When I make martinis, the recipe I use is 2.5 fluid ounces1 of gin2 and 1/2 fluid ounce dry vermouth3, shaken or stirred4 with 7 ice cubes5, then strained into a cocktail glass (I mix until it's cold enough, and not for some specific length of time.)

My Observation

If I use less ice, the martini is more diluted; if I use more ice, the martini is stronger (too strong in fact—I prefer some dilution.)

My Hypothesis

With a lot of ice, the drink chills very fast, not giving the ice as much time to melt, and vice-versa.

My Friends

think I am crazy about the amount of ice6. They claim that (a) the ice has to melt to chill the drink so no matter how much ice I start with, the same amount is melted to reach the desired temperature, and (b) I had to retake Physics 2A in college7, so what do I know anyway?

I pointed out that I could chill the drink with really cold rocks (a.k.a. whiskey stones) and it could get just as cold with no melting and maybe they are missing something.

The answers to these questions seem to support my friends' argument that the melting of the ice is the overwhelming contributor to the cooling:

However my experiment seems to demonstrate otherwise.

My Experiment

I tested this by making six martinis, two each with 4, 7, and 10 ice cubes. I used the same amount of gin and vermouth in each. I weighed8 the ingredients before adding them to the mixing cup. I stirred until the desired temperature9 was achieved. Then I weighed the amount after straining.

My Result

The 4-cube martinis gained more weight than the 7-cube martinis, which in turned gained more weight than the 10-cube martinis. I assume the weight gain was the melted ice. My friends happily drank all the martinis, but remained unconvinced of my Physics acumen, either theoretical or experimental10.

My Question

Am I exhibiting confirmation bias and my stupid friends are right? Or is there an explanation for my hypothesis and observation and they should finally shut up about my having to retake Physics 2A because come on it was like 30 years ago already and besides, I'm not using Planck's constant to calculate how long it takes atoms to slow to a halt, I'm just making martinis!

Is the ratio of nearly 2:1 ice-to-liquid a factor? How about the constant mixing? Or is this dependent on the starting temperature of the ice and even though the warming of the ice contributes very little, it is enough to affect the outcome?


  1. Apologies for the use of American measurements in this scholarly context, but that's what my utensils are labeled with.

  2. No apologies for the insistence on gin. If you want to use vodka or make some other cocktail and call it a martini, we have nothing further to discuss.

  3. I use a 5:1 ratio. Others may quibble. They are wrong.

  4. Seriously, I do not care. Can I please continue?

  5. For my ice trays, this is 5 fluid ounces of water, prior to freezing, in a plain old kitchen freezer.

  6. Although they are all too happy to drink the martinis I make.

  7. These are friends that I went to college with so I can't argue that point, but there were three intramural softball playoff games the weekend before finals, so when was I going to study? But I digress...

  8. Postal scale, accurate to 1/10 (American again) ounce, sorry.

  9. 28 degrees American Fahrenheit on my kitchen thermometer that measures to 1/10 degree, but accuracy unknown.

  10. I am willing to rerun that experiment as long as necessary until I get it right!


I appreciate the answers and I am accepting the answer from @cyberx86 because (a) it independently supports my insistence that 7 ice cubes is The Right Number, and (b) What kind of physics site would we have if we didn't reward showing your work?

However no one really "put a bow on it" so I'll add

My Conclusions

Since the final temperature of the martini drops below 0° C and there is also ice melting, there must be both melting and absorption occurring together during the mixing.

The measurable change in the amount of melting in this case is due to the addition of the heat absorption capacity of more ice, but only because it is below 0° C to start with.

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    $\begingroup$ I love physics (ex-scientist, PhD) and I cry over how physics is taught at schools. It is not a surprise that a lot of children do not like it because of just that: examples of logs slipping on a slope instead of real-life problems like this one. They end up with a learnt-by-heart Schrödinger equation and do not understand the kWh on their electricity bill. This is to say that the question is great on its own, but the reasoning and experimental effort behind is even greater. $\endgroup$
    – WoJ
    Commented Jul 28, 2017 at 6:38
  • 2
    $\begingroup$ Wait, let me understand: when you say that the 4-cube martinis gained more weight you mean after removing what remained of the ice cubes? $\endgroup$
    – valerio
    Commented Jul 28, 2017 at 9:09
  • $\begingroup$ Everyone is leaving out a huge factor in how ice behaves in drinks: surface area. Larger ice cubes with smaller surface area will cool a drink more with less dilution than shaved ice, which will melt quite quickly by comparison. This is why high end bars go through the extra effort to make and offer spherical ice that just barely fits in a tumbler. Not practical for a martini glass, but the size and shape of the ice is a factor. $\endgroup$ Commented Jul 28, 2017 at 12:33
  • 1
    $\begingroup$ @valerio92, yes the gain was after removing the ice by straining the martini into a cocktail glass, but before adding an olive. $\endgroup$
    – bmb
    Commented Jul 28, 2017 at 15:20
  • $\begingroup$ more ice would be a larger heat sink and therefore could provide more cooling before the phase change at that point cooling occurs as the latent heat of fusion melts the ice. $\endgroup$
    – pete
    Commented Nov 14, 2020 at 3:37

2 Answers 2



  • If the ice is at 0 degrees Celsius, so that the energy to chill the drink is absorbed by the latent heat of fusion, then it seems to me that your friends are right. Cooling the gin by a certain amount requires you to melt a certain amount of ice, independently of how many ice cubes you use.
  • If the ice is so cold that a single cube can chill the entire drink without melting, then again it doesn't matter how many you use (because there will be no melting).
  • In the intermediate case, which is probably the realistic one if you're getting your ice from a home freezer, the ice will first cool the drink until it reaches its melting point, and then begin to melt. More ice cubes will allow the first part of this process to account for more of the cooling, so the drink will indeed be less watery.
  • You should probably just also keep your gin in the freezer if you like martinis enough to go through all this.
  • $\begingroup$ Thanks, AGML. The key part of your answer, "the ice will first cool the drink until it reaches its melting point, and then begin to melt" seems to be merely restating my hypothesis, "with a lot of ice, the drink chills very fast, not giving the ice as much time to melt." My friends, and the links to other questions, want to refute that by claiming that the warming of the ice does not contribute enough. Why do you think it does? $\endgroup$
    – bmb
    Commented Jul 28, 2017 at 1:22
  • 3
    $\begingroup$ It's not the same because it has nothing to do with time. It depends on how much energy is required to bring the ice to its melting point. Whether the warming of the ice contributes "enough" depends on the concentration of water you feel is significant, the temperatures of the ice and gin, the shape and volume of the ice cubes, and maybe the purity of the ice. Very roughly, heating a gram of ice by 1 degree will heat 0.75 grams of gin by the same. Supposing the ice starts at -20C that could make a difference. $\endgroup$
    – AGML
    Commented Jul 28, 2017 at 1:54
  • 1
    $\begingroup$ If it's not about time, why write "...first cool the drink until it reaches its melting point, and then begin to melt" Are you suggesting the ice just sits there not melting until the entire cube reaches 0 degrees C? $\endgroup$
    – bmb
    Commented Jul 28, 2017 at 4:32
  • 2
    $\begingroup$ Yes, the entire cube will reach zero, or very near, quickly. When you drop them in water the surface does not start flaking off. They crack right through, and very quickly. It does not just all go liquid because of the great heat of fusion. It takes a lot of energy for the water to change from solid to liquid, and it does so without changing temperature. So, it has to melt from the outside in as heat becomes available. Ice starts at -20. Warms to zero. Stays at zero until it melts. I agree with AGML. With ice from a typical freezer, more cubes means less melt. Cool some cubes with dry ice. $\endgroup$ Commented Jul 28, 2017 at 6:56
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    $\begingroup$ Arguably gin from the freezer is too cold for martinis. Unless you're putting in a frankly unreasonable quantity of room-temperature vermouth. Of course, if it's too cold you can sit there stirring it until it's drinkable, whereas if it's warm you're ruined. An alternative approach, when you're concerned about dilution by ice, is to use overproof gin. $\endgroup$ Commented Jul 28, 2017 at 8:53

In order for the martini to cool, heat is transferred from it, into the ice. This results in the martini's temperature dropping and the ice's temperature rising. Once the ice reaches its melting point, it's temperature will not rise, but a state change will occur (this is the latent heat). It is only if the ice melts that water is added to your martini and dilutes it.

Here is an attempt and expressing this idea mathematically, there are lots of assumptions made.

$1 fl. oz = 0.0295735 mL$

$c_{water} = 4184 J/kg^{\circ}C$

$D_{water} = 1 kg/L$

$c_{ice} = 2108 J/kg^{\circ}C$

$L_{ice} = 333550 J/kg^{\circ}C$

$c_{ethanol} = 2460 J/kg^{\circ}C$

$D_{ethanol} = 0.789 kg/L$

Additionally, the following assumptions are made:

  • Once melted, the water is ignored in the temperature calculations
  • Gin is assumed to be 45% ethanol and 55% water, other components are ignored
  • Vermouth is assumed to be 18% ethanol and 82% water, other components are ignored
  • All specific heat capacities of mixtures are calculated as weighted averages
  • Freezer temperature is assumed to be -18 degrees Celsius
  • Room temperature is assumed to be 20 degrees Celsius

${T_{inital}}_{martini} = 20^{\circ}C$

${T_{final}}_{martini} = -2.22^{\circ}C$

${T_{inital}}_{ice} = -18^{\circ}C$

${T_{final}}_{ice} = 0^{\circ}C$

$V_{gin} = 2.5 fl. oz = 0.0739 L $

$D_{gin} = 0.905 kg/L $

$m_{gin} = 0.0669 kg $

$c_{gin} = 3408.2 J/kg^{\circ}C $

$V_{vermouth} = 0.5 fl. oz = 0.0148 L $

$D_{vermouth} = 0.962 kg/L $

$m_{vermouth} = 0.0142 kg $

$c_{vermouth} = 3873.68 J/kg^{\circ}C $

$V_{martini} = 3 fl. oz = 0.0887 L $

$D_{martini} = 0.915 kg/L $

All of this, gives us the following numbers for the martini:

$m_{martini} = 0.0811 kg $

$c_{martini} = 3485.78 J/kg^{\circ}C $

The energy lost by the martini to bring it to its final temperature is:

$Q_{martini} = m_{martini}c_{martini}({T_{final}}_{martini}-{T_{inital}}_{martini})$

$Q_{martini} = (0.0811 kg)(3485.78 J/kg^{\circ}C)(-2.22^{\circ}C-25^{\circ}C)$

$Q_{martini} = -6285 J $

This means that the ice needs to absorb 6285 J of energy. Part of this will occur by an increase in the temperature of the ice, any additional energy will go into melting the ice.

The mass of an ice cube is based on 5 fl. oz for 7 ice cubes:

$m_{icecube} = 0.02112 kg $

The amount of energy each ice cube can absorb as its temperature rises to 0 degrees Celsius is:

$Q_{ice} = m_{ice}c_{ice}({T_{final}}_{ice}-{T_{inital}}_{ice})$

$Q_{ice} = (0.02112 kg)(2108 J/kg^{\circ}C)(0^{\circ}C-(-18^{\circ}C))$

$Q_{ice} = 801.5 J/cube $

The amount of energy each ice cube would absorb if it completely melted is:

$Q_{melt} = m_{ice}L_{ice}$

$Q_{melt} = (0.02112 kg)(333550 J/kg^{\circ}C)$

$Q_{melt} = 7046 J/cube $

This means, that if we have enough ice cubes, assuming all absorb heat from the martini equally, that more ice cubes will be able to remove more heat before needing to melt. Additionally, this means that after a certain point, additional ice cubes shouldn't make an appreciable difference. (Assuming that the ice cubes are removed once the target temperature is reached, and not allowed to melt).

Ice Cubes    Energy from Melting    Water Added (mL)     Estimated Final Mass (g)
1             -5483                  16                        97.6
2             -4681                  14                        95.2
3             -3880                  12                        92.8
4             -3078                   9.2                      90.4
5             -2277                   6.8                      88.0
6             -1475                   4.4                      85.6
7              -673.8                 2.0                      83.2
8               127.7                 0                        81.1
9               929.2                 0                        81.1
10             1731                   0                        81.1

You should be able to verify the basic trend (mass decreases with additional ice cubes until a point which it plateaus) with the experimental setup you used. You can compare your final masses to those estimated in the table above, although I would expect your measured values to be somewhat higher than these. These numbers do not include the energy needed to cool the mixing vessel.

  • 3
    $\begingroup$ Nice work, but also notice that this calculation ignores that ice cubes might melt at the outside while still beeing well below the melting point at the inside. However this doesn't change the trend. $\endgroup$
    – Anedar
    Commented Jul 28, 2017 at 8:53
  • 1
    $\begingroup$ That is true - would be interested to know how much the ice actually melts. I would guess that a thin layer of martini, surrounding the ice, reaches an equilibrium temperature with it, and then slowly sinks due to its increased density. It seems that you might get varying amounts of ice melting depending on how the drink is mixed. $\endgroup$
    – cyberx86
    Commented Jul 28, 2017 at 9:05
  • $\begingroup$ So it seems those other questions apply if the ice is close to 0 C, but each degree below 0 C increases the energy a cube can absorb before melting by about 44.5 J. This is .6 % of the contribution, so at -18 C, 11% of the cooling is from energy absorption. This is not as negligible as described in those other questions. Excellent answer; thanks for showing your work! $\endgroup$
    – bmb
    Commented Jul 28, 2017 at 17:38

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