# Best way to chill a cup of coffee with cold water and 5 minutes [duplicate]

## Initial data

• 1 x 3/4 full cup of hot coffee / tea / your favorite morning beverage
• cold water
• 5 minutes

Considering that it's starting to get hot outside, and we all want to drink reasonably cold coffee / tea (well, in any case, not scalding hot), I have the following options:

1. Wait for 5 minutes

So the hot coffee releases as much energy as possible in the 5 minutes, then add cold water to fill the coffee cup

So the temperature is lowered as fast as possible in the start, then let the air do the rest.

## Question

Which option cools the coffee more ? Are there other, better options ?

From what I can figure out, waiting 5 minutes first is the most obvious choice, but I noticed that physics isn't always as intuitive as we may think.

PS: Would a spoon to stir the beverage have any impact on the difference between the two ?

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## marked as duplicate by Qmechanic♦Jun 17 '13 at 13:07

You tagged this "home-experiment," so how about trying it and telling us what happens! :) Otherwise this sounds like a standard homework question. –  Michael Brown Jun 17 '13 at 5:03
well I always use three ice cubes to bring my tea to drinkable temperature. takes 30 seconds. –  anna v Jun 17 '13 at 5:14
@MichaelBrown: I did try this, but didn't really notice a difference :) It's not a homework btw, it's just a curiosity. @ anna: unfortunately at work I don't have ice cubes, but it's a good tip –  Vlad Preda Jun 17 '13 at 6:12
Possible duplicates: physics.stackexchange.com/q/5265/2451 and links therein. –  Qmechanic Jun 17 '13 at 7:34
I had a somewhat related question cooking.stackexchange.com/q/25510 –  Vixen Jun 17 '13 at 7:49

Lets take each case and make some calculation. So, the first case, waiting for 5 minutes than adding some cold water. Assume the following values:

The initial "hot" temperature of the coffee $T_H=80^{\circ}C$

The temperature of the surrounding medium $T_m=23^{\circ}C$

Using Newtons cooling law

$$\frac{dT}{dt}=-k(T-T_m)$$

and after a simple integration we get

$$T_c=T_m+(T_H-T_m)\mathrm{e}^{-kt}$$

Taking $k=0.05$ we find $T_c$ to be $T_c\simeq63^{\circ}C$. Now we have to add some cold water. Lets say we add $1/4^{th}$ of a cup at $T_{cold}=10^{\circ}C$.

$$-Q_{coffee}=Q_{water}$$ $$-cm_{c}\Delta T_{coffe}=cm_{w}\Delta T_{water}$$ $$-\frac{3}{4}(T_f-T_{c})=\frac{1}{4}(T_f-T_{cold})$$

And we find $T_f$ to be $T_f\simeq49^{\circ}C$

Now lets look at the second method, mixing them from the start. We just have to replace the numerical value for $T_c$ in the above formula with the initial temperature of the coffee $T_H$. Doing this, we find that the temperature after mixing is $T_{f}'\simeq 62^{\circ}C$.

Thus, as Neuneck already said, the first method is the best one. These are at best some approximate calculations, but even so, the difference is clearly visible.

Edit:

As a response to the comments, if you add hotter and hotter water to your coffee the final temperature after mixing will be higher (it will grow linearly with the temperature of the water you add.) Here is a plot for it.

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Very nice, exactly what I was looking for –  Vlad Preda Jun 17 '13 at 7:12
Wouldn't the cold water get warmer in 5 minutes? So, for example, instead of adding water with 10°C at the beginning, we would add water with 14°C after 5 minutes. Does this affect the result? –  Uooo Jun 17 '13 at 10:46
@w4rumy: interesting point, in my case though that's not an issue, either the cold water is at room temperature, or I get it at a fixed temperature from the fridge. But would love to also see how this impacts the overall result :) –  Vlad Preda Jun 17 '13 at 13:03
@VladPreda I've edited my question with a "nice" plot. –  nijankowski Jun 17 '13 at 13:22

Option 1 is the way to go. To a good approximation the cooling process has an exponentially decaying temperature, so the rate of cooling is proportional to the temperature itself. If you add the cool water first, you will reduce the cooling effect of waiting 5 minutes.

We did this experiment back in high school. The assumption that enters here is that the temperature drop caused by the addition of cold water is approximately independent of the beverage's temperature. This is justified as long as the temperature difference between cold water and hot beverage is large. I guess one could put more detail into the simulation by calculating the mixing temperature explicitly, though...

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The temperature after mixing would in good approximation be the weighted average of the two liquids before mixing. –  Bernhard Jun 17 '13 at 6:07
I figured that something similar would happened, but couldn't quite explain it properly, thanks –  Vlad Preda Jun 17 '13 at 6:14

If you have an additional mug, then the most effective way to cool the coffee will be to run the empty mug under cold water to chill it, dry it, then pour your coffee into that mug. Then repeat with cooling the empty mug with water and pouring your coffee back into it.

Or you could just pour the coffee between the two cups continuously to aid cooling by exposing it to their air.

Stirring will help too, as it will keep the coffee at a constant temperature rather than the top layer being a colder temperature than the rest (which will reduce the rate of cooling.)

For actually mixing cold water with your drink, this should be done last for the reasons mentioned in Nijankowski V's answer; you will lose more heat in the same time by being at a higher temperature.

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## protected by Qmechanic♦Jun 17 '13 at 8:48

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