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Energy is not temperature. The lack of kelvins in the definition of the joule is not weird when you consider that every material requires a different amount of joules to heat up by 1 kelvin, and during a phase change, there is no change in temperature (kelvins) at all even as you add or remove energy (joules). You can add a bunch of joules to a material, and ...
5
This is because negative "magnetic orientation temperatures" properly reflect the change in number of available magnetic states with energy increase, but only of the magnetic states. The entropy of the system as a whole remains on the increase as does the thermal temperature.
Your comment on "indefinite number of inversions" misses the ...
5
Is heat kinetic energy or changes in electron levels? or both?
It is neither.
Heat is energy transfer due solely to temperature difference. It is a mechanism for transferring energy. The other mechanism is work.
Therefore heat can cause a change in internal kinetic energy of a substance or cause a change in electron energy level, but it is not the energy ...
4
I would consider the process you described as quasi static. In fact, in that particular process, both the system and the surroundings experience reversible changes. However, for the low thermal conductivity medium in-between, the process is not reversible, and entropy is generated within this medium. This generated entropy is transferred to the system. So ...
4
In branches of physics other than thermodynamics, the word "adiabatic" is used as synonym of "infinitesimally slow." This creates some confusion.
In equilibrium thermodynamics, as you correctly intuit, you need to be more specific. Infinitesimally slow processes are introduced as a useful concept in order that the equation of state be ...
4
Because ice is (usually) at a lower temperature than steel. It also absorbs energy when it melts.
You can calculate the amount of energy absorbed by either steel/ice with $mc \Delta T$. With ice you add an extra amount due to it melting. This yields the maximum amount of energy that can be absorbed. If you have enough steel at sufficiently low temperatures, ...
4
The partition function $Z$ is defined as
$$Z(T,\mathbf{x})=\sum_{\{\mu\}}e^{-\beta \mathcal{H}(\mu)}$$
where sum is over microstate ($\mu$).
For an ideal gas,
$$Z(T,V,N)=\int \frac{1}{N!} \prod_{i=1}^N\frac{d^3q_id^3p_i}{h^3}\exp\left[-\beta \sum_{i=1}^N\frac{p_i^2}{2m}\right]$$
$$Z(T,V,N)=\frac{1}{N!}\left(\frac{V}{\lambda(T)^3}\right)^N$$
$$\Rightarrow \...
4
In my kitchen the background radiation is even around 293K, and yet in the fridge there is 273K and below. The second law of thermodynamics does not say that it is impossible to cool down anything below ambient level, but rather only that this does not happen spontaneously, i.e. without supplying energy that feeds a thermodynamic machine.
4
Temperature is not a measure of energy. It seems what you are doing is you are interpreting temperature as a kind of "energy per unit mass", but that is not the case - otherwise we'd just use that: energy per unit mass (joules per kilogram, say), and we would not need a separate unit (kelvin) for temperature. It's far from unrelated to energy, and ...
3
The nichrome wire undergoes a phase transition at that mysterious temperature which changes its unit cell structure. The two different unit cells have slightly different bulk resistivities. Superimposed on that shift is the usual increase in resistivity with temperature for metallic solids.
3
The fact that the cosmic microwave background has a nearly perfect blackbody distribution does not rule out the possibility that the universe has an upper-bounded energy, whatever "energy" is supposed to mean in cosmology. To avoid ambiguities in the meaning of "energy," we can say it like this: The fact that the cosmic microwave ...
3
With thanks to 'Chet' in the comments it appears this problem (at least without convection) isn't really hard.
$$\partial_t T=\alpha \partial_{xx}T$$
Boundary conditions:
$$T(x,0)=T_i$$
$$T(0,t)=T_s$$
Introduce a Similarity Variable $\eta$:
$$\eta=\frac{x}{2\sqrt{\alpha t}}$$
This transforms the original PDE into an ODE:
$$\frac{\mathrm{d}^2T(\eta)}{\mathrm{...
3
You do it by heating the gas. If you consider the gas in a cylinder where the piston applies constant pressure, then by heating the gas, the volume will expand while the pressure stays the same.
However, if you want to do it without adding energy, the energy conservation law prohibits it.
3
I have experienced rather similar effects on several shower systems. The problem typically occurs when the mixer is working below its intended pressure or flow range.
My experience is that high water pressures and flow rates are less problematic than gentle flow. Where I have fitted a shower pump, the problem has gone away.
I once tried a "turbo" ...
3
So, either negative temperatures can never be achieved in isolated systems and therefore there is no violation, or is there something else going on here?
It is the word "isolated" that makes the difference. Entropy is decreasing when a diamond is formed, or any crystal out of the solution, but the system to be considered for the second law is the ...
3
The kinetic energy (in Joules), for each atom in a gas, is $\frac{3kT}{2}$, where $k$ is Boltzmann's constant and $T$ is the temperature in Kelvin.
The number of atoms in each gram depends on the atomic mass; for example, 12 g of carbon has Avogadro's number ($6.02\times 10^{23}$) atoms of carbon.
The total heat energy is then the kinetic energy times the ...
2
You are mixing two things:
What is the average kinetic energy of the system in $n$-dimensions? Thermodynamics states that
$$\bar E = \frac{f}{2}k_B T$$
where $f$ is the number of degree of freedom.
How do we convert between the energy unit "Joule" and the temperature unit "Kelvin"? Since $k_B T$ has the dimension of an energy, we use
$T ...
2
The neutrino background cools as the universe expands. The reason for this is the de Broglie relation which relates the de Broglie wavelength to the momentum $p = h/\lambda$. As the universe expands, so does the de Broglie wavelength and thus the momentum decreases as the inverse of the scale factor of the universe.
As the universe expands, the neutrinos ...
2
In thermal equilibrium, if we assume that the entropy conserved by expanding the universe, we can obtain the temperature of the photon gas:
$$T_\gamma = T_\gamma^0 (\frac{4}{11})^{\frac{1}{3}}(1+z)$$
where $T_\gamma^0$ is the initial temperature of photon gas and z is the redshift. Therefore, over time, the gas temperature decreases.
Another hand we can ...
2
Heat is a transfer of energy from an object with higher temperature to an object with lower temperature. I think what you actually wanted to know is if thermal energy is associated with vibrations or electronic excitations.
Thermal energy is energy that is stored in the internal degrees of freedom that we cannot keep track of individually. This includes ...
2
The gas is not at thermal equilibrium because heat is being conducted across the gas. If gravity is not present, then the pressure will be constant, and the temperature will vary as you say. In this case, you have $$\rho (x)=\frac{P}{kT(x)}$$ where $\rho(x)$ is the local "number density" of molecules. The total number of molecules in the ...
2
If the pressure was higher on the hot end than on the cold end, the total force would also be higher on the hot end than on the cold end, and hence, the container would accelerate without external directed force being applied, which would contradict momentum conservation. Therefore the density must adapt accordingly, as you have already suspected. This ...
2
A molecule with $N$ atoms will in general have $3N$ degrees of freedom, i.e., $3$ per every atom (see the table in this wikipedia article). Indeed, at low temperatures the vibrational degrees of freedom are frozen out, i.e., the motion of the molecule can be considered as the motion of a rigid body with $3$ translational and $3$ rotational degrees of freedom....
2
This document provides a formula for the vapour pressure of $N_2O$:
$$ log_e \left(\frac{p}{p_c}\right) = \frac{1}{T_r}\left[b_1(1-T_r) + b_2(1-T_r)^{3/2} + b_3(1-T_r)^{5/2} + b_4(1-T_r)^5\right]$$
where $p$ is the pressure you want to solve for, $p_c = 7251$kPa, $b_1 = -6.71893$, $b_2 = 1.35966$, $b_3 = -1.3779$, $b_4 = -4.051$, $T_r = T / 309.57$K, and $T$ ...
2
There is no universe before the Big Bang, it is the point at which space and time come into existence. There is no "where" that could have a temperature and no "when" you could call before matter. The question is meaningless.
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$\newcommand{\mean}[1] {\left< #1 \right>}$
$\DeclareMathOperator{\D}{d\!}$
$\DeclareMathOperator{\pr}{p}$
Proof that $\beta = \frac{1}{k T}$ and that $S = k \ln \Omega$
This proof follows from only classical thermodynamics and the microcanonical ensemble.
It makes no assumptions about the analytic form of statistical entropy, and does not involve the ...
2
I think you are referring to the temperature of the cosmic microwave background? In which case, the derivation arises from Wien's law for a blackbody distribution, which connects the temperature of the blackbody with the peak wavelength of the intensity vs wavelength distribution. As the universe expands, this wavelength increases along with all other length ...
2
Here is one way of looking at this which might help you.
Imagine an old-school bulb-type thermometer. You stick the bulb into the water bath, the red alcohol meniscus inside the thermometer takes up a position along the length of the thermometer, and then you read the temperature of the bath off the scale point nearest the meniscus position.
Now imagine the ...
2
Fahrenheit and Centigrade scale are related via a linear transformation:
$$
T_F = a T_C + b \Leftrightarrow T_C=\alpha T_F + \beta,
$$
so they are really the: once we know one, we automatically know the other. Note that this would be also true for more general functions (e.g., if we had two thermometers based in substances that expand with temperature in a ...
2
There are several factors (temperature, specific heat, volumetric heat capacity, thermal conductivity, and latent heat of fusion for ice) to consider. See the data below for steel and ice. Their relative importance depends on exactly what your goal is (lowest final water temperature and/or time to reach that temperature), how much water, ice, and steel you ...
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