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:
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let the smaller amount of boiling water cool down first then add the cold water
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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.
Best Answer
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.