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.
Mobile phones transmit at microwave frequencies so they can induce currents in metals and other conductors. The energy dissipates as heat. The principle is the same used by microwave ovens. The phone would not itself get hot because (presumably) the microwave radiation would be directed away from metal components in the phone. Your key must have been close to the antenna and was perhaps aligned with it in a way that maximized the effect, or perhaps its length was just right for a resonance.
Here is a paper that seems relevant http://www.springer.com/about+springer/media/springer+select?SGWID=0-11001-6-1296921-0
Edit: Let me add a small calculation. Suppose a phone were outputting 1 Watt and 50% was absorbed by the key for 3 minutes. This puts 1W*180s*0.5 = 90 J of heat into the key. If the key weighs 10 g and is made of iron with a heat capacity of 450 J/Kg/K then the temperature goes up by 90J/0.01Kg/450J/Kg/K = 20K So this could raise the temperature of the key from 25C to 45C which would make it feel quite hot to the touch. Actual result depends on how fast the heat is conducted away and how much of the power is absorbed but it is not beyond the bounds of possibility that enough heating is possible to account for the observation. I would not blame anyone for being skeptical though. It would perhaps be unusual for the phone to transmit on full power for that long.
Best Answer
I think there is some interesting physics to be had here. The rate of change of temperature depends on the rate of heat flow in from the electric heating element and the rate of heat flow out as heat is lost to the air. If we write the heat capacity of the hotplate as $C$ ($C$ is the traditional symbol for heat capacity) then:
$$ \frac{dT}{dt} = C \left( \frac{dH_{in}}{dt} - \frac{dH_{out}}{dt} \right) $$
where $dH_{in}/dt$ is the rate of heat flow in and $dH_{out}/dt$ is the rate of heat loss.
$dH_{in}/dt$ is simply the power being supplied to the hotplate. You can measure the current the hotplate draws from the mains and calculate the power that way, or you could simply use a power meter. The power in watts, call this $W$, is simply the energy in joules per second, so it's exactly what you need for $dH_{in}/dt$.
The rate of heat loss, $dH_{out}/dt$ is harder because it depends on how the hotplate is cooled. If the cooling is dominated by convention (it probably is) then the cooling will obey Newton's law of cooling and the heat loss will be given by:
$$ \frac{dH_{out}}{dt} = A \space (T - T_0) $$
where $T_0$ is the ambient temperature and $A$ is some constant to be determined experimentally. Put all this together and you'll get:
$$ \frac{dT}{dt} = C \left( W - A \space (T - T_0) \right) $$
You'll need to measure $C$ and $A$ experimentally. If you have a copy of excel to hand you can use its Solver to fit values of $C$ and $A$. Alternatively, you can get $C$ from the initial rate of temperature rise. When $T \approx T_0$ the heat loss is small and:
$$ \frac{dT}{dt} \approx CW $$
so if you know $W$ you can calculate C. You can calculate $A$ by heating the hotplate then turning the power off and letting it cool. As it cools the temperature variation is:
$$ \frac{dT}{dt} = -C A \space (T - T_0) $$
so if you know $C$ you can calculate $A$.