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This is a question about nondimensionalization in the heat equation $$\rho c \frac{\partial u}{\partial t}= \kappa \frac{\partial^2 u}{\partial x^2}~? $$

How could one devise a nondestructive experiment to find the thermal diffusivity i.e $\frac{\kappa}{\rho c}$?

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    $\begingroup$ In the current form it is not clear what your first question means... may be you could reformulate it. As of your second question, you could check en.wikipedia.org/wiki/Specific_heat_capacity#Measurement $\endgroup$
    – Pavlo. B.
    Commented Oct 4, 2021 at 2:35
  • $\begingroup$ I'm sorry. The question isn't clear. I didn't write correctly. I changed it $\endgroup$
    – Jama
    Commented Oct 4, 2021 at 2:37
  • $\begingroup$ @Jama Hello and welcome to Physics.SE! Please ask only one question per post. $\endgroup$ Commented Oct 4, 2021 at 3:50
  • $\begingroup$ What are your thoughts so far on how to do what you want? $\endgroup$ Commented Oct 4, 2021 at 3:56
  • $\begingroup$ Well I know I can write it as $u_t=D u_{tt}$ where D is the quantity I'm looking for. But that's not helping. Now I'm thinking that the units of D are $L^2 M T^-5$ and I should equate that to $[\rho]^a[c]^b[\kappa]^c$ and solve for $a,b,c$? $\endgroup$
    – Jama
    Commented Oct 4, 2021 at 4:22

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I would advise simply finding the individual properties $\rho, c, k$, as these can all be found via steady state experiments which are much easier to set up and would have less uncertainty.

As the equation implies, diffusivity as a property only really shows itself in transient cases, when time derivatives are non zero.

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    $\begingroup$ There are experimental techniques yielding directly the diffusivity. That's much simpler and precise than measuring these 3 quantities separately... $\endgroup$ Commented Oct 5, 2021 at 4:38
  • $\begingroup$ Can you give a little more detail into these techniques? $\endgroup$
    – Jama
    Commented Oct 13, 2021 at 6:03

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