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When you stick ice in a drink, AFAICT (the last physics I took was in high school) two things cool the drink:

  • The ice, being cooler than the drink, gets heat transferred to it from the drink (Newton's law of cooling). (This continues after it melts, too.)
  • The ice melts, and the cool liquid mixes with the drink, which makes the mixture feel cooler.
My first question is, to what extent does each of these cool the drink: which has a greater effect, and how much greater?

Secondly, how much of the cooling by the first method (heat transfer) is without melting the ice? That is, is there any significant amount of heat transfer to each speck of ice before it's melted, and how much does that cumulatively affect the drink's temperature?

I suppose all this will depend on a bunch of variables, like the shape and number of ice cubes and various temperatures and volumes. But any light that can be shed, let's say for "typical" situations, would be appreciated.

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3 Answers 3

up vote 8 down vote accepted

There are three processes to take into account:

  1. The warming of ice towards the melting point if it was originally below $0^{\circ} C$.
  2. The melting of ice itself
  3. The warming of the resulting water

The 1. and 3. part is addressed by heat capacity of ice and water respectively and the amount of heat will be directly proportional to temperature difference and weight of the water/ice. The proportionality constant (actually it also depends on the temperature but not very strongly so let's just ignore that) is called specific heat. For water it is about twice as large as that of ice at temperatures around $0^{\circ} C$.

As for the 2. part, this has to do with latent heat. Simply put, this is an amount of heat you need to change phases without changing temperature. Less simply put, when warming you are just converting the heat into greater wiggling of water molecules around their stable positions in the crystal thereby increasing their temperature. But at the melting point that heat will instead go into breaking chemical bonds between molecules in the ice lattice.

Now, latent heat is really big (you need lots of energy to break those bonds). To get a hang on it: you would need the same amount of heat to warm water from $0^{\circ} C$ to $80^{\circ} C$ as you would need to melt the same amount of ice.

Now, presumably you want your drink cold in the end so that temperature for 3. will be close to $0^{\circ} C$ and also the ice cubes should be pretty warm (no use in producing ice cubes of e.g. $-50^{\circ} C$, right?). This means that these processes won't contribute much cooling. It's fair to say that melting of the ice takes care of everything.

Note: we can also quickly estimate how much ice you need by neglecting the processes 1. and 3. Say you are starting with a warm drink of $25^{\circ} C$ and you want to get it to $5^{\circ} C$. So, reusing the argument about the $80^{\circ} C$ difference being equivalent to a latent heat of the same mass, we see that you need four times less ice than water to get the job done.

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Thank you, and other respondents. –  msh210 Dec 20 '10 at 20:14
  1. If you want to cool the drink by heat transfer from the drink to ice then you should not put ice into it. You should surround the glass of liquid with ice.
  2. Suppose you pour cold water(melted ice) into the drink then

the heat difference between the drink and the mixture is
$H_h = m_h c_h (T_h - T_m)$

similarly, for cold water
$H_c = m_c c_c (T_m - T_c)$

Here, $H_h = H_c$

$T_m = \frac{m_h c_h T_h + m_c c_c T_c}{m_h c_h + m_c c_c}$.

So, whenever $T_m < T_h$ we say resultant mixture is cooled.

--copied shamelessly from

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Suppose we can an ice cube at 0 °C with mass mi kg, and some water at temperature T with mass mw kg.

1.
The specific latent heat of ice is Li = 334 000 J kg-1, and the specific heat capacity of water is 4200 J K-1 kg-1. Therefore, to completely melt the ice cube, the water will need to release heat of (334000 m_i) J, i.e. the temperature drop is $$334000 m_i = 4200 m_w \Delta T \implies \Delta T = \frac{334000 m_i}{4200 m_w}.$$

As discussed in the previous question, if the drink is not pure water, the heat capacity can become less than 4200 J K-1 kg-1, and this only makes ΔT larger.

2.
When the completely melted ice is mixed with water. The final temperature will be the weighted average of the individual mixture after all heat transfer. $$ m_w T + m_i 0 = (m_w + m_i) (T - \Delta T) \implies \Delta T = \frac{m_i}{m_w+m_i}T $$

During the melting of ice, the temperature will not be a constant (T is decreasing), and thus the ΔT by this mixing process will also become less. The result thus only gives an upper bound of temperature decrease.

 

Consider an example case where your drink is at room temperature 20 °C, the mass of water is 100 g and the mass of ice cube is 10 g. Then the ΔT just by melting is 8 C°, and that by mixing is at most 2 C°, this shows the cooling is mainly due to melting.

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