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## Main Question or Discussion Point

Hi all,

I do an calorimeter experiment at temperatures around 40 mK. To get more grip on the time constants that are associated with the heat flows, I calculate the thermal resistance as well as the heat capitance of the materials in the set-up.

We assume that Newton's law of cooling is the only relevant process, as the compartiments of the cryostat are all in a vacuum (no convection) and the nearest plate from the dilution refrigerator (14 mK) has a temperature of 50 mK (so the importance of radiative heat transfer is negligible).

My question is: what is C

Let's assume that I use the common 70/30 brass. I found somewhere (lost the link) that

C

This is somewhat comparable to that of copper, as

C

However, the experiments for copper are done for the submillikelvin temperatures, which yield

C

There, on a log(C)-log(T) scale, almost all curves decrease in a linear way. Should I assume that this is also the case for C

C

Your help is much appreciated!

Xavier

I do an calorimeter experiment at temperatures around 40 mK. To get more grip on the time constants that are associated with the heat flows, I calculate the thermal resistance as well as the heat capitance of the materials in the set-up.

We assume that Newton's law of cooling is the only relevant process, as the compartiments of the cryostat are all in a vacuum (no convection) and the nearest plate from the dilution refrigerator (14 mK) has a temperature of 50 mK (so the importance of radiative heat transfer is negligible).

My question is: what is C

_{brass}@ mK temperatures? I can't find it anywhere.Let's assume that I use the common 70/30 brass. I found somewhere (lost the link) that

C

_{brass}= 3*10^-2 J / (Kg*K) at T = 1 K.This is somewhat comparable to that of copper, as

C

_{Cu}= 1.3*10^-2 J / (Kg*K) at T = 1 K.However, the experiments for copper are done for the submillikelvin temperatures, which yield

C

_{Cu}= 6*10^-4 J / (Kg*K) at T = 40 mK, (Fig. 3.12 from 'Matter and Methods at Low Temperatures by Pobel).There, on a log(C)-log(T) scale, almost all curves decrease in a linear way. Should I assume that this is also the case for C

_{brass}to extrapolate to T = 40 mK? With a slope that is comparable to that of the heat capacity of copper? I probably make a lot of mistakes then, but it would give meC

_{brass}= 1.4*10^-3 J / (Kg*K) at T = 40 mK.Your help is much appreciated!

Xavier

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