# Quicker way to cool liquid

I need to prepare a bottle of baby milk from formula quickly. To prepare it I must use some boiling water to sterilise the powder however it must be served at just above room temperature for the baby to accept it so I top up the bottle with cold water and leave it until it reaches the desired temperature.

I'm curious to know what the faster method would be:

1. let the smaller amount of boiling water cool down first then add the cold water

2. add the cold water as soon as the powder is sterile

I'm not sure if quantities are important, but there is normally $~60mL$ of boiling water and $~150mL$ of room temperature water.

• You have two methods, why not test them and find out? Sep 21, 2015 at 16:46
• Also, I have never heard of sterlizing the formula powder (certainly the bottles, though). Are you sure you are reading the instructions correctly? Sep 21, 2015 at 16:47
• One could bubble liquid nitrogen through it (or liquid helium!) I suppose. On the other hand, optimizing the time to prepare never seemed very useful to me - the prep was always shorter than the time to make sure it wasn't too hot. Go with what works easily, particularly at 2am when very groggy (so complexity is right out). Sep 21, 2015 at 18:00
• While this is certainly off-topic... the adding of cold tap water destroys the sterility. You certainly couldn't use procedure b) either in a biology lab or in an industrial food production process without getting into major trouble. Sep 21, 2015 at 20:03
• If you take away the red herring of sterilization, it's a good question, similar to the coffee/creamer problem. Sep 22, 2015 at 17:44

Surprisingly (it surprised me!) it doesn't make much difference. Well, if I do an approximate calculation it makes no difference but real life will differ a bit from my approximation. This is how it's done.

The cooling will be approximately described by Newton's law of cooling. This tells us that the rate of heat loss will be:

$$W \approx kA\Delta T \tag{1}$$

where $\Delta T$ is the temperature difference, i.e. the temperature of the water minus the ambient temperature, $A$ is the surface area of the object and $k$ is some constant that has to be determined experimentally.

Take the case where you keep the boiling and cold water seperate. We'll use $T_h$ and $T_a$ for the temperature of the hot water and the ambient temperatures, and we'll assume the cold water is at ambient temperature. We'll take the mass of the hot water to be $M_h$ and the mass of the cold water to be $M_c$. The rate of heat loss is then simply:

$$W_1 \approx kA(T_h - T_a)$$

Now suppose we mix the hot and cold water. We'll use $x_h$ to indicate the fraction of hot water i.e. the ratio:

$$x_h = \frac{M_h}{M_h + M_c}$$

where $M_h$ and $M_c$ are the masses of the hot and cold water. The fraction of cold water $x_c$ is just equal to $1-x_h$. Now, the temperature of the mixture, $T_m$, will be:

$$T_m = T_hx_h + T_ax_c = T_hx_h + T_a(1 - x_h)$$

The area $A$ changes as well. If you're pouring the water into a roughly cylindrical bottle the area will be proportional to the total mass of the water, so the area of the mixed water will be roughly:

$$A_m \approx A \frac{M_h + M_c}{M_h} \approx A \frac{1}{x_h}$$

Now we have $T_m$ and $A_m$ we can plug them into equation (1) to get the rate of heat loss for the mixture:

\begin{align} W_2 &= k A_m (T_m - T_a) \\ &= k A \frac{1}{x_h} \left( T_hx_h + T_a(1 - x_h) - T_a \right) \\ &= k A \frac{1}{x_h} \left( T_hx_h + T_a - T_ax_h - T_a \right) \\ &= k A \left( T_h - T_a \right) \\ &= W_1 \end{align}

and we end up with the surprising result that the two rates of heat loss are the same. This happens because although the hot water cools faster, the mixture has a greater surface area to cool through and the two effects cancel out.

Now, I did start out by saying this was just an approximation. Newton's law of cooling tends to fail at smaller temperature difference i.e. cold things cool more slowly that it would predict. So my guess is that keeping the water separate would actually cool it faster. However the difference probably won't be that great.

• There could be a noticeable difference due to more evaporation from the hotter surface. Dec 18, 2016 at 20:37

I'm going to give an answer that is more practical, rather than physics related. In my opinion, it would make sense to make up several bottles of milk with water as hot as you think you need. Then, let them cool off a bit and put them in the refrigerator. When you need them, put them in the microwave oven for 10-20 seconds, depending on the volume, take them out and shake them, and try a drop of warm milk on your wrist. After trying this with just one or two bottles, you'll know exactly how long to reheat them before using them, but remember to ALWAYS test a drop on your wrist before giving it to the baby.

I've dealt with newborns too, and the sleep deprivation that is involved makes parents a bit "punchy" after several days. Strictly following a procedure is a very good idea, because of the consequences of getting things wrong. For my final disclaimer - all of the nice physics equations in the world will not come close to alleviating the guilt involved if you scald your infant son or daughter because you made a physics omission or miscalculation!