# Physics of the electric hot plate

For an electronics experiment, I began wondering about the electric hot-plates (specifically the temperature dependence over time). If I were to measure the given temperature over time, I assume that the temperature would not increase linearly. In fact, I suspect the initial heating rate will only decrease.

Do anyone have some intuition regarding a model for this? An answer giving an ODE is fine too...

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If it can be analitically solved, it will be in Carslaw's Conduction of Heat in Solids. If you can work out the details of the model you have in mind, I can look it up tonight. – Jaime Aug 29 '12 at 15:35
@Jaime - The problem is, I don't know what model to use. I assume It's like a giant resistor, and since the resistance increases with temperature, the rate should drop, but that's about it. – nbubis Aug 29 '12 at 16:29

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 $A$.

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Perfect! So this is an exponential decay, rather like a capacitor (which makes sense). – nbubis Aug 29 '12 at 16:44
Yes, when the power is turned off the cooling is exponential. But only if there is good convection so Newton's law applies. – John Rennie Aug 29 '12 at 16:51
I think there will be, As I plan to place a pot of water on top. – nbubis Aug 29 '12 at 16:56