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I used an electronic device during a certain time. Due to the powering of this device, the temperature, measured with an NTC, increased. After a certain time, I switched off the power of the device but still measured the temperature, with the NTC.

When drawing the evolution of temperature with time, I realized that the increase of temperature followed a $A(1-Be^{-Cx})$ law while the decrease of temperature followed a $De^{-Ex}$ law.

Is there a fundamental reason for this kind of analytic evolution of the temperature?

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    $\begingroup$ Note that the decrease only looks like that, if the external temperature is zero degrees (in whatever unit system). Otherwise, it will also tend to the (non-zero) external temperature exponentially. $\endgroup$
    – oliver
    Jun 7, 2021 at 19:40
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    $\begingroup$ Differential equations, maybe? Exponentials often have a relation to differential equations. $\endgroup$ Jun 8, 2021 at 7:43
  • $\begingroup$ Ultimately it comes from Fick's law with just a single interface for heat flux. Fick's law is not actually a law contrary to the name, it is just a model. It is only really supported from first principles in the limit of infinitesimal difference between the temperatures. $\endgroup$
    – Ian
    Jun 8, 2021 at 14:43
  • $\begingroup$ For larger temperature differences, a wide class of heat flux equations are possible in principle, with basically the only constraint being the second law (and the first law, but the first law is essentially neglected anyway by assuming the surroundings have so much more heat capacity than the system that it is safe to model that heat capacity as infinite). $\endgroup$
    – Ian
    Jun 8, 2021 at 14:57
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    $\begingroup$ As a non-physicist, that's an unusual use of the word "evoution". $\endgroup$
    – RonJohn
    Jun 9, 2021 at 18:45

5 Answers 5

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This behavior is basically described by Newton cooling with heat generation, using the equation: $$MC\frac{dT}{dt}=G-k(T-T_{\infty})$$where T is the temperature, t is time, M is the mass, C is the heat capacity, G is the heating rate, k is the Newton cooling coefficient (convective heat transfer coefficient times surface area), and $T_{\infty}$ is the surrounding room temperature. For the heating portion of the cycle, the initial temperature is $T_{\infty}$, and the solution to the equation for the temperature increase is $$(T-T_{\infty})=\frac{G}{k}(1-e^{-\frac{kt}{MC}})$$

For the cooling portion, G is zero, and the starting temperature $T_0$ is the final temperature from the heating portion. The solution for the temperature decrease in this portion is $$(T_0-T)=(T_0-T_{\infty})e^{-\frac{kt}{MC}}$$

Both these variations match the functionality of what you have observed. The analysis also shows that the coefficients C and E in your measurements should be roughly equal to one another.

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    $\begingroup$ ... is of course the RIGHT answer (sadly you beat me to it! ;-) ) $\endgroup$
    – Gert
    Jun 7, 2021 at 21:30
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    $\begingroup$ It seems to me that this answer is substantially similar to the accepted one. (Not that there's anything wrong with that. This is more fleshed out. It's just that I don't see how this answer is more correct.) $\endgroup$
    – d_b
    Jun 7, 2021 at 22:28
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    $\begingroup$ @d_b I guess you didn't notice that I provided a physical interpretation for all the parameters in the equation and that I showed how to determine the temperature variation for the portion of the process where the heater is on. Both of these features of my answer are absent from the "accepted" answer; jeez, it doesn't even include the heating part. $\endgroup$ Jun 7, 2021 at 22:38
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Yes there is. Provided that the temperature of the (single) heat sink can be considered constant during cooling, and the heat capacity of the components between the device and the heat sink are negligible, the cooling part is described by the differential equation $$\frac{dT}{dt}=-A\cdot (T-T_{sink})$$ no matter what geometry connects the device and the sink. This equation derives from the heat equation. The value of $A$ is the determined by geometry, heat conductivity and the heat capacity of the device itself.

Of course, separating the arrangement into a heat carrying object (device), an infinite heat sink (environment) and a transfer path inbetween, is just an approximation, albeit often a pretty good one. The advantage of this discrete model is that you don't have to take into account the heat capacity of the transfer path.

For example, if you want a basic understanding of the thermal insulation of a house, you just have to use the thermal conductivity of the (usually very light) insulation material in the walls, while the heat capacity of the walls (lightweight construction) may be negligible for the calculation.

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  • $\begingroup$ You are right, but the question essentially remains: what is the fundamental reason for the heat equation to look like it looks? I suppose you might want to add that this is a result of linearization of differential equations assuming (relatively) small perturbations. $\endgroup$
    – sleepy
    Jun 7, 2021 at 19:47
  • $\begingroup$ Oliver kindly explained from the basis the reason of my observation. Thanks $\endgroup$ Jun 7, 2021 at 19:51
  • $\begingroup$ This equation derives from the heat equation. The value of $A$ is the determined by geometry.. No, it derives from Newton's Law of Cooling. And the value of $A$ depends on far more than just geometry. $\endgroup$
    – Gert
    Jun 7, 2021 at 21:17
  • $\begingroup$ heat capacity (which is a kind of "heat memory") What do you mean by by "heat memory"? $\endgroup$
    – Gert
    Jun 7, 2021 at 21:27
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    $\begingroup$ @MathieuKrisztian The answer provided by Chet Miller is in fact the correct one, NOT this answer by Oliver. $\endgroup$
    – Gert
    Jun 7, 2021 at 21:28
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Very fundamentally, the reason is that the heat flow is proportional to the temperature difference, which in turn depends linearly on the amount of heat flown. This — the rate of change is (inversely) proportional to the overall change — describes an exponential relationship: The steepness of the exponential curve is proportional to the y value at that point.

The physics of heat flow resemble those for electric current or general diffusion processes; the equations for a mass acquiring room temperature resemble those for a charging capacitor.

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Can one go more fundamental than "heat flow is proportional to temperature difference, and this then leads to exponentially slowing approach to equilibrium"? Of course one can.

Newton's law of cooling $$Q_{flow} = h\Delta T = h(T_{body}-T_{environment})$$ is of the same form as many other equations of conduction, where there is a transport coefficient $\gamma$ (diffusion constants, electric conductivity, shear viscosity, thermal conductivity) that relate a flux $J$ of something subject to a conservation law to gradient forces $X$: $J = \gamma X$, or $$J=\gamma \Delta U$$ for the values $U$ that the property has in the two parts of the system-. This is valid for disturbing forces small enough that things remain linear.

The value for $\gamma$ is given by Green-Kubo's formula $$\gamma = \frac{V}{k_B T}\int_0^\infty \langle J(0)J(t)\rangle dt.$$ That is, it comes from the autocovariance at equilibrium flux (and this applies even for quantum systems). Conductivity is linked to how long the "memory" lasts in the matter: a material that can retain it long can transfer heat more efficiently than a "forgetful" material. Which is why removing sources of phonon scattering increases heat conductivity.

And why does this behaviour hold? It seems to follow from the celebrated fluctuation-dissipation theorem. Systems with detailed balance link dissipation to thermal fluctuations. Detailed balance just means that at equilibrium each elementary process is in equilibrium with its reverse process: if there is no temperature difference there will be as many phonons going left as are going right, as many exciting atoms in the material or getting emitted by excited atoms.

In a sense detailed balance is what you get when there is time reversal symmetry: in an equilibrium state the microscopic dynamics doesn't change if you reverse time.

So maybe the fundamental reason for the cooling curve is: microscopically physics is time reversal symmetric, there are a vast number of degrees of freedom that can vibrate, they interact in sufficiently random ways that the system ergodically averages out to a macroscopic state. From this you get detailed balance and a thermodynamic concept of temperature and heat, the Green-Kubo relations hold, and you get the linear Newton's law of cooling that implies the exponential approach to equilibrium.

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We can derive this exponential behavior from the laws of thermodynamics as follows: on the one hand, heat is a quantity that is subject to a conservation law (first law of thermodynamics): locally, the amount of heat $H$ stored in a body can only change due to a flow of heat $J$ out of it. On the other hand, the second law of thermodynamics tells us that the flow of heat between two bodies is driven by a difference in temperature $T_2-T_1$. Let's say you have a reservoir at some constant temperature $T_0$ (temperature of the room your device is in), then the flow of heat out of your device is to first approximation given by $J\propto (T_{\textrm{device}} - T_0)$. Temperature (in Kelvin) of the device is proportional to the amount of heat stored in it, $T\propto H$, and thus the change of temperature of your device after some short time $\delta t$ is $$\delta T_{\textrm{device}} \propto \delta H = -c J \delta t = -c' (T_{\textrm{device}}-T_0) \delta t,$$ where $c$ and $c'$ are some proportionality constants which will involve the specifics of your device and your room. Divide both sides by $\delta t$ and take the limit $\delta t \to 0$, and you obtain the standard differential equation for an exponential, $$\frac{dT_\textrm{device}}{dt} \propto -(T_\textrm{device} - T_0)$$.

What happens if you turn on the power? Lets for simplicity assume that your device is fairly useless and does no work besides heating up. In keeping with the first law of thermodynamics, its power consumption will be an additional flow of heat which we will have to subtract from $J$ (we defined it as outward flow). We will have to be a bit more careful than before because temperature is what is called an intrinsic quantity (not dependent on the size, mass, volume of the object) whereas the total heat stored in the device and its power consumption are extrinsic (they increase as the object becomes bigger). Instead of thinking about how the various proportionality constants are affected, lets consider average power consumption per volume of your device $w=W/V$ and let's think of the heat flow $J$ as the heat flow per unit volume. Then we will have $J = k(T_\textrm{device}-T_0) - w$, where $k$ is a device-dependent constant that encodes how much work it takes to heat up a unit volume of "device" by one degree. We see that if we define $T_\textrm{equilibrium} \equiv T_0 + w/k$, we recover the same differential equation, but with $T_0$ replaced by $T_\textrm{equilibrium}$, $$\frac{dT_\textrm{device}}{dt} \propto -(T_\textrm{device} - T_\textrm{equilibrium}),$$ and the system exponentially approaches this temperature: if you turn on power, your system will exponentially approach a temperature above room temperature, if you turn it off again, it will exponentially return to room temperature.

Note that you need nothing besides the laws of thermodynamics to reach these conclusions. The approximations that we make are that the temperature change is small enough to neither affect the way heat is transferred (no change in convection, e.g.) nor to affect the heat capacity of the device, i.e. $k$ is constant.

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