What you need to apply is the engineering equations for a heat exchanger. In the below equation, $\dot{Q}$ is the heat transfer rate, $UA$ is the heat transfer coefficient times the area and $\Delta T_m$ is the log mean temperature difference (LMTD).
$$\dot{Q}=U A \Delta T_m$$
For this problem there are 3 barriers to the heat transfer in series. You have the convective heat transfer coefficient of the syrup $h_c$, the convective heat transfer of the steam $h_s$, and that of the pipe metal $h_R$. Take $n$ to be the number of pipes, $r_i$ to be the inner radius of the pipe, and $r_0$ to be the outer radius of the pipe, and $R$ the heat transfer coefficient of the Copper.
$$UA = \frac{2\pi n}{\frac{1}{h_s r_i}+\frac{1}{R L} ln \frac{r_o}{r_i} +\frac{1}{h_c r_0} }$$
The LMTD is the average temperature difference, but for the specific case where one part of the heat exchanger is saturated, and thus constant temperature, just know that it is the following, where $T_{s}$ is the saturation temperature of the steam, or 100 degrees C. Then $T_h$ and $T_c$ are the temperatures after and before preheating respectively.
$$\Delta T_m = \frac{T_h-T_c}{ln\frac{T_h-T_s}{T_h-T_s} }$$
The hard part of the above equations is the $h_c$ and the $h_s$. You will probably use things like the Dittus-Boelter correlation. But it's more important that for the moment we address $\dot{Q}$ itself. For the described hood, I see two possibilities.
- The steam is being provided at a faster rate than it is being condensed
- The steam is being provided at a slower rate that it is being condensed
In the first case, you will see steam leaking out of the hood. In this case, the equilibrium operation see the air mostly evacuated because it is pushed out. In the other case, steam is present with air and the air reduces the heat transfer. In case #2 the given is the rate of steam condensation, which directly determines $\dot{Q}$ and $h_c$ adjusts to compensate. In case #1 $h_c$ is a given based on the assumption of your geometry and the atmospheric steam heat transfer properties.
For $h_s$ use Wikipedia:
$$h_s={{k_w}\over{D_H}}Nu$$
where
$k_w$ - thermal conductivity of the liquid
$D_H$ - $D_i$ - Hydraulic diameter (inner), Nu - Nusselt number
$Nu = {0.023} \cdot Re^{0.8} \cdot Pr^{n}$ (Dittus-Boelter correlation)
Pr - Prandtl number, Re - Reynolds number, n = 0.4 for heating (wall hotter than the bulk fluid) and 0.33 for cooling (wall cooler than the bulk fluid).
What I don't have right now are the numbers for applying the above for Maple Syrup and the approach for the steam side of the tubes. This is all I have time for right now, but I think this amount will still be helpful. I can try to look up these things later, maybe you can specify what you understand the least first.
The container on Earth will be cooled by convection currents i.e. it transfers heat to the air around it, and also by black body radiation. By contrast the container in space can only cool by black body radiation, and obviously it will cool down more slowly. You can calculate the cooling in space using the Stefan-Boltzmann law assuming you know the emissivity (if you paint the container black the emissivity will be close to unity). Calculating the cooling in air is harder; typically you'd use Newton's law with empirically derived constants.
The final temperature in air is obviously just the temperature of the air around your container. The final temperature in space depends on where your container is. Just as the container can lose heat by emitting radiation it can gain heat by absorbing radiation, and space is full of radiation. For example the Moon is just a lump of inert rock with little or no internal generation of heat, however by absorbing sunlight the daytime temperature can rise to over 100ºC. However at night, when there is no sunlight the temperature can fall to -150ºC. So the final temperature of your container would be different during the lunar night and day, even though it's in a vacuum in both cases. If you took your container into intergalactic space, well away from any radiation sources, then it would indeed cool to the 2.7K of the cosmic microwave background.
Best Answer
From the wikipedia article,
From the design one can see that the metal inside container can transfer from conductivity heat to the outside surface through the small neck, from where also the radiative and convective heat losses can happen.
If you have ever cooked you will know that even if you stir the pot with a metal spoon the heat of the pot does not transfer to the handle since other mechanisms than conductivity keep the temperature low (air convective cooling for one). Thus the neck ring of metal through which any heat transfer must pass is too small to heat the outside larger metal mass and it is lost to air conduction, air convection and metal radiation at the neck , which are the reason such dewars will lose their internal heat after some time.