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3

Since your plasma is in a vacuum environment, the only way for it to loose energy is by radiation (conduction transfer through the magnets are neglected). You have thus to consider which bodies are surrounding your plasma and which radiative model is the best ton consider for them. I guess you can consider a black body with the simple Stefan equation. The ...


0

The zeroth law posits the existence of temperature by stating that if A is in equilibrium with B and A is in equilibrium with C, then B is in equilibrium with C. We can then assign an intensive property to A, B and C that we call "temperature". They are in equilibrium == they have the same temperature. As soon as they are NOT in equilibrium, the zeroth law ...


0

To have an equivalence relation ~ on a set $X$ you need three things: reflexivity, symmetry, and transitivity: Reflexivity - for any $x \in X$, $x$~$x$ Symmetry - for any $x, y \in X$, if $x$~$y$, then $y$~$x$ Transitivity - for any $x, y, z \in X$, if $x$~$y$, and $y$~$z$, then $x$~$z$ Do do this right, we need to first understand that thermal ...


0

The rational behind the definition of temperature is exactly what is stated in the 0th law of thermodynamics. As such: Since objects into contact will reach a thermal equilibrium (and heat transfers from hotter to colder), this means one can use a device that can measure this difference of this quantity that changed as follows: We have a device (called a ...


1

As it was done quickly, the mass of the mercury remained at room temperature for a short period of time, even as the thermometer itself (including the metal bulb) began to heat. I suppose it is possible that even though it is a tiny portion, the metal bulb expanded sufficiently so that the volume inside the bulb increased and the level of the mercury went ...


1

If you stick with the Statistical Mechanics formalism, then temperature is discrete. Entropy S is $ S = k_B\ln(\Gamma) $, where $\Gamma$ is the number of available quantum states, and temperature is $T = \frac{dE}{dS} $ , where E is the total kinetic energy of the system. Since every quantity there is discrete, so is temperature.


1

The usual hydraulic analogy for a capacitor is an elastic membrane: A capacitor doesn't allow current to flow across it, but you can push charge onto it by applying a potential. In the hydraulic analogy an elastic membrane across the pipe doesn't allow water to flow through it, but you can push some water through the pipe by elastically deforming the ...


1

I suggest the same answer for a different reason out of personal experience, as I am rather fussy: If you pour cold water first, when you pour warm water (unless the jet is powerful) it will stay above and the drink will have an unpleasant difference of temperature. If you pour warm water first, when you pour cold water it will sink to the bottom, creating ...


1

I think that it makes little difference... but anyway... You should pour the room-temperature water first, and then the cold water until you get the target temperature. Why? Because if you pour the cold water first, it will immediately start to warm. If you leave it enough time it will actually reach equilibrium at room temperature. So the amount of cold ...


0

From seatemperature.org, the sea surface temperature at Shizuoka-shi is currently 25°C. Most people would find that temperature to be just about perfect. Okinawa is a better place if you like warmer waters.


1

I would strongly disagree with your statement that entropy has always been quite a mysterious quantity. Quite the contrary. What makes water (without something else interfering) always flow down the hill? Gravity. One does not have to know anything further about gravity to make good use of this statement, and people have used this fact for thousands of ...


1

What it means is that the number of bits required to specify the exact physical state the system is in, increases by 100/log(2) bits after the gas is heated. I think measuring the temperature in energy units is a step in the right direction, but what is even better is to do without any units. I.e. while units may be introduced for convenience, the formalism ...


1

Today, stars don't emit the CMB – they emit much more energetic (hotter, higher-frequency, shorter-wavelength) radiation in general. CMB was emitted by "everything" that could emit radiation at the relevant time, around 400,000 years after the Big Bang. At that time, the radiation was in thermal equilibrium with everything else, so its distribution to ...


2

Things like flower and lamb leg become brittle because of large portion of water contained in them. Water gets frozen into ice upon cooling, which is brittle. Basically, brittleness is related to the directionality of chemical bonds. Materials form by more directional bonds tend to be more brittle. Meanwhile, they tend to be harder.


0

The name of the property is itself a clue here : enthalpy of vaporization. By nature, enthalpy does take into account the work required to push the atmosphere. You can see the impact of increasing the pressure on the enthalpy of vaporization on a Mollier diagram. Increasing the pressure has the overall the effect of reducing the enthalpy of vaporization, ...


5

Many organic substances will become brittle at the temperature of liquid nitrogen, but there are plenty of material choices left for e.g. vacuum seals, pipes, containers, etc., which are not. Indeed, we have constructed entire rocket fuel systems at the temperature of liquid hydrogen, which retain most of their mechanical properties. If you go down further, ...


5

As far as I remember, yes, everything becomes brittle at low enough temperatures. This is due to the brittle-to-ductile transition (BDT - or sometimes referred to in reverse as DBT, ductile-to...). This transition is temperature dependent (amongst others (strain-rate ...)). Can every composition actually reach low enough temperatures, or do some have a ...


3

As far as dimensional analysis goes, temperature and energy are separate and independent physical dimensions. However, there is a more or less unique way to translate temperatures into energies and vice-versa, which is by means of Boltzmann's constant $$k_\text{B}=1.380×10^{−23}\:\text{J}/\text K.$$ Any given temperature $T$ has an associated ...


0

Viscosity is the material's resistance to shearing stress. According to the Wikipedia entry, water has a viscosity that is exponential: $$ \mu\sim A\cdot10^{B/(T-C)} $$ If we then look at the Navier-Stokes equations (which holds for Newtonian fluids, something grease is not (instead a non-Newtonian fluid)), $$ \rho\left(\color{blue}{\frac{\partial\mathbf ...


0

TL;DR: Feynman was talking about a gas, so he was correct: molecules of a gas are largely independent, they don't have any quantum zero point kinetic energy. Their mean (random) kinetic energy is temperature. If he had been talking about any other particles such as atoms in a solid he would not have been correct - they do have kinetic energy even at ...


1

I am very ignorant here, but.. Surely the question is whether you want to do statistical mechanics or not. Finite temperatures will allow you to ignore the energy sources - but still enable you to calculate the production of particles. I don't see how you would ever use it for scattering problems involving small numbers of particles. On Noldig's answer: ...


4

You have to check if the temperature is small compared to the chemical potential. In heavy ion collisions, the chemical potential is rather low but the temperature is very high, so one have to stick with thermal field theory. In compact stars the densities and therefore the chemical potential is at the MeV scale whereas the temperature is at the keV scale so ...


6

Your question, I think, is best answered by separately addressing its many directly and indirectly implied questions. You are mixing up a lot of different notions together in your question as you've phrased it. So to clarify, I thought I would pull all of the pieces apart. To start: Do atoms have nonzero kinetic energy at absolute zero? Yes, they still ...


7

If carefully interpreted and converted to mathematics, the passages are not contradicting one another. The uncertainty principle guarantees that there is some zero-point energy that can't be eliminated (the second passage) – it is the energy of the ground state of the physical object (atom or a macroscopic piece of a material). On the other hand, the ...


0

I'll give brief answers to your questions. If you need more detail, you should ask your questions separately. What's the difference between heat and work at the atomic level? Isn't heat simply work between particles colliding with different momentum against each other? Treating a substance semi-classically, one can say that at the atomic level, the ...


4

As a visual demonstration of Luboš Motl's answer, this: Image obtained using Climate Reanalyzer (http://cci-reanalyzer.org), Climate Change Institute, University of Maine, USA. is the average surface temperature on earth for 2013. This: is the solar flux by Luboš' formula. And here: I've tried to replicate their crazy color scheme. The point is, ...


0

Heat is energy in transit. It only comes into play when one body is transferring heat to other. Feynman best explained in a video on youtube, it is just the vibrations of atoms at an atomic scale. So, if a more vibrating solid object comes in contact with solid with relatively low "jiggling", we say heat is transferred. [Ref]: Feynman lectures on physics ...


0

as in simple word or theory.....according to energy conservation and distribution law. In the universe there is fix energy for it. Now universe is expanding so the volume of universe is increasing so respectively temperature gets low.


3

According to the Section 1.1 of the ENDF-6 Formats Manual, each header of the raw ENDF/B-VII.1 Incident-Neutron Data file available in the LANL Data area has the following format: ZA, AWR, LRP, LFI, NLIB, NMOD ELIS, STA, LIS, LISO, 0, NFOR AWI, EMAX, LREL, 0, NSUB, NVER TEMP, 0.0, LDRV, 0, NWD, NXC where the TEMP field denotes ...


3

For the most part, the temperature of the medium doesn't matter. Thermal energies are typically around $kT=25\,\mathrm{meV}$, while nuclear reactions typically have energies of a few MeV. A factor of a billion in energy is a big difference. A skim of the explanatory text in the datafile corresponding to your plots reveals no mention of temperature. If ...


3

For those of you are perhaps a little behind on the background, these cross section libraries need to be processed before they can be applied to a realistic material. This would be appropriate for situations like a simple source term with the neutron tracks propagating throughout the material. You know the temperature, and you have the library specific to ...


5

This is the basic confusion: but nevertheless it will start emitting visible light, that is thermal photons with a temperature in the excess of 400°. Temperature is an intensive variable, it characterizes the whole sample. Examples of intensive properties are the temperature, refractive index, density and the hardness of an object. No matter how ...


28

If you heat a metal (or anything else) up to a temperature $T$ then the average energy of any degree of freedom of the metal will be of order $kT$. At 600ºC this is about 0.075eV, and as you say the energy of visible light is around 2 - 3eV, which is a factor of 30 or so higher. The reason that visible light can be produced is because the thermal energy is ...


0

The temperature of the blackbody radiation predicted from the event horizon of a black hole with mass $M$ is $$ T = \frac{\hbar c^3}{8\pi k G M} = \frac{62\,\mathrm{nK}}{M/M_\text{sun}}. $$ So a supermassive black hole with $M=10^9M_\text{sun}$ would have a horizon temperature in the neighborhood of $10^{-16}\,\mathrm K$.


0

The absolute zero of temperature is, well, absolute zero. This is the situation in which everything in is its ground state and there are no excitations in the system. Most places in the universe are hotter than this: the 4 degree background radiation e.g. keeps them hot. The coldest places we know of are all artificial and the current record keeps ...


0

A very simplified back of the envelope calculation would be as follows: The amount of energy in joules necessary to heat up the air: $$W = m c_{air} \Delta T$$ With $m$ the mass of the air enclosed in the room, i.e. $m = \rho V$, so you need to take a value for density $\rho$ if you're looking for a numerical estimate. $c_{air}$ is the specific heat ...


1

The answer given by Kyle refers of course only to the surface or photospheric temperature of the neutron star - the temperature of the layer from which photons can escape to reach an observer. In these outer layers the relationship between temperatures and particle motions is more-or-less consistent with the "everyday" Maxwell-Boltzmann picture referred to ...


0

Question 1. The efficiency only depends on temperatures because all reversible engines operating between the same two heat baths are equally efficient and the temperatures is the only quantity that specifies a heat bath. Question 2. We indeed use definitions and units in which $f(T_2,T_1)=T_1/T_2$ but this simple form is only possible because we have ...


20

First, strictly speaking a neutron star is not a nucleus since it is bound together by gravity rather than the strong force. Measuring a surface temperature for any star is deceptively simple. All that is needed is a spectrum, which gives the luminous flux (or similar quantity) as a function of photon wavelength. There will be a broad thermal peak somewhere ...


0

A specific temperature corresponds to an equilibrium population of available states. At normal temperatures the atoms (and quarks) are too tightly bound to have translational degrees of freedom and therefore have no kinetic energy states. Instead the molecules (in gas) have kinetic states. Atomic movements are captured in vibrational (solid) or rotational ...


0

The thermodynamic definition of temperature has been found to be emergent from the underlying particulate nature of matter. It is connected with an average over the kinetic energy of individual particles. Here v is velocity of a molecule, m its mass, k_B the Bolzman constant and T the temperature The kinetic energy requires to have a degree of freedom, ...


3

Taking an intuitive guess here: The pressure above the water column is indeed very low, and water molecules at the surface may escape - but they are also held back by the surface tension of the water (is your meniscus concave or convex?). There is an equilibrium here, and the temperature is low enough that the water won't boil off all at once. So at room ...


3

I believe that in the top you'll find it's the saturated vapor pressure of water. http://en.wikipedia.org/wiki/Vapour_pressure_of_water (80 F ~27 C vapor pressure of about 27 mmHg or 27mm/760 mm * 14.7 = 0.52 psi.. not bad.


0

The ideal gas equation you try to use for understanding does not provide a full description. To completely describe a thermodynamical system, you need the relevant thermodynamic potential, which here is the internal energy if I understood well your formulation (is a bit vague). Processes like the isobaric expansion occuring here, cannot be explained through ...


-3

The average kinetic energy of the particles is directly related to the temperature of the object.


0

You're on the right track here. For convection between plates to 'stop' (it won't really stop, just get wicked slow) you want viscosity to be very high relative to buoyancy, so you want low Grashof. Also, in the case of very narrow plates, you'd want to use the gap width as your characteristic length $L$. $10^8$ is the transition point at which the flow ...


0

I humbly submit that aside from any evaporative benefits, a simple lawn sprinkler on the roof would help prevent ignition from floating embers from burning trees. I'd think that these embers could float for great distances and readily ignite tinder dry grasses, foliage, house siding or roofing material, way before combustibility from heat alone.


0

Why would the temperature of the gas in the balloon be fixed? Note that it will depend, among other things, on the thermal conductivity of the rubber of the balloon: if the thermal conductivity were zero, the gas in the balloon would expand isentropically, rather than isothermally.


1

For a system of atoms to have a meaningful temperature, I would say there has to be uncertainty in their state, and there must be a way (at least hypothetically) for the system to exchange energy with some other system. The first example you give of the two beams is an idealized state with no uncertainty, so assigning a temperature would be impossible. You ...



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