Does too little ice make my martinis watery?

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.

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.