Does too little ice make my martinis watery?

Question

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:

• What cools a drink?
• Is an ice globe the worst possible way to cool a drink (with ice)?
• Calculate mass of ice needed to cool water $\Delta$T degrees

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?

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Footnotes:

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!

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Edit

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.

Answer 1

Thoughts:

• 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.

Answer 2

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.