
Because only molecules at the surface have the option to fall off, surface to volume ratio is quite important. This ratio goes as 1/radius. The rate of melting will be directly related to this (or maybe related to some positive power of it, sorry I didn't check the units of other quantities involved or actually do any calculations).
But honestly, easier than all this babble, would be to have you go get two sodas. Put crushed ice in one, and normal ice cubes in the other. Which melts faster?
posted by nat at 12:28 PM on December 31, 2008
Instead individual molecules either remain attached to the ice matrix, or release and become part of the liquid mass around it. In fact this happens even if both the ice and liquid are exactly at freezing temperature (which as you know varies depending on the pressure, and yes will also be lower because of all that alcohol), it's just that if no more ice is melting, the molecules move off the ice and back onto the ice at the same rate, so we macroscopic beings think nothing is happening.
If the ice is still busy melting (and it will be, because the glass isn't going to be at room temperature, much less hand temperature, until the ice is melted, and the liquid is brought up to room temp--- the drink is much closer to being in a heat bath (the room) than being a closed system)...
So the when the ice is still busy melting, the number of molecules falling off the solid and into the liquid is greater than the number reattaching. (If it's freezing, we're going the other way). As you can imagine, only molecules at the surface can do this, and the process takes time (as well as needing heat input, from your hands or the room temp. environment, and as always, increasing the entropy of the system).
Because only molecules at the surface have the option to fall off, surface to volume ratio is quite important. This ratio goes as 1/radius. The rate of melting will be directly related to this (or maybe related to some positive power of it, sorry I didn't check the units of other quantities involved or actually do any calculations).
But honestly, easier than all this babble, would be to have you go get two sodas. Put crushed ice in one, and normal ice cubes in the other. Which melts faster?
posted by nat at 12:28 PM on December 31, 2008
Yep, the surface area does it. Think of it this way, a cube of ice has 6 sides. A cube of ice cut in half has 12.
posted by Ironmouth at 12:28 PM on December 31, 2008 [1 favorite]
posted by Ironmouth at 12:28 PM on December 31, 2008 [1 favorite]
Ooh I should also mention that while liquid is going to be all at the same temperature (for several reasons, mostly that it moves around so much), solids don't have to be. Heat flow takes time through a solid.
posted by nat at 12:29 PM on December 31, 2008
posted by nat at 12:29 PM on December 31, 2008
Yeah, surface area. Imagine a cube, 4 inches in width, depth, height. The surface are is the area of a side, times six, or 4*4*6 = 96 square inches.
Now divide the cube in half, a single cut through the centerline of any one face. You now have two hexahedrons, each of dimensions 4*4*2, each with surface area of (4*4) * 2 (two "back" and "front" faces) + (4*2) * 4 (4 sides) = 32 + 32 = 64 square inches each, = 128 square inches surface area.
Now cut the two hexahedrons by bisecting the front faces, and you have four 4*2*2 hexhedrons, with individual surface area of 4*2 * 4 + 2*2 * 2 = 40 square inches, times four of them = 160 square inches total.
Now bisect every four inch face, you have eight 2*2*2 cubes, each with surface area 2*2 * 6 sides = 24 sq. inches, times eight cubes = 192 sq. inches total of surface area.
Heat transfers, ice melts at the surface. Eight smaller cubes gives you 192/96 = 2, twice the surface area for the same (internal) volume of ice.
posted by orthogonality at 12:30 PM on December 31, 2008
Now divide the cube in half, a single cut through the centerline of any one face. You now have two hexahedrons, each of dimensions 4*4*2, each with surface area of (4*4) * 2 (two "back" and "front" faces) + (4*2) * 4 (4 sides) = 32 + 32 = 64 square inches each, = 128 square inches surface area.
Now cut the two hexahedrons by bisecting the front faces, and you have four 4*2*2 hexhedrons, with individual surface area of 4*2 * 4 + 2*2 * 2 = 40 square inches, times four of them = 160 square inches total.
Now bisect every four inch face, you have eight 2*2*2 cubes, each with surface area 2*2 * 6 sides = 24 sq. inches, times eight cubes = 192 sq. inches total of surface area.
Heat transfers, ice melts at the surface. Eight smaller cubes gives you 192/96 = 2, twice the surface area for the same (internal) volume of ice.
posted by orthogonality at 12:30 PM on December 31, 2008
Response by poster: I guess I should break the question down into two parts even more clearly than I already did:
Initial cooling: Over whatever time T it takes to cool the drink to 0° C, without considering environmental factors, either the same amount of ice melts into liquid or it doesn't. Just saying that increasing the surface area increases the melt doesn't address this, because we're talking about a [theoretical] number of joules E transferred from the drink into the ice, which has equal mass either way (and which, presumably, isn't actually large enough within the constraints of a kitchen that the velocity of heat transfer from surface to core is a factor). How is it possible that the same number of joules would produce more liquid water from the same mass of ice broken into smaller pieces? I don't care about which one's faster, I'm curious how it can be claimed that you get more melt one way than the other just in cooling the drink.
Maintenance: Once the liquid has been cooled to something of a steady state temperature (0° C in theory, presumably not exactly 0° C in practice) how much more melting is there? This was the extra credit question and Nat's answer does seem to satisfy it.
What's the answer for the initial cooling? Hand-waving about surface area isn't enough here.
posted by fedward at 12:49 PM on December 31, 2008
Initial cooling: Over whatever time T it takes to cool the drink to 0° C, without considering environmental factors, either the same amount of ice melts into liquid or it doesn't. Just saying that increasing the surface area increases the melt doesn't address this, because we're talking about a [theoretical] number of joules E transferred from the drink into the ice, which has equal mass either way (and which, presumably, isn't actually large enough within the constraints of a kitchen that the velocity of heat transfer from surface to core is a factor). How is it possible that the same number of joules would produce more liquid water from the same mass of ice broken into smaller pieces? I don't care about which one's faster, I'm curious how it can be claimed that you get more melt one way than the other just in cooling the drink.
Maintenance: Once the liquid has been cooled to something of a steady state temperature (0° C in theory, presumably not exactly 0° C in practice) how much more melting is there? This was the extra credit question and Nat's answer does seem to satisfy it.
What's the answer for the initial cooling? Hand-waving about surface area isn't enough here.
posted by fedward at 12:49 PM on December 31, 2008
Your simplifying assumption of "ignoring environmental effects" isn't a valid one. During the period in which the ice is cooling the drink, the drink in its turn is cooling the environment.
Or rather, to be more precise, heat is flowing into the drink from the environment at the same time as it is flowing from the drink into the ice. A bar tender isn't interested in physics; he's being practical about serving drinks in a bar.
Small ice increases the interface between the ice and drink, permitting more heat to flow. The size and shape of the glass controls the interface between the drink and the environment. Given equal glasses, the one with small ice will cool faster than the one with big ice.
And because of that, given equal glasses, the one with big ice will produce more melt water by the time it reaches zero C, because it took longer, absorbed more heat from the environment, and thus required more ice melting to achieve the same overall cooling.
posted by Chocolate Pickle at 12:56 PM on December 31, 2008
Or rather, to be more precise, heat is flowing into the drink from the environment at the same time as it is flowing from the drink into the ice. A bar tender isn't interested in physics; he's being practical about serving drinks in a bar.
Small ice increases the interface between the ice and drink, permitting more heat to flow. The size and shape of the glass controls the interface between the drink and the environment. Given equal glasses, the one with small ice will cool faster than the one with big ice.
And because of that, given equal glasses, the one with big ice will produce more melt water by the time it reaches zero C, because it took longer, absorbed more heat from the environment, and thus required more ice melting to achieve the same overall cooling.
posted by Chocolate Pickle at 12:56 PM on December 31, 2008
4th the others, surface area is increased on a large scale. More surface area = more heat transfer.
posted by damiano99 at 1:08 PM on December 31, 2008
posted by damiano99 at 1:08 PM on December 31, 2008
Given equal masses of ice, large cubes will melt more slowly than equal masses of small cubes, due to the larger mass to surface area ratio. None of your assumptions change this fact.
There will be equal quantities of melt, if the cubes melt completely. But they're banking on the fact that you drink quickly enough so that this is not the case.
posted by electroboy at 1:11 PM on December 31, 2008
There will be equal quantities of melt, if the cubes melt completely. But they're banking on the fact that you drink quickly enough so that this is not the case.
posted by electroboy at 1:11 PM on December 31, 2008
Response by poster: The simplifying assumption is valid enough in that it presents a system that can be solved conclusively. I'm aware that the environment constantly provides a quantifiable amount of heat energy and that this is not a closed system in a perfect vessel, but for the sake of this discussion the time-and-environment element isn't to be a factor. If one knows the amount of heat going into the system during the maintenance phase (simple enough to determine), one can then factor that against the cooling time T determined above to get a reasonable calculation of the extra time-and-environment melt. I just don't think it's enough of an issue over the duration of shaking/stirring/agitation/what-have-you to matter in any way other than aesthetics.
posted by fedward at 1:11 PM on December 31, 2008
posted by fedward at 1:11 PM on December 31, 2008
If you assume the liquid is completely still, this comes down to a case of conduction. If you want to assume a moving liquid, you have convection but that only serves to complicate the proof here.
The point I'm trying to illustrate is that heat transfer is a function of contacting surfaces. In both of these cases, the transfer of heat from material A to material B is based on the surfaces that are in contact.
posted by phrakture at 1:19 PM on December 31, 2008
The point I'm trying to illustrate is that heat transfer is a function of contacting surfaces. In both of these cases, the transfer of heat from material A to material B is based on the surfaces that are in contact.
posted by phrakture at 1:19 PM on December 31, 2008
If one knows the amount of heat going into the system during the maintenance phase (simple enough to determine), one can then factor that against the cooling time T determined above to get a reasonable calculation of the extra time-and-environment melt.
Not true... the heat being transfered from the room to the drink depends on their relative temperatures, which you could reasonably well approximate with Netwon's law of cooling. You end up having to solve a pretty nasty set of differential equations once you take the environment into account.
BUT, you don't want to do that. If I understand you, you're positing a drink thermally isolated from everything else in the universe, and asking if the initial physical configuration of the ice, all else being equal, affects the amount of ice which has melted once it reaches an equilibrium state. YOU'RE RIGHT, it doesn't -- the same amount of ice will melt whether it's in small or large cubes (or, for that matter, whether it's one big block frozen in the bottom of the glass of whether it's a scoop of snow).
All bartender and mefite assertions to the contrary (which are factually and intuitively correct, in the real world) assume thermal interaction between the drink and the environment.
(aside: note that, in an equilibrium state for this system, everything will be at the same temperature, so a surface-to-core thermal gradient in the ice itself contradicts the hypothesis.)
posted by 7segment at 1:28 PM on December 31, 2008
Not true... the heat being transfered from the room to the drink depends on their relative temperatures, which you could reasonably well approximate with Netwon's law of cooling. You end up having to solve a pretty nasty set of differential equations once you take the environment into account.
BUT, you don't want to do that. If I understand you, you're positing a drink thermally isolated from everything else in the universe, and asking if the initial physical configuration of the ice, all else being equal, affects the amount of ice which has melted once it reaches an equilibrium state. YOU'RE RIGHT, it doesn't -- the same amount of ice will melt whether it's in small or large cubes (or, for that matter, whether it's one big block frozen in the bottom of the glass of whether it's a scoop of snow).
All bartender and mefite assertions to the contrary (which are factually and intuitively correct, in the real world) assume thermal interaction between the drink and the environment.
(aside: note that, in an equilibrium state for this system, everything will be at the same temperature, so a surface-to-core thermal gradient in the ice itself contradicts the hypothesis.)
posted by 7segment at 1:28 PM on December 31, 2008
"melts more slowly" is where you started at, and where you should stay. changing the basic premise of the question from "which melts faster" to "which melts more" is only going to confuse you as people will tend to answer the question you started with, not the follow up.
posted by nomisxid at 1:35 PM on December 31, 2008
posted by nomisxid at 1:35 PM on December 31, 2008
we're talking about a [theoretical] number of joules E transferred from the drink into the ice, which has equal mass either way (and which, presumably, isn't actually large enough within the constraints of a kitchen that the velocity of heat transfer from surface to core is a factor). How is it possible that the same number of joules would produce more liquid water from the same mass of ice broken into smaller pieces?
Someone who knows more about physics can feel free to correct me, but here it is as I understand it: It takes E amount of energy to melt the ice regardless, but the surface area changes that rate at which energy can be transfered. When more ice is left, it's because less energy (heat) has been transfered.
posted by burnmp3s at 1:39 PM on December 31, 2008
Someone who knows more about physics can feel free to correct me, but here it is as I understand it: It takes E amount of energy to melt the ice regardless, but the surface area changes that rate at which energy can be transfered. When more ice is left, it's because less energy (heat) has been transfered.
posted by burnmp3s at 1:39 PM on December 31, 2008
one big block of ice has less surface area than the equivalent amount of cubes.
posted by thinkingwoman at 1:58 PM on December 31, 2008
posted by thinkingwoman at 1:58 PM on December 31, 2008
The simplifying assumption is valid enough in that it presents a system that can be solved conclusively. I'm aware that the environment constantly provides a quantifiable amount of heat energy and that this is not a closed system in a perfect vessel, but for the sake of this discussion the time-and-environment element isn't to be a factor.
The point is that the bartenders you've been talking to are not talking about ideal systems. They're talking about real glasses containing real drinks being served to real people, which means the environment can't be ignored.
Yes, your ideal case can be solved conclusively. But the resulting solution won't have anything to do with reality, which makes it a pointless intellectual exercise. Even if it contradicted what the bartenders said, that wouldn't prove the bartenders to be wrong.
posted by Chocolate Pickle at 2:12 PM on December 31, 2008 [2 favorites]
The point is that the bartenders you've been talking to are not talking about ideal systems. They're talking about real glasses containing real drinks being served to real people, which means the environment can't be ignored.
Yes, your ideal case can be solved conclusively. But the resulting solution won't have anything to do with reality, which makes it a pointless intellectual exercise. Even if it contradicted what the bartenders said, that wouldn't prove the bartenders to be wrong.
posted by Chocolate Pickle at 2:12 PM on December 31, 2008 [2 favorites]
The glass-to-environ point is a non-issue. The solution to the original question is simply that heat transfer is a function of contacting surfaces. The liquid has more contact with the smaller ice, and can thus transfer more heat, causing it to melt quicker. How the liquid interacts with the environment is important, sure, but not to the original question of why larger ice cubes melt slower.
posted by phrakture at 2:16 PM on December 31, 2008
posted by phrakture at 2:16 PM on December 31, 2008
For what it's worth, I asked a similar question, albeit different in structure, some time back about soup of differing viscosities and how they'd come to room temperature.
In fact, it's somewhat the opposite issue, in a way. Except, you know, not really.
posted by disillusioned at 2:19 PM on December 31, 2008
In fact, it's somewhat the opposite issue, in a way. Except, you know, not really.
posted by disillusioned at 2:19 PM on December 31, 2008
The simplifying assumption is valid enough in that it presents a system that can be solved conclusively. I'm aware that the environment constantly provides a quantifiable amount of heat energy and that this is not a closed system in a perfect vessel, but for the sake of this discussion the time-and-environment element isn't to be a factor.
Sure, if you assume the drink is a closed system, then it doesn't matter what size the ice is. But the drink isn't a closed system.
Let's consider two extreme cases. Assume we are using enough ice to bring the entire system to equilibrium. If we pour the booze over well crushed ice and stir, equilibrium will be reached very quickly. If we pour the booze over the same quantity of ice in 80mm spheres and stir, it will take much longer to reach equilibrium. This is fairly intuitive.
There are two different questions we can ask: 1) how much ice melts during the time to reach equilibrium and 2) how much ice melts in a fixed time. If we consider how much ice melts during the time to reach equilibrium, clearly there will be more melting in the ice-sphere drink, because it takes much longer for the ice-sphere drink to reach equilibrium and thus there is much more energy transferred to the drink from the environment. If we consider how much ice melts over a fixed period of time, more ice has melted in the crushed-ice drink. Since the crushed-ice drink cools faster, it has a lower average temperature over the fixed period. This lower average temperature means more energy is transferred from the environment (the greater heat difference between the drink and the environment causes a greater rate of heat transfer) and thus more ice is melted.
posted by ssg at 2:20 PM on December 31, 2008 [1 favorite]
Sure, if you assume the drink is a closed system, then it doesn't matter what size the ice is. But the drink isn't a closed system.
Let's consider two extreme cases. Assume we are using enough ice to bring the entire system to equilibrium. If we pour the booze over well crushed ice and stir, equilibrium will be reached very quickly. If we pour the booze over the same quantity of ice in 80mm spheres and stir, it will take much longer to reach equilibrium. This is fairly intuitive.
There are two different questions we can ask: 1) how much ice melts during the time to reach equilibrium and 2) how much ice melts in a fixed time. If we consider how much ice melts during the time to reach equilibrium, clearly there will be more melting in the ice-sphere drink, because it takes much longer for the ice-sphere drink to reach equilibrium and thus there is much more energy transferred to the drink from the environment. If we consider how much ice melts over a fixed period of time, more ice has melted in the crushed-ice drink. Since the crushed-ice drink cools faster, it has a lower average temperature over the fixed period. This lower average temperature means more energy is transferred from the environment (the greater heat difference between the drink and the environment causes a greater rate of heat transfer) and thus more ice is melted.
posted by ssg at 2:20 PM on December 31, 2008 [1 favorite]
Big snow piles last longer than small snow piles, too. Not a truly closed system, though it feels that way in February.
posted by theora55 at 2:25 PM on December 31, 2008
posted by theora55 at 2:25 PM on December 31, 2008
"Shaken not stirred" is the way James Bond insists on his martinis being made. And this thread helps to show why. Shaking the drink rapidly brings the warmer liquid in touch with more ice surface in a shorter time, resulting in quicker cooling of the liquid with less dilution resulting from melting.
Now, let us ask. Big ice or small ice? Wouldn't the greater surface area that transfers heat faster also melt the ice (that donates the heat) faster? Is it a wash?
posted by megatherium at 2:29 PM on December 31, 2008
Now, let us ask. Big ice or small ice? Wouldn't the greater surface area that transfers heat faster also melt the ice (that donates the heat) faster? Is it a wash?
posted by megatherium at 2:29 PM on December 31, 2008
Okay, let's back up a second and check what our abstractions are based on.
Joules are a measure of the mysterious - thing? property? - we call energy. Energy is not a thing in the sense of having a position. "Ah, here we are, a few joules of energy off the starboard bow!". Energy is an abstraction used to describe (in this case) the motion of particles with mass (in this case, water molecules, the solutions of air and glass, etc).
Temperature is the average amount of energy for a group of particles. You can imagine a big collection of beach balls bouncing around and measuring the oomph each one has, which will depend on how fast it's moving (velocity) and how much it weighs (mass).
Now we get in the the murkier waters (no pun intended) of phase. The appearance (macroscopically) of something as a liquid or solid (in fact very nebulous terms, with many states between) is a manifestation of the behavior of it's constituent particles. Densely packed particles are solid, spread out a bit more they become liquid, and then in to various gaseous forms. It gets a little trickier though because most types of particle behave differently at different scales. Water is strongly polar and cohesive, so some complicated things happen between those beachballs when they're close to one another (they form crystals, for one).
So anyway, if we imagine the icecube situation as a bunch of microscopic beach balls, it becomes clearer why smaller cubes melt faster. The greater the surface area, the more beach balls are knocking up against the air/glass particles, and in turn knocking against water particles further inside the cube (heat conduction). The greater number of water particles interacting with air/glass particles relative to the total number, the higher the temperature (this is a near-tautology... the more particles are moving faster, the higher the average speed of all the particles under consideration). These things knock around and grow further apart from one another, and we call this melting and the resultant spread apart molecules a liquid.
posted by phrontist at 2:33 PM on December 31, 2008
Joules are a measure of the mysterious - thing? property? - we call energy. Energy is not a thing in the sense of having a position. "Ah, here we are, a few joules of energy off the starboard bow!". Energy is an abstraction used to describe (in this case) the motion of particles with mass (in this case, water molecules, the solutions of air and glass, etc).
Temperature is the average amount of energy for a group of particles. You can imagine a big collection of beach balls bouncing around and measuring the oomph each one has, which will depend on how fast it's moving (velocity) and how much it weighs (mass).
Now we get in the the murkier waters (no pun intended) of phase. The appearance (macroscopically) of something as a liquid or solid (in fact very nebulous terms, with many states between) is a manifestation of the behavior of it's constituent particles. Densely packed particles are solid, spread out a bit more they become liquid, and then in to various gaseous forms. It gets a little trickier though because most types of particle behave differently at different scales. Water is strongly polar and cohesive, so some complicated things happen between those beachballs when they're close to one another (they form crystals, for one).
So anyway, if we imagine the icecube situation as a bunch of microscopic beach balls, it becomes clearer why smaller cubes melt faster. The greater the surface area, the more beach balls are knocking up against the air/glass particles, and in turn knocking against water particles further inside the cube (heat conduction). The greater number of water particles interacting with air/glass particles relative to the total number, the higher the temperature (this is a near-tautology... the more particles are moving faster, the higher the average speed of all the particles under consideration). These things knock around and grow further apart from one another, and we call this melting and the resultant spread apart molecules a liquid.
posted by phrontist at 2:33 PM on December 31, 2008
megatherium: Big icecubes will lower the temperature more slowly, smaller ones more quickly, but in the end they reach the same temperature (given the same amount of ice). The big cube will "last longer", and "cool longer", but it will cool commensurately less.
posted by phrontist at 2:37 PM on December 31, 2008
posted by phrontist at 2:37 PM on December 31, 2008
As others have hinted, this question must be addressed from the perspective of the drinker, not the physicist.
One does not consume the whole drink at a gulp (except where required by the circumstances, of course). There are a couple of practical consequences of this:
First, the individual amount actually being consumed is not a representative sample of the entire beverage. Whether sipped or taken through a straw, it passes through the ice at the top of the glass to the drinker's mouth. In either case, the majority of the ice is engaged in cooling that particular sip immediately before it is consumed - in that exquisite moment when the lips touch the glass and the beverage in question sluices over the ice to meet them. Therefore, it is desirable to have a larger mass of ice remaining to effect this final cooling, rather than to imbibe a liquid that is slightly cooler overall through a smaller final barrier of ice.
Second, there is not a finite amount of liquid in the glass over time. If most of the small ice melts before the drink is half consumed, then that first half-drink might indeed be cooler than half-a-drink served with big ice, but the remaining portion of the drink will just get warmer and warmer. If most of the big ice remains after half a drink, then the cooling effect on the second half will only be more pronounced.
"Shaken not stirred" is the way James Bond insists on his martinis being made. And this thread helps to show why. Shaking the drink rapidly brings the warmer liquid in touch with more ice surface in a shorter time...
Bond is a philistine in this respect. Everyone ought to know that vigorously shaking a martini results in ice chips which seriously dilute the carefully crafted mixture. If you want an ice-cold martini, serve it in a chilled glass and drink it fast.
posted by Urban Hermit at 3:11 PM on December 31, 2008
One does not consume the whole drink at a gulp (except where required by the circumstances, of course). There are a couple of practical consequences of this:
First, the individual amount actually being consumed is not a representative sample of the entire beverage. Whether sipped or taken through a straw, it passes through the ice at the top of the glass to the drinker's mouth. In either case, the majority of the ice is engaged in cooling that particular sip immediately before it is consumed - in that exquisite moment when the lips touch the glass and the beverage in question sluices over the ice to meet them. Therefore, it is desirable to have a larger mass of ice remaining to effect this final cooling, rather than to imbibe a liquid that is slightly cooler overall through a smaller final barrier of ice.
Second, there is not a finite amount of liquid in the glass over time. If most of the small ice melts before the drink is half consumed, then that first half-drink might indeed be cooler than half-a-drink served with big ice, but the remaining portion of the drink will just get warmer and warmer. If most of the big ice remains after half a drink, then the cooling effect on the second half will only be more pronounced.
"Shaken not stirred" is the way James Bond insists on his martinis being made. And this thread helps to show why. Shaking the drink rapidly brings the warmer liquid in touch with more ice surface in a shorter time...
Bond is a philistine in this respect. Everyone ought to know that vigorously shaking a martini results in ice chips which seriously dilute the carefully crafted mixture. If you want an ice-cold martini, serve it in a chilled glass and drink it fast.
posted by Urban Hermit at 3:11 PM on December 31, 2008
for the sake of this discussion the time-and-environment element isn't to be a factor.
I'm no physicist, but you're asking why bartenders like larger ice - because it melts slower. The fact that the same amount of heat transfer has occurred at the end of all melting matters not at all - the time element is the factor that is important, and the factor that makes drinkers prefer the larger cubes of ice. Ideally you want to have finished your drink before the melting has completed, leaving still-large chunks of ice in the glass. It may be quite true that shape doesn't matter to total heat transfer, but it still matters when you care about time to consumption and avoiding a watered down drink.
posted by Miko at 3:29 PM on December 31, 2008
I'm no physicist, but you're asking why bartenders like larger ice - because it melts slower. The fact that the same amount of heat transfer has occurred at the end of all melting matters not at all - the time element is the factor that is important, and the factor that makes drinkers prefer the larger cubes of ice. Ideally you want to have finished your drink before the melting has completed, leaving still-large chunks of ice in the glass. It may be quite true that shape doesn't matter to total heat transfer, but it still matters when you care about time to consumption and avoiding a watered down drink.
posted by Miko at 3:29 PM on December 31, 2008
I could swear this was called "buckling" when I took thermodynamics - the ratio of the surface area to the volume. It greatly affected the specific rate of heat transfer in or out of a material
Now a Google search only brings up what you would normally think of for that word, even when combined with other likely clues. Stupid military education.
posted by ctmf at 3:31 PM on December 31, 2008
Now a Google search only brings up what you would normally think of for that word, even when combined with other likely clues. Stupid military education.
posted by ctmf at 3:31 PM on December 31, 2008
. If you want an ice-cold martini, serve it in a chilled glass and drink it fast.
Well...The glass alone won't chill the drink enough. You also have to pre-chill the alcohol by gently swirling it around large ice cubes in a pre-chilled metal or glass shaker or mixing glass for a few seconds. Otherwise, the room-temp liquor going into the chilled glass will immediately warm the glass and the drink will be tepid.
posted by Miko at 3:32 PM on December 31, 2008
Well...The glass alone won't chill the drink enough. You also have to pre-chill the alcohol by gently swirling it around large ice cubes in a pre-chilled metal or glass shaker or mixing glass for a few seconds. Otherwise, the room-temp liquor going into the chilled glass will immediately warm the glass and the drink will be tepid.
posted by Miko at 3:32 PM on December 31, 2008
Also, the rate of heat transfer = UA (delta)t, where U is a property of the material (fixed), A is heat transfer surface area, and (delta)t is the temp difference between the ice and the drink.
So, a warmer drink will melt the ice faster. Ice with more surface area will melt faster but cool the drink down more quickly. Using some other material, like dry ice, may affect "U" and thus, the melting time. Might also make your drink taste terrible.
posted by ctmf at 3:34 PM on December 31, 2008
So, a warmer drink will melt the ice faster. Ice with more surface area will melt faster but cool the drink down more quickly. Using some other material, like dry ice, may affect "U" and thus, the melting time. Might also make your drink taste terrible.
posted by ctmf at 3:34 PM on December 31, 2008
If you want an ice-cold martini, serve it in a chilled glass and drink it fast.
Well...The glass alone won't chill the drink enough. You also have to pre-chill the alcohol by gently swirling it around large ice cubes in a pre-chilled metal or glass shaker or mixing glass for a few seconds. Otherwise, the room-temp liquor going into the chilled glass will immediately warm the glass and the drink will be tepid.
Well, I was being a bit glib there, but the point is that shaking is out.
posted by Urban Hermit at 3:41 PM on December 31, 2008
Well...The glass alone won't chill the drink enough. You also have to pre-chill the alcohol by gently swirling it around large ice cubes in a pre-chilled metal or glass shaker or mixing glass for a few seconds. Otherwise, the room-temp liquor going into the chilled glass will immediately warm the glass and the drink will be tepid.
Well, I was being a bit glib there, but the point is that shaking is out.
posted by Urban Hermit at 3:41 PM on December 31, 2008
Once the drink has been properly cooled, does large ice melt more slowly than an equal mass of small ice over the 30 minutes to one hour a drink might be nursed?
One point that that the simple heat/energy analysis misses is that water doesn't freeze at 0 degrees C any more, once it is mixed with alcohol. (Thus that famous invention, antifreeze.)
Thus a huge difference between the large ice cubes and the small ice chips emerges: It is on that boundary between the ice cube and the liquid that the frozen water can potentially mix with alcohol, lowering its freezing point so that it can become liquid without changing temperature.*
Just to give a specific example:
Let's say you have a situation (in your perfect vessel and without any outside heat transfer) in which both the liquid (which for our purposes consists of water mixed with ethanol) and the ice cubes (100% water) are at exactly -2C.
At the temperature the water/ethanol mixture is liquid whereas the ice cubes (100% water in theory) are below their freezing point.
Now in simplistic theory once everything had reached this steady state at -2C there it would stay, permanently.
But in reality something different will happen. Shaking, stirring, brownian motion, etc., will lead the water molecules at the edge of the ice to mix with the liquid (water/ethanol mixture). The result of that will be some minuscule amount of the solid ice water at -2C becoming liquid water/ethanol mix at -2C.*
In short, the ice cubes will be melting (ie slowly turning into liquid) even though they are not melting (reaching the normal melting point of water at 0C).
This melting phenomenon obviously happens at the boundary between the ice & the liquid.
Almost all of the "chip ice" is on this boundary and so what you will find is in fairly short order the chip ice will convert from solid ice (100% water) at -2C to liquid water/ethanol solution at -2C.*
By contrast, almost all of the large ice cube is internal to the cube, completely removed from the opportunity to mix with the ethanol. So it will just remain there as pure water ice at -2C.
In short, the chip ice will convert to -2C liquid much quicker than the large ice cubes, which will remain solid frozen -2C ice much longer.
Call this whatever you want but the result of chip ice will be more liquid water mixed in with your ethanol--in short, a more watery drink.
*In reality saying that it "won't change temperature" and will go from -2C water ice to -2C water/ethanol mix with no thermal consequences is a simplification. I confess I'm not knowledgeable to work out the details, what with the ethanol, mixtures, and all. I *think* that you'll go from a state with -2C water/ethanol liquid and -2C water ice cubes to a state where all is liquid but at a slightly lower temperature. Or maybe it's a slightly higher temperature. What's important for this point though is it can all be done *below* the freezing temperature of water. Maybe it goes -2C to -3C, or maybe -2C to -1C. Either way, in a 100% water ice cube should stay frozen "in theory" in any below zero temperature. But in reality it's going to eventually convert to water/ethanol mix at a temp. below freezing, because of mixing, brownian motion, etc.
And it's in the "eventually" that your difference between the small ice chips and large ice cubes will be found.
posted by flug at 10:36 PM on December 31, 2008
One point that that the simple heat/energy analysis misses is that water doesn't freeze at 0 degrees C any more, once it is mixed with alcohol. (Thus that famous invention, antifreeze.)
Thus a huge difference between the large ice cubes and the small ice chips emerges: It is on that boundary between the ice cube and the liquid that the frozen water can potentially mix with alcohol, lowering its freezing point so that it can become liquid without changing temperature.*
Just to give a specific example:
Let's say you have a situation (in your perfect vessel and without any outside heat transfer) in which both the liquid (which for our purposes consists of water mixed with ethanol) and the ice cubes (100% water) are at exactly -2C.
At the temperature the water/ethanol mixture is liquid whereas the ice cubes (100% water in theory) are below their freezing point.
Now in simplistic theory once everything had reached this steady state at -2C there it would stay, permanently.
But in reality something different will happen. Shaking, stirring, brownian motion, etc., will lead the water molecules at the edge of the ice to mix with the liquid (water/ethanol mixture). The result of that will be some minuscule amount of the solid ice water at -2C becoming liquid water/ethanol mix at -2C.*
In short, the ice cubes will be melting (ie slowly turning into liquid) even though they are not melting (reaching the normal melting point of water at 0C).
This melting phenomenon obviously happens at the boundary between the ice & the liquid.
Almost all of the "chip ice" is on this boundary and so what you will find is in fairly short order the chip ice will convert from solid ice (100% water) at -2C to liquid water/ethanol solution at -2C.*
By contrast, almost all of the large ice cube is internal to the cube, completely removed from the opportunity to mix with the ethanol. So it will just remain there as pure water ice at -2C.
In short, the chip ice will convert to -2C liquid much quicker than the large ice cubes, which will remain solid frozen -2C ice much longer.
Call this whatever you want but the result of chip ice will be more liquid water mixed in with your ethanol--in short, a more watery drink.
*In reality saying that it "won't change temperature" and will go from -2C water ice to -2C water/ethanol mix with no thermal consequences is a simplification. I confess I'm not knowledgeable to work out the details, what with the ethanol, mixtures, and all. I *think* that you'll go from a state with -2C water/ethanol liquid and -2C water ice cubes to a state where all is liquid but at a slightly lower temperature. Or maybe it's a slightly higher temperature. What's important for this point though is it can all be done *below* the freezing temperature of water. Maybe it goes -2C to -3C, or maybe -2C to -1C. Either way, in a 100% water ice cube should stay frozen "in theory" in any below zero temperature. But in reality it's going to eventually convert to water/ethanol mix at a temp. below freezing, because of mixing, brownian motion, etc.
And it's in the "eventually" that your difference between the small ice chips and large ice cubes will be found.
posted by flug at 10:36 PM on December 31, 2008
"Over whatever time T it takes to cool the drink to 0° C, without considering environmental factors, either the same amount of ice melts into liquid or it doesn't."
There are two fallacies embedded here that you might want to think about.
Fallacy number one is that the drink will cool to 0° C. (And what--stop? What's keeping a water/ethanol liquid from dropping in temperature below 0° C?) Given a sufficient quantity of ice at sufficiently low temperature, the drink (ethanol/water mixture, with freezing point below 0° C) will end up significantly below 0° C. So behavior of the ice cubes both above and below zero is important.
Fallacy number two is that ice can cool the drink only by melting. Or as you put it another way, "same amount of heat be transferred into the ice, resulting in an equal quantity of melt". Melting isn't the only way to transfer heat!
Imagine a 100 ccs of ice at -50° C dropped into 100 ccs of water at 5° C. You'll end up with 200 ccs of frozen ice. Lots of cooling, no melting (well, there might be some brief intermediate melting, which would almost immediately re-freeze).
(And you'd better think about intermediate melting. In the ice/cocktail scenario, that intermediate melting won't re-freeze, it will all end up as liquid water. So is the large or small ice cube scenario more likely to result in some parts of the ice temporarily becoming warmer than some other parts, resulting in them melting?)
Now imagine 100ccs of ice at -0° C mixed with 100 ccs of water at 5° C. Now a whole bunch of the cooling that happens will come from the melting.
Or to put it another way, the ice can take joules out of the liquid either to raise its own temperature (say from -15° C to -5° C) or to melt (at 0° C).
Assuming complete lack of stirring/shaking/mixing it's quite possible to assume that a very cold ice cube almost immediately cools the liquid immediately surrounding itself to below 0° C. From then all the energy transfer is by conduction and none by melting (because both the ice cube and immediately surrounding liquid are below 0° C).
Note that a large ice cube will be more efficient at this (cooling immediately surrounding liquid) than a small one. That is because the higher mass-to-surface-area ratio of the large ice cube means that it has more coldness "in reserve" to cool the liquid immediately surrounding it to a colder temperature quicker.
A large ice cube is likely to bring the liquid in its immediate area close to the ice cube's temperature quickly. So if the large ice cube starts out well below zero the temperature of the liquid around it will soon be below zero.
A small ice cube will be the opposite--it will be surrounded by large masses of liquid in comparison to its own mass and will be relatively more influenced to take on the temperature of the surrounding liquid. In other words, it will get warmer faster. There will be higher chance that some portion of it will reach 0° C where it can start melting.
In short, the final steady-state temperature may be the same but the in-between stages will be quite different for small vs large ice cubes. What happens in those in-between stages could very well influence how much of the ice cubes melt in the process.
Assuming sufficient and cold enough ice cubes, the steady state would be reached when both ice cubes and liquid reach some point below 0° C. In this scenario, with a large enough and cold enough ice cube, you could have essentially no melting of the ice cube at all. The ice cube almost immediately generates a layer of liquid around it below 0° C which continues until the system reaches steady state. At steady state the temperature of both ice cube and liquid is below 0° C. So throughout that whole process, the ice cube is only just briefly (when first put into the cocktail) in a state where it's boundary is at 0° C and the immediately surrounding liquid is above 0° C.
The small ice cubes, on the other hand would have more of their surface area spend more time at or near 0° C, meaning they have more potential to produce more melted water in the process.
Any type of shaking or stirring only exacerbates the problem for the smaller ice cubes. When you shake or stir the cocktail, the inside of the large ice cube remains intact and separate/insulated from the warmer liquid.
When you crush the large ice cube into small ice cubes however, the inside has now become exposed at the surface. When you stir, all that surface is exposed to the (warmer) liquid, un-insulated, and that means those areas could be (at least temporarily) at 0° C, giving them the chance to melt.
posted by flug at 12:10 AM on January 1, 2009
There are two fallacies embedded here that you might want to think about.
Fallacy number one is that the drink will cool to 0° C. (And what--stop? What's keeping a water/ethanol liquid from dropping in temperature below 0° C?) Given a sufficient quantity of ice at sufficiently low temperature, the drink (ethanol/water mixture, with freezing point below 0° C) will end up significantly below 0° C. So behavior of the ice cubes both above and below zero is important.
Fallacy number two is that ice can cool the drink only by melting. Or as you put it another way, "same amount of heat be transferred into the ice, resulting in an equal quantity of melt". Melting isn't the only way to transfer heat!
Imagine a 100 ccs of ice at -50° C dropped into 100 ccs of water at 5° C. You'll end up with 200 ccs of frozen ice. Lots of cooling, no melting (well, there might be some brief intermediate melting, which would almost immediately re-freeze).
(And you'd better think about intermediate melting. In the ice/cocktail scenario, that intermediate melting won't re-freeze, it will all end up as liquid water. So is the large or small ice cube scenario more likely to result in some parts of the ice temporarily becoming warmer than some other parts, resulting in them melting?)
Now imagine 100ccs of ice at -0° C mixed with 100 ccs of water at 5° C. Now a whole bunch of the cooling that happens will come from the melting.
Or to put it another way, the ice can take joules out of the liquid either to raise its own temperature (say from -15° C to -5° C) or to melt (at 0° C).
Assuming complete lack of stirring/shaking/mixing it's quite possible to assume that a very cold ice cube almost immediately cools the liquid immediately surrounding itself to below 0° C. From then all the energy transfer is by conduction and none by melting (because both the ice cube and immediately surrounding liquid are below 0° C).
Note that a large ice cube will be more efficient at this (cooling immediately surrounding liquid) than a small one. That is because the higher mass-to-surface-area ratio of the large ice cube means that it has more coldness "in reserve" to cool the liquid immediately surrounding it to a colder temperature quicker.
A large ice cube is likely to bring the liquid in its immediate area close to the ice cube's temperature quickly. So if the large ice cube starts out well below zero the temperature of the liquid around it will soon be below zero.
A small ice cube will be the opposite--it will be surrounded by large masses of liquid in comparison to its own mass and will be relatively more influenced to take on the temperature of the surrounding liquid. In other words, it will get warmer faster. There will be higher chance that some portion of it will reach 0° C where it can start melting.
In short, the final steady-state temperature may be the same but the in-between stages will be quite different for small vs large ice cubes. What happens in those in-between stages could very well influence how much of the ice cubes melt in the process.
Assuming sufficient and cold enough ice cubes, the steady state would be reached when both ice cubes and liquid reach some point below 0° C. In this scenario, with a large enough and cold enough ice cube, you could have essentially no melting of the ice cube at all. The ice cube almost immediately generates a layer of liquid around it below 0° C which continues until the system reaches steady state. At steady state the temperature of both ice cube and liquid is below 0° C. So throughout that whole process, the ice cube is only just briefly (when first put into the cocktail) in a state where it's boundary is at 0° C and the immediately surrounding liquid is above 0° C.
The small ice cubes, on the other hand would have more of their surface area spend more time at or near 0° C, meaning they have more potential to produce more melted water in the process.
Any type of shaking or stirring only exacerbates the problem for the smaller ice cubes. When you shake or stir the cocktail, the inside of the large ice cube remains intact and separate/insulated from the warmer liquid.
When you crush the large ice cube into small ice cubes however, the inside has now become exposed at the surface. When you stir, all that surface is exposed to the (warmer) liquid, un-insulated, and that means those areas could be (at least temporarily) at 0° C, giving them the chance to melt.
posted by flug at 12:10 AM on January 1, 2009
"isn't actually large enough within the constraints of a kitchen that the velocity of heat transfer from surface to core is a factor"
Sorry to go on & on, but this is fallacy #3.
If there is indeed a difference between the large & small ice cubes, it is exactly in the transient effects like this that the difference lies.
posted by flug at 12:18 AM on January 1, 2009
Sorry to go on & on, but this is fallacy #3.
If there is indeed a difference between the large & small ice cubes, it is exactly in the transient effects like this that the difference lies.
posted by flug at 12:18 AM on January 1, 2009
Response by poster:
[Here, I'm sure, is where somebody will chime in with an anecdote about some molecular mixologist who uses some science lab technology to do it exactly that way for a particular drink. I'd like to see that in action, but it's also not how I'm going to make a Manhattan at home].
Again, the question isn't, "what could be done in a lab," it's, "are these effects pronounced enough to matter in the quantities in use and temperatures expected in the average kitchen?"
posted by fedward at 5:35 AM on January 1, 2009
Fallacy number one is that the drink will cool to 0° C. (And what--stop? What's keeping a water/ethanol liquid from dropping in temperature below 0° C?) Given a sufficient quantity of ice at sufficiently low temperature, the drink (ethanol/water mixture, with freezing point below 0° C) will end up significantly below 0° C. So behavior of the ice cubes both above and below zero is important.Not so much a fallacy as a shorthand gone wrong. I've never assumed that the equilibrium temperature would be exactly 0, but I didn't do a good job of explaining that. In hindsight, one of the questions should have been, "what would the equilibrium temperature of such a mixture be?"
Fallacy number two is that ice can cool the drink only by melting. Or as you put it another way, "same amount of heat be transferred into the ice, resulting in an equal quantity of melt". Melting isn't the only way to transfer heat!This isn't a fallacy, it's a ground rule. The question wasn't "what can be done in a lab in order to produce a particular result," it was all about what can happen in a kitchen with the ingredients starting at certain temperatures. When mixing cocktails, one should start with the liquid ingredients at room temperature, and one is also unlikely to have a fridge that cools to -50 ° C at home. The rule of thumb is that after shaking or stirring, the drink will have gained enough water through melting to increase its volume by about a third. If you're cooling all your liquid ingredients to 5 ° C before mixing a cocktail, you're doing it wrong.
Imagine a 100 ccs of ice at -50° C dropped into 100 ccs of water at 5° C. You'll end up with 200 ccs of frozen ice. Lots of cooling, no melting (well, there might be some brief intermediate melting, which would almost immediately re-freeze).
[Here, I'm sure, is where somebody will chime in with an anecdote about some molecular mixologist who uses some science lab technology to do it exactly that way for a particular drink. I'd like to see that in action, but it's also not how I'm going to make a Manhattan at home].
If there is indeed a difference between the large & small ice cubes, it is exactly in the transient effects like this that the difference lies.I'm happy to admit to the third fallacy, but not just on the basis of hand-waving. What's the applicable theory on heat transfer in a solid? To what degree would this effect actually change the result? I took college calculus so it's not like I'm afraid of the math. I just had no idea how to look something like that up.
Again, the question isn't, "what could be done in a lab," it's, "are these effects pronounced enough to matter in the quantities in use and temperatures expected in the average kitchen?"
posted by fedward at 5:35 AM on January 1, 2009
Response by poster:
I'm also not asking for answers to the age-old question, "shaken or stirred?" There are too many factors to consider for there to be one right answer for all cocktails, and in the end if Nick Charles wants the bartender to shake his Martini to waltz time, then the bartender would be well advised to keep Mr. Charles happy.
posted by fedward at 6:04 AM on January 1, 2009
I'm no physicist, but you're asking why bartenders like larger ice.No, I'm not. I have a pretty good idea why they like it already. When you stir or shake a cocktail with ice, some of the ice is going to melt into the drink, and the drink is eventually going to reach a temperature where stirring or shaking any longer isn't going to make it any colder. At that point for many cocktails the ice is strained out anyway, putting an end to both cooling and melting.
I'm also not asking for answers to the age-old question, "shaken or stirred?" There are too many factors to consider for there to be one right answer for all cocktails, and in the end if Nick Charles wants the bartender to shake his Martini to waltz time, then the bartender would be well advised to keep Mr. Charles happy.
posted by fedward at 6:04 AM on January 1, 2009
Response by poster: To recap:
1. Many people have cited the increased surface area as if, due to that one factor, the rest is just hand waving. These people didn't read the question closely enough.
2. In a similar vein, many people keep trying to invalidate (or just ignore) the closed system, insisting that environmental factors trump all others. To put this to rest: I don't deny the environment plays a factor, but we're talking about a cocktail with somewhere between 100ml and 200ml of liquid, cooled by ice to a near-equilibrium in a matter of seconds no matter whether it's stirred or shaken. Once you solve for the closed system (the shaker or pitcher) you can account for the extra melt caused by the heat bath that is the world. When it comes to this cooling phase, the extra melt caused by the environment is outside the scope of my question. You wanna know whether shaking or stirring leads to more melt in the real world, ask your own question.
3. There have been several answers that get at the meat of the question:
Nat wrote:
Also, I'll add a new extra credit question: using the mix described in the question, and the temperatures given (ice at -15° C, liquids at 20° C), how does one predict the equilibrium temperature, and what would that temperature be given, say, 200g of ice? What about 400g?
posted by fedward at 10:02 AM on January 1, 2009
1. Many people have cited the increased surface area as if, due to that one factor, the rest is just hand waving. These people didn't read the question closely enough.
2. In a similar vein, many people keep trying to invalidate (or just ignore) the closed system, insisting that environmental factors trump all others. To put this to rest: I don't deny the environment plays a factor, but we're talking about a cocktail with somewhere between 100ml and 200ml of liquid, cooled by ice to a near-equilibrium in a matter of seconds no matter whether it's stirred or shaken. Once you solve for the closed system (the shaker or pitcher) you can account for the extra melt caused by the heat bath that is the world. When it comes to this cooling phase, the extra melt caused by the environment is outside the scope of my question. You wanna know whether shaking or stirring leads to more melt in the real world, ask your own question.
3. There have been several answers that get at the meat of the question:
Nat wrote:
So the when the ice is still busy melting, the number of molecules falling off the solid and into the liquid is greater than the number reattaching. (If it's freezing, we're going the other way). As you can imagine, only molecules at the surface can do this, and the process takes time (as well as needing heat input, from your hands or the room temp. environment, and as always, increasing the entropy of the system).7segment wrote:
Because only molecules at the surface have the option to fall off, surface to volume ratio is quite important. This ratio goes as 1/radius. The rate of melting will be directly related to this (or maybe related to some positive power of it, sorry I didn't check the units of other quantities involved or actually do any calculations).
the heat being transfered from the room to the drink depends on their relative temperatures, which you could reasonably well approximate with Netwon's law of cooling. You end up having to solve a pretty nasty set of differential equations once you take the environment into account.ctmf wrote:
Also, the rate of heat transfer = UA (delta)t, where U is a property of the material (fixed), A is heat transfer surface area, and (delta)t is the temp difference between the ice and the drink.Before addressing the fallacies in my question, flug wrote:
But in reality something different will happen. Shaking, stirring, brownian motion, etc., will lead the water molecules at the edge of the ice to mix with the liquid (water/ethanol mixture). The result of that will be some minuscule amount of the solid ice water at -2C becoming liquid water/ethanol mix at -2C.*So. The question of large ice vs small ice in a drink served on the rocks is pretty well answered, I think, since it can be illustrated that the amount of continued melting in the drink at room temperature is a function of the surface area of the ice. The question of how much difference there is during the shaking, stirring, etc. is still open.
In short, the ice cubes will be melting (ie slowly turning into liquid) even though they are not melting (reaching the normal melting point of water at 0C).
This melting phenomenon obviously happens at the boundary between the ice & the liquid.
Also, I'll add a new extra credit question: using the mix described in the question, and the temperatures given (ice at -15° C, liquids at 20° C), how does one predict the equilibrium temperature, and what would that temperature be given, say, 200g of ice? What about 400g?
posted by fedward at 10:02 AM on January 1, 2009
The question of how much difference there is during the shaking, stirring, etc. is still open.
It's open if what you're interested in is the quantification. Empirically, drinks are in a shaker for between 10-30 seconds. If you use crushed ice, you need less time in the shaker. If you use cubes, you can take a little more time before getting a wet drink. Again, sure they end up equally cold, but the bartender adjusts his or her technique so as to pour at the maximum coldness with minimum melted-ness, so they change the time variable based on the desired outcome. Crushed ice cools a drink a lot faster but you can't let the drink stay in long, or as the melting begins it will begin to warm faster, too. Cubes mean you can leave the liquor in longer, which is nice for drinks that you are stirring or mixing end-over-end rather than shaking.
posted by Miko at 1:32 PM on January 1, 2009
It's open if what you're interested in is the quantification. Empirically, drinks are in a shaker for between 10-30 seconds. If you use crushed ice, you need less time in the shaker. If you use cubes, you can take a little more time before getting a wet drink. Again, sure they end up equally cold, but the bartender adjusts his or her technique so as to pour at the maximum coldness with minimum melted-ness, so they change the time variable based on the desired outcome. Crushed ice cools a drink a lot faster but you can't let the drink stay in long, or as the melting begins it will begin to warm faster, too. Cubes mean you can leave the liquor in longer, which is nice for drinks that you are stirring or mixing end-over-end rather than shaking.
posted by Miko at 1:32 PM on January 1, 2009
flug, I have a difficult time believing your argument about conduction and ice size. Given that energy transferred from the drink to the ice must either raise the temperature of some of the ice or melt some of the ice and that we know the amount of energy transferred to the ice must be equal no matter what size the ice is, we know that for there to be less melting if there is more conduction and vice versa. So, your argument that there will be less melting of larger ice cubes, means that there must be more energy transferred by conduction. Thus we must believe that the average temperature of the remaining ice would need to be higher for the large ice cubes versus the crushed ice. I think it would be much more likely that the crushed ice will be warmed to near the equilibrium temperature much faster than the large ice cubes.
Maybe in some ideal world, if one were to leave a drink on the rocks sitting for long periods, the liquid ethanol/water layer that you write about might have some effect. In the real world, the drink is going to be stirred by the motion of the glass when raised and lowered by the drinker and the temperature of the liquid is going to be pretty much the same throughout.
Finally, keep in mind the heat capacity and heat of fusion of ice. Gram for gram, the energy needed to melt ice is an order of magnitude larger than the energy needed to raise the temperature of ice from -15C to 0C.
posted by ssg at 4:29 PM on January 1, 2009
Maybe in some ideal world, if one were to leave a drink on the rocks sitting for long periods, the liquid ethanol/water layer that you write about might have some effect. In the real world, the drink is going to be stirred by the motion of the glass when raised and lowered by the drinker and the temperature of the liquid is going to be pretty much the same throughout.
Finally, keep in mind the heat capacity and heat of fusion of ice. Gram for gram, the energy needed to melt ice is an order of magnitude larger than the energy needed to raise the temperature of ice from -15C to 0C.
posted by ssg at 4:29 PM on January 1, 2009
It's the same reason why fat people are hot. Lower ratio of surface area to volume, so heat transfer is slower. This kind of reasoning can be learned from The Search for Solutions.
posted by neuron at 10:08 PM on January 1, 2009
posted by neuron at 10:08 PM on January 1, 2009
These people didn't read the question closely enough.
A good general rule is that if everyone is misunderstands the question you've written, the problem is not with the reader.
posted by electroboy at 8:40 AM on January 2, 2009 [5 favorites]
A good general rule is that if everyone is misunderstands the question you've written, the problem is not with the reader.
posted by electroboy at 8:40 AM on January 2, 2009 [5 favorites]
The question of how much difference there is during the shaking, stirring, etc. is still open.
Okay, consider the extreme, then: ice broken into pieces the size of slush. If you were to take a single ice cube and shake it in a cocktail shaker for 1 minute, you would expect a good proportion of that ice cube to still be present. If you bash that same ice cube into snowflake-sized pieces (increasing the surface area), placed that snow in the drink, and shook for 1 minute, you would probably not expect any of the slush to be present at the end of the minute.
There is nothing magical about slush-sized pieces of ice; if you blew them up with a microscope they would look just like tiny cubes of ice. The fact that the same mass of ice now has a much greater surface area is the reason that it melts faster.
Let's take heat out of the equation: what would you expect to dissolve faster in a cup of water: a large piece of rock candy, or the same mass of superfine sugar? It's a different process but the principle--that greater surface area affects the rate of change for the given substance--is the same.
I barely remember my high school science at all, so I may be completely wrong here. But that is how I understand the "it's the surface area" handwaving to come about.
Also, how exactly is the shaker supposed to be a "closed environment" if one of the conditions is that it is being shaken externally?
posted by Deathalicious at 1:19 AM on January 3, 2009
Okay, consider the extreme, then: ice broken into pieces the size of slush. If you were to take a single ice cube and shake it in a cocktail shaker for 1 minute, you would expect a good proportion of that ice cube to still be present. If you bash that same ice cube into snowflake-sized pieces (increasing the surface area), placed that snow in the drink, and shook for 1 minute, you would probably not expect any of the slush to be present at the end of the minute.
There is nothing magical about slush-sized pieces of ice; if you blew them up with a microscope they would look just like tiny cubes of ice. The fact that the same mass of ice now has a much greater surface area is the reason that it melts faster.
Let's take heat out of the equation: what would you expect to dissolve faster in a cup of water: a large piece of rock candy, or the same mass of superfine sugar? It's a different process but the principle--that greater surface area affects the rate of change for the given substance--is the same.
I barely remember my high school science at all, so I may be completely wrong here. But that is how I understand the "it's the surface area" handwaving to come about.
Also, how exactly is the shaker supposed to be a "closed environment" if one of the conditions is that it is being shaken externally?
posted by Deathalicious at 1:19 AM on January 3, 2009
whoa, a lot of people care a lot about this question!
posted by lostburner at 5:53 PM on January 13, 2009
posted by lostburner at 5:53 PM on January 13, 2009
Mod note: Final update from the OP:
I now consider this article to be authoritative on the matter.posted by mathowie (staff) at 3:40 PM on June 4, 2013
This thread is closed to new comments.
posted by iamabot at 12:10 PM on December 31, 2008 [12 favorites]