The full details based on Wikipedia Heat Loss From Vacuum Flask
This is a pretty comprehensive math description, but from other sources, and when you do your experiment, you can check, it should take about 4/5 hours to drop in temperature.
The rate of heat (energy) loss through a vacuum flask can be analyzed thermodynamically, starting from the second $TdS$ relation
$${\displaystyle TdS=dH-VdP}$$
$${\displaystyle T_{surr}\Delta S=mc_{p}\delta T-Vdp}$$
Assuming constant pressure throughout the process,
$${\displaystyle T_{surr}\Delta S=c_{p}\left(T_{b'}-T_{c}\right)}$$
Rearranging the equation in terms of the temperature of the outside surface of the vacuum flask's inner wall,
$${\displaystyle T_{b'}=T_{c}+{\frac {T_{surr}\Delta S}{c_{p}}}}$$
Where
$T_{surr}$ is the temperature of the surrounding air
$∆S$ is the change in specific entropy of stainless steel
$c_p$ is the specific heat capacity of stainless steel
$T_c$ is the temperature of the liquid contained within the flask
$T_b'$ is the temperature of the outside surface of the vacuum flask's inner wall
Now consider the general expression for heat loss due to radiation:
$${\displaystyle Q'_{0}=A\epsilon \sigma \left(T^{4}-T_{0}^{4}\right)}$$
In the case of the vacuum flask,
$${\displaystyle Q'_{0}=A_{in}\epsilon _{s.s.}\sigma \left(T_{b}'^{4}-T_{surr}^{4}\right)}$$
Substituting our earlier expression for $T_b'$,
$${\displaystyle Q'_{0}=A_{in}\epsilon _{s.s.}\sigma \left[\left(T_{c}+{\frac {T_{surr}\Delta S}{c_{p}}}\right)^{4}-T_{surr}^{4}\right]}$$
Where
$Q'_0$ is the rate of heat transfer by radiation through the vacuum portion of the flask.
$A_{in}$ is the surface area of the outside of the inner wall of the flask
$ε_{s.s}$. is the emissivity of stainless steel
$σ$ is the Stefan–Boltzmann constant
Assuming that the outer surface of the inner wall and the inner surface of the outer wall of the vacuum flask are coated with polished silver to minimize heat loss due to radiation, we can say that the rate of heat absorption by the inner surface of the outer wall is equal to the absorptivity of polished silver times the heat radiated by the outer surface of the inner wall,
$${\displaystyle \alpha Q'_{0}=Q'_{in}}$$
In order for energy balance to be maintained, the heat lost through the outer surface of the outer wall must be equal to the heat absorbed by the inner surface of the outer wall,
$${\displaystyle Q'_{in}=Q'_{out}}$$
Since the absorptivity of polished silver is the same as its emissivity, we can write
$${\displaystyle Q'_{out}=\epsilon _{s.s.}Q'_{0}}$$
We must also consider the rate of heat loss through the lid of the vacuum flask (assuming it is made of polypropylene, a common plastic) where there is no vacuum inside the material. In this area, the three heat transfer modes of conduction, convection, and radiation are present. Therefore, the rate of heat loss through the lid is,
$${\displaystyle Q'_{lid}=Q'_{cond}+Q'_{conv}+Q'_{rad}}$$
$${\displaystyle Q'_{lid}=kA_{lid}\left({\frac {T_{b}-T_{surr}}{\Delta x}}\right)+hA_{lid}\left(T_{b}-T_{surr}\right)+A_{lid}\epsilon _{p.p.}\sigma \left[\left(T_{c}+{\frac {T_{surr}\Delta S_{p.p.}}{c_{p}^{p.p.}}}\right)^{4}-T_{surr}^{4}\right]}$$
Where
$k$ is the thermal conductivity of air
$h$ is the convective heat transfer coefficient of free air
$ε_{p.p}$ is the emissivity of polypropylene
$A_{lid}$ is the outer surface area of the lid
$c_{pp.p}$. is the specific heat capacity of polypropylene
$∆S_{p.p}$. is the specific entropy of polypropylene
$∆x$ is the distance over which conduction across the temperature gradient takes place
Now we have an expression for the total rate of heat loss, which is the sum of the rate of heat loss through the walls of the vacuum flask and the rate of heat loss through the lid,
$${\displaystyle Q'_{total}=Q'_{out}+Q'_{lid}}$$
where we substitute each of the expressions for each component into the equation.
