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I would like to calculate the steady state temperature of a small cube of any material heated by electromagnetic radiation. Suppose the cube floats in air and doesn't touch anything. The air is at 20 °C.

What I have understood so far

By knowing the complex dielectric constant of the material, which is $$\underline{\varepsilon_r}=\varepsilon_r'-j\varepsilon_r''$$ and the electric field strength $E_\mathrm{rms}$, I can calculate the power dissipated in the cube like $$\frac{P}{V}=\omega\varepsilon_0\varepsilon_r''E_\mathrm{rms}^2$$ where $V$ is the volume of the cube. From there I can calculate the heat rise rate $\Delta T/t$ together with the density $\rho$ and the specific heat capacity $c_\mathrm{p}$ of the cubes material like $$\Delta T/t=\frac{P}{V}\cdot\frac{1}{\rho c_\mathrm{p}}$$ I know that these formulas are only valid for a homogenous electrical field inside the cube, which isnt the case in reality. But this simplification is good enough for me.

What I would like to know

How can i calculate the steady state temperature of the cube from the heat rise rate? What additional Information do I need to know (like emission coefficient? or something like that?)

I have read a few things about newtons law of cooling but didnt really understand how to use it in my case. I would have somehow incorporated the heat rise rate in the law of cooling but i dont even know how to find the coefficients required to use the law of cooling alone (like in the case of an already heated cube floating in the air that cools down)...

I would be very thankful for any help!

Edit: What I have learned from the answers

So from the answer of @Newbie I have found the following solution for a time dependent description of the temperature increase $$T(t)=T_\mathrm{air}+\frac{\omega\varepsilon_0\varepsilon_\mathrm{r}''E_\mathrm{rms}^2a}{6k}\left[1-e^{-\frac{6ka^2}{mc_\mathrm{p}}t}\right]$$ which looks like it makes sense. (It behaves similar to the charges on a capacitor over time when charging it).

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    $\begingroup$ What do you mean by "heat rise rate"? Based on your description, you already know the heat generated in the cube. For steady state this should be equal to the heat transferred to the surrounding air using heat conduction from the 6 surfaces of the cube. Does this help? $\endgroup$
    – Newbie
    Commented Jan 13, 2022 at 13:53
  • $\begingroup$ Is conduction really important here? Because the cube doesnt touch anything. Shouldn't I take a look at convection? $\endgroup$
    – x3b7z99
    Commented Jan 13, 2022 at 14:28
  • $\begingroup$ It touches air. $\endgroup$
    – Newbie
    Commented Jan 13, 2022 at 14:30
  • $\begingroup$ Please spell check your newly added section. $\endgroup$
    – Newbie
    Commented Jan 14, 2022 at 14:38
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    $\begingroup$ @Newbie Now I understand what you mean. Well I didnt write it but I assumed that in the beginning the cube has the same temperature as the air, before the radiation is "turned on" at $t=0$ $\endgroup$
    – x3b7z99
    Commented Jan 14, 2022 at 15:12

2 Answers 2

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I wouldn't recommend calculating the rate of temperature variation $\Delta T/t$ as you attempted. Instead just start with the definition of steady state; that the input and output heat to the cube should cancel out. As you already mentioned the power per unit volume dissipated in the cube is $$P^{\rm in}_{V}=\omega\epsilon_{0}\epsilon_{\rm r}^{''}E^{2}_{\rm rms}$$ On the other hand, the heat transferred to the surrounding air may be calculated via Newton's law of cooling via $$P^{\rm out}=kA(T_{\rm ss}-T_{\rm air})$$ Note that this power is not per unit volume and will be lost through each 6 surfaces of the cube; there is a surface area $A$ in $P^{\rm out}$. Assuming the cube has dimension $a$ we have $$P^{in}_{V}a^{3}=P^{\rm out}\rightarrow \omega\epsilon_{0}\epsilon_{\rm r}^{''}E^{2}_{\rm rms}a^{3}=6ka^{2}(T_{\rm ss}-T_{\rm air})$$ Thus, $$T_{\rm ss}=T_{\rm air}+\frac{1}{6}\frac{\omega\epsilon_{0}\epsilon_{\rm r}^{''}E^{2}_{\rm rms}a}{k}$$

Considering the convection, conduction, and radiation mechanisms of heat transfer and the context of the problem, i.e., the fact that the cube is in direct contact with air, I think the process of heat transfer should be conduction through the cube surfaces.

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  • $\begingroup$ Thank you for this answer. I wonder if a time dependent equation can be derived to calculate the temperature at any point in time. Do you have any idea on how to estimate $k$ from material parameters? Or can this value be looked up in some tables? $\endgroup$
    – x3b7z99
    Commented Jan 13, 2022 at 15:02
  • $\begingroup$ @x3b7z99 I think this is where specific heat capacity comes into your equations! Something like $mcdT/dt=\omega\epsilon_{0}\epsilon_{\rm r}^{''}E^{2}_{\rm rms}a^{3}-6ka^{2}[T(t)-T_{\rm air}]$. $m$ is the cube's mass and $c$ is the specific heat capacity. This should be easy to solve if you know $T(t=0)$ for the cube. I doubt if $k$ is a geometrical parameter. As such, you may be able to find the $k$ for a material in some table online possibly. $\endgroup$
    – Newbie
    Commented Jan 13, 2022 at 15:12
  • $\begingroup$ Thank you. With your comment and the procedure here en.wikipedia.org/wiki/… I have found the solution to be $T(t)=T_\mathrm{air}+[T(0)-T_\mathrm{air}]e^{-\frac{6ka^2}{mc_\mathrm{p}}t}+\frac{\omega\varepsilon_0\varepsilon_\mathrm{r}''E_\mathrm{rms}^2a}{6k}$ Now i only need to figure out $k$ and a way to somehow check the result! $\endgroup$
    – x3b7z99
    Commented Jan 13, 2022 at 16:13
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The question is that of heat balance: the object obtains energy at a rate $R_{in}$ and loses it at rate $R_{out}(T)$, which typically depends on the temperature. One then could find the steady state temperature by solving $$R_{in}=R_{out}(T)$$ for $T$.

The cooling can proceed via different mechanisms (see, e.g., the discussion in this answer, and particularly the calculations in the linked article): radiation, convection, heat conductance. Heat conductance is usually approximated as proportional to the temperature difference (which is essentially a discrete form of the Fourier's law): $$R_{conductance}=q(T_{object} - T_{environment}),$$ whereas the heat lost via radiation is well described by the Stefan-Boltzmann law: $$R_{radiation}=\alpha T^4.$$

Radiation is probably the best first try in your case, judging by how you try to address the problem via dielectric constant - here it is perhaps possible estimate $\alpha$ in terms of the dielectric constant.

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  • $\begingroup$ It seems that the dielectric here is mentioned in the context of heat absorption by the cube as opposed to radiation of heat. This is essentially how microwaves heat up food. $\endgroup$
    – Newbie
    Commented Jan 13, 2022 at 13:55
  • $\begingroup$ @Newbie it could be also an object placed outside on a sunny day. Or objects in front of a burning fireplace. $\endgroup$
    – Roger V.
    Commented Jan 13, 2022 at 13:59
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    $\begingroup$ @RogerVadim Can you spell check the last sentence of your answer please? $\endgroup$
    – Newbie
    Commented Jan 13, 2022 at 14:02
  • $\begingroup$ From your answer and the comments I understand that I would have to find the rate of energy lost due to radiation and convection, wouldnt I? I wonder if conduction is really important here, as I stated above, because the cube doenst touch anything. $\endgroup$
    – x3b7z99
    Commented Jan 13, 2022 at 14:30
  • $\begingroup$ In principle, oen could heav heat conductance through the air - the difference with convection is absence of macroscopic flows. I strongly recommend you checking this page (if you haven't yet done so) - the calculation is very instructive, even if tedious. $\endgroup$
    – Roger V.
    Commented Jan 13, 2022 at 14:36

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