# (Specific) Heat capacity of brass at milliKelvin temperatures

• xavier_Ghz
In summary: Assuming your brass is 70/30 Cu/Zn, that yields:Cbrass = 3.0*10^-2 J / (Kg*K) at T = 1 K.This is somewhat comparable to that of copper, asCCu = 1.3*10^-2 J / (Kg*K) at T = 1 K.However, the experiments for copper are done for the submillikelvin temperatures, which yieldCCu = 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(
xavier_Ghz
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 Cbrass @ 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

Cbrass = 3*10^-2 J / (Kg*K) at T = 1 K.

This is somewhat comparable to that of copper, as

CCu = 1.3*10^-2 J / (Kg*K) at T = 1 K.

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

CCu = 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 Cbrass 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 me

Cbrass = 1.4*10^-3 J / (Kg*K) at T = 40 mK.

Xavier

Last edited by a moderator:
A clue about the heat capacity of any type of brass is good, although I assumed the most common type of brass (Cu with 30 or 37% Zinc).

xavier_Ghz said:
A clue
Having absolutely zero experience at anything less than 20 K, my input doesn't count for a whole lot. I'm not aware of any pathological behavior of any of the brasses, but there're no guarantees that niobium-tin type phenomena don't occur. i.e., for estimating properties/behavior of your calorimeter, what you've proposed sounds good.

Perhaps anyone else?

There are some papers by "J. A. Rayne" on the "heat capacity" of "brasses" (GoogleScholar).

xavier_Ghz

## 1. What is the specific heat capacity of brass at milliKelvin temperatures?

The specific heat capacity of brass varies depending on the temperature, but at milliKelvin temperatures (less than 1 Kelvin), it is typically around 0.2 Joules per gram-Kelvin.

## 2. How does the specific heat capacity of brass change at different milliKelvin temperatures?

As the temperature of brass decreases, its specific heat capacity also decreases. This is because at lower temperatures, there is less energy available to excite the particles in the brass, resulting in a lower heat capacity.

## 3. Can the specific heat capacity of brass at milliKelvin temperatures be measured accurately?

Yes, the specific heat capacity of brass at milliKelvin temperatures can be measured accurately using specialized equipment and techniques, such as adiabatic calorimetry or the heat-pulse method.

## 4. How does the specific heat capacity of brass at milliKelvin temperatures compare to other materials?

Brass has a relatively low specific heat capacity at milliKelvin temperatures compared to some other materials, such as copper or aluminum. This is due to its lower atomic mass and structure.

## 5. Why is it important to understand the specific heat capacity of brass at milliKelvin temperatures?

Understanding the specific heat capacity of brass at milliKelvin temperatures is important for various applications, such as in cryogenics and low-temperature experiments. It also helps in the design and optimization of cooling systems and materials for use at extremely low temperatures.

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