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The equation for thermal resistance R is:

$$R=\frac{\Delta T}{q}=\frac{L}{k}$$

with

  • T being the temperature difference (in kelvins)
  • q being the heat flow rate (in W/m²)
  • L being the thickness of the material (in metres)
  • k being the thermal conductivity (in W/(mK), watts per metre kelvin)

Whether you use:

  • temperature and q, kelvin / (watts × metres²) ⇒ (metres² × kelvin) / watts or

  • L and k, metres / (watts / (metre × kelvin)) ⇒ (metres² × kelvin) / watts,

you obtain (metres² × kelvin) / watts. This makes sense, as (m² × K) / W is the units for thermal resistance.

So why do datasheets for electronic components give thermal resistance with units of °C/W? I understand °C/W likely means for every watt of power dissipated by the device, the device heats up by that temperature. But how can this parameter have two different units?

Is 8°C/W the same as 8 (m² × K) / W?

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  • $\begingroup$ Hello and welcome to Physics Stack Exchange! It is preferable to use MathJax (LaTeX) to write equations – see math.meta.stackexchange.com/q/5020. You can edit your question. Thanks! $\endgroup$
    – jng224
    Commented May 18, 2021 at 16:08

3 Answers 3

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The datasheets for electronic components have already incorporated the relevant area, turning the heating flux in W/m² into a heating rate in W. This saves you the effort of looking up the area for that component and correcting for it.

No, 8°C/W is not the same as 8 K-m²/W, but it is the same as 8 K/W because the intervals of the Celsius and Kelvin scales are identical.

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  • $\begingroup$ Thank you for your explanation, this helps a lot. How would I convert between $8K-m^2/W$ and °C/W? Would I calculate the heating flux per $m^2$ and use a ratio to convert to the area of the surface I have? (e.g. $8K-m^2/W$ and my surface is half a square metre, therefore the heating rate would be 4°C/W?) $\endgroup$
    – MRB
    Commented May 19, 2021 at 9:30
  • $\begingroup$ Yes, but I calculate the result here as 16°C/W, as you’re dividing by the area. $\endgroup$
    – Chemomechanics
    Commented May 19, 2021 at 15:41
  • $\begingroup$ silly mistake on my part! But it is great to know that is how you convert between the two units (though I suppose knowing this I now understand they are actually the same unit) Thank you for your help, much appreciated. $\endgroup$
    – MRB
    Commented May 19, 2021 at 16:27
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To add to Chemomechanics' answer, yu can find two definitions of the thermal resistance, if you read papers related to this notion. Of is the one you show, with q being the heat flux and the other one with q being the just the thermal power, in watts. They seem to be both used by people working in the fields related to heat transfer. You just have to be careful to be consistent. I had the same problem when I satrted in a thermal properties project.

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The defining equation with consistent units is

$$ R = \frac{\Delta T}{\dot{q}} = \frac{L}{k A} $$

where $\Delta T$ is a temperature difference $^o$C or $K$, $\dot{q}$ is heat flow (W) (not heat flux), $L$ is distance (m), $k$ is thermal conductivity (W/m $^o$C), and $A$ is area (m$^2$). The analogy is Ohm's law with $\Delta V = i R$ where $i$ is current flow not current flux.

The units become

$$ \frac{^oC}{W} = \frac{m}{(W/m\ ^oC)\ m^2} $$

and are thereby consistent.

An alternative form of thermal resistance is $R' = R A$ with units of m$^2$ $^o$C/W. The alternative form would be $R' = L/k$. The two forms for thermal resistance are not to be set equal to each other ($R \neq R'$). This is the root of the confusion on units. You can however take a basis unit of 1 in the area units (e.g. 1 m$^2$ or 1 ft$^2$) to compare values for $R$ and $R'$.

Finally, this conversion set shows how the units on temperature scales can be considered.

$$\Delta T = 0\ ^oC = 0\ K\ \mathrm{as\ a\ difference} \neq 0\ K\ \mathrm{as\ an\ absolute}$$

$$ T = 0\ ^oC = 273.15\ K$$

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  • $\begingroup$ There is no such thing as °K: the degree Kelvin has lost the degree symbol a long time ago, and the unit name is now just Kelvin. The thermodynamic temperature $T$ and the Celsius temperature $t$ are two different physical quantities related by the relation $t = T-273.15\,\mathrm{K}$, such that $\Delta t = \Delta T$, and the units degree Celsius and Kelvin are defined so that $\Delta T/\mathrm{K} = \Delta t/^\circ\mathrm{C}$. $\endgroup$
    – Massimo Ortolano
    Commented May 18, 2021 at 18:18
  • $\begingroup$ So, the first of your last relations can be better written as $\Delta T = 0\,^\circ\mathrm{C} = 0\,\mathrm{K}\neq T = 0\,\mathrm{K}$, but the second one doesn't make sense. $\endgroup$
    – Massimo Ortolano
    Commented May 18, 2021 at 18:22
  • $\begingroup$ It is nor uncommon to use low case q to label heat flux, heat per time and area. There is nothing wrong as long as you indicate what your symbol is. Which he did. $\endgroup$
    – nasu
    Commented May 18, 2021 at 18:54
  • $\begingroup$ Your second equation is irrelevant for the question. The two units mentioned in the OP are different and they correspond to two different definitions. $\endgroup$
    – nasu
    Commented May 18, 2021 at 18:58
  • $\begingroup$ @MassimoOrtolano I cannot verify the loss of the degrees sign on the K scale but accept the notation change. $\endgroup$
    – Jeffrey J Weimer
    Commented May 18, 2021 at 20:03

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