How to experimentally find the thermal conductivity of a material without an expensive thermal conductivity meter?

Question

Physics

I am looking for an opportunity to evaluate the thermal conductivity of a heater for the walls of a private house, if it is not possible to get any thermal conductivity meter (for example, ITP-MG4).
Found this on a forum:

"We take a material with a known thermal conductivity and a new material. We connect them together. We place a temperature sensor at the contact boundary. Well, we organize the temperature difference on each side of the resulting pie. Then, using the formula, we determine the thermal conductivity of the new material."

But from the heat flux formulas and Fourier's law, I don't understand how to calculate it. Can someone explain to me the essence of such an experiment and the formula that was supposed to be used?
Thank you.

Answer 1

Sometimes, when I get a question that follows the pattern of "I'm an idiot, can someone explain something to me?", I'm tempted to answer. There are no guarantees here, because a person who has designated himself this way probably knows himself better, so the risk of wasting time is great. On the other hand, you can choose such simple words for an answer that he will be ashamed to admit that he did not understand anything. There is probably something cheating in this, but, on the other hand, what will I be to blame for? In general, such a temptation exists.
So, the author asked a question in which the formula is mentioned, but not named. This reinforces my suspicions. But I'll try not to give in. I will not name the formula either, but I will say that it is most likely correct. My confidence is based on the fact that a reasonable design of the experiment is proposed and the relevant terms are mentioned. I will also explain the essence of the experiment and the secret formula. You are welcome.
For illustrative purposes, we will conduct an auxiliary thought experiment. So:
House, window, frame with double glazing. For simplicity, let's assume that the wood (or plastic) frame has zero thermal conductivity. Outside -20°, in the room +20°, what temperature will be established between the panes? For symmetry reasons, it appears to be 0°. How about really? And so it is. According to the law of conservation of energy and in view of the fact that the street / inter-pane space / room system is in dynamic equilibrium, in order to maintain a stable inter-pane temperature, it is necessary that the flow of thermal energy entering through the inner glass is equal to the flow outgoing through the outer glass to the street. According to the Fourier law (no formulas!) the temperature drop on each of the identical glasses will also be the same.
And now we will replace our double window with the proposed sandwich frommaterial with a known thermal conductivity and a new material . It seems that the law of conservation of energy should not be canceled. Let's make an equation (no formulas!), describing the equality of heat fluxes through the sandwich layers. By simple mathematical transformations, we derive the same statement (no!).
Something like this.

Answer 2

Thermal conductivity is equivalent to conductivity in electricity
Thermal resistance = resistance in electricity
I*(Rout + Rx) =Uob
I*Rout = Uout -> I=Uout/Rout
Rx = Uob/(Uout/Rout) -Rout
conductivity reciprocal of
Px = 1/Rx = 1/(Uob/(Uout/Rout) -Rout)
================================
Px = 1/Rx = 1/(Urev/(Uout*Pout) -1/Pout
============================== ==
where
Uob - total temperature difference
Uiz - temperature difference on a known material
Piz - thermal conductivity of a known material (not specific but total !!)