A resistor at a temperature T has a fluctuating voltage. This is a consequence of the fluctuation dissipation theorem which you can use to calculate the spectrum of the voltage. The wikipedia article on the Fluctuation Dissipation Theorem has a section on resistor thermal noise. When measured over a bandwidth $\Delta\nu$, the average squared voltage is:
$$\langle V^2\rangle = 4Rk_BT\Delta\nu$$
where $k_B= 1.38\times 10^{-23}$J/K is Boltzmann's constant.
Suppose we have a resistor $R$ maintained at a temperature T, and hooked up to a resistor $R_0$ initially at absolute zero. The thermal noise of the warm resistor will be as given above. When you apply a voltage $V$ to a resistor $R_0$, the dissipation (in watts) will be given by $IV = V^2R_0$, so the watts applied to the resistor initially at absolute zero, over a bandwidth $\Delta\nu$ will be
$$\langle V^2\rangle = 4R_0Rk_BT\Delta\nu.$$
As that resistor warms up to temperature $T_0$, it will apply a fluctuating voltage on the warm resistor. Following the above, but with the two resistors swapped, the power applied to the warm resistor by the colder resistor at temperature $T_0$ will be:
$$\langle V^2\rangle = 4RR_0k_BT_0\Delta\nu.$$
The system will be in balance when the above two powers are equal. This happens algebraically when $T=T_0$.
A possible source of paradoxical confusion is that the above calculation was done over a limited bandwidth range. But the calculation does not depend on frequency; instead the power transmitted is simply proportional to the range of bandwidths.
For the usual physical system, we consider frequencies that run from 0 to infinity. Thus the total bandwidth is infinite. This suggests that the power flow in the above should be infinite. This paradox is avoided by noting that physical resistors have a limited bandwidth. There is always a parasitic capacitance so that the bandwidth is limited on the high side. Thus the power transfer rate depends on how ideal your resistors are.
As an example calculation, suppose that a resistor has a maximum frequency of 100 GHz $= 10^{11}$ Hz, a (room) temperature of 300K, and a resistance of 1000 ohms. Then the power transfer rate is:
$$ 4\times 1000 \times 1.38\times 10^{-23} 10^{11} \times 300 = 1.66\;\;\textrm{uWatts}$$
Given the heat capacity of the resistor, you can compute the relaxation time with which the colder resistor exponentially approaches an equal temperature.
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.
Best Answer
Temperature is not a concept that has a lot of utility at the level of single atoms because it represents the mean kinetic energy of a group of particles (to within a coefficient).
You can define it, it just doesn't help much.
At the level of two atoms you revert to a more fundamental model such as the forces between them.
One atom transfers energy to another through electromagnetic forces between them. When that energy manifests as randomized kinetic energy at the microscopic scale we refer to it as "heat" at the macroscopic scale.
In a solid it is usually reasonable to treat the forces between individual pairs of atoms as being spring-like (i.e. they obey $F_{i,j} = -k(r_{i,j} - r_0)$). Starting from there you can build various models of solid behavior. For instance the Einstein model of a crystal.