Should the thermos flask better be half full or half empty? Every evening I am preparing hot water for my two year old son wakes up in the night to get his milk. We use a rather bad isolation can for this. It is a typical metal cylinder shaped can holding half a liter. If I put cooking hot water into it, I know that about 5 hours later it will have room temperature already, but it does the job as my son typically wakes up 2 or three hours after I go to bed, and so he gets his milk temperated.
As I need only about 200ml then to mix up his milk, I was asking myself if it is better to only fill in that amount of hot water or to fill up the whole can.
I guess losing temperature has much to do with the amount of water but also with its surface touching the (colder) room air outside.
With no idea anymore of what my old physics teacher told me twenty five years ago I hope you could share some wisdom for my little story here. Thanks in advance ;)
 A: The idealised formula (lumped thermal analysis) for a cooling object, according to Newton's Cooling Law is:
$$T(t)=T_{\infty}-(T_{\infty}-T_0)\exp\Big(-\frac{t}{\tau}\Big)$$
where $\tau$ is the characteristic time:
$$\frac{1}{\tau}=\frac{U A}{m c_p}$$
with:


*

*$T(t)$ is the temperature of the object in time $t$

*$T_0$ is the object's starting temperature and $T_{\infty}$ the surroundings' temperature

*$A$ is the surface area of the object exposed to the surroundings

*$m$ is the mass of the object, with specific heat capacity $c_p$

*finally, $U$ is the overall heat transfer coefficient. Better quality thermos flasks will generally have lower values of $U$
From this we can conclude that for larger $m$ the rate of cooling will be slower. Note however that greater mass usually also implies greater $A$, thereby somewhat offsetting the mass-effect.

Following in the footsteps of @probably_someone (comments) I'll explore $\frac{A}{m}$ for the following shape:


With $m=\rho V$ and $\rho$ (density) a constant we can evaluate $\frac{A}{V}$ instead:
$$V=\frac{\pi D^2}{4}H+\frac12 \frac 43 \pi \Big(\frac{D}{2}\Big)^3=\frac{\pi D^2 H}{4}+\frac{\pi D^3}{12}=\frac{\pi D^2(3H+D)}{12}$$
$$A=\frac12 4\pi \Big(\frac{D}{2}\Big)^2+2 \pi \Big(\frac{D}{2}\Big)H=\frac12 \pi D^2+\pi D H=\frac{\pi D(2H+D)}{2}$$
This gives us a result for $\frac{A}{V}$ of:
$$\frac{A}{V}=\frac{6(2H+D)}{D(3H+D)}$$
Here $D$ is a constant and only $H$ increases with $m$. Because the coefficient $3$ in the denominator, instead of $2$ in the numerator, as $H$ increases $\frac{A}{V}$ does indeed decrease which points to lower cooling rate, as expected. 

Another consideration. Assume we half-fill the flask with boiling water. Due to steam formation it is reasonable to expect the entire flask to reach approximately the same temperature. In that case, again as an approximation, the total surface area $A$ could be used.
A: It's a metal can so the heat from the water will spread across the whole surface and be lost at approximately the same rate however much water you use.
Now, since a larger volume of water will hold more heat to start with, a full can will keep its temperature better, as the temperature loss will be shared across a greater mass of water.
A: Actually the heat of liquid you have poured in flask is half of the volume of Thermos so the heat of liquid will get conventionally transfer to air which is another half of Thermos so better to fill it to full to get long time to put water hot.
