# Tag Info

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The temperature of a true vacuum would be a measure for the energy distribution of the photon gas in that vacuum. You can derive the occupation of the electromagnetic modes in a volume with Bose-Einstein statistics, which is essentially what Planck did to describe the emission spectrum of a black body. However, you don't need to do understand the details of ...

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The simplest answer, if I did not misunderstood your question, is to adiabatically compress the gas, both pressure and temperature will raise.

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If you take a bottle of gas and carry it with you on a supersonic plane, then the molecules will go much faster without the temperature changing. If you let pressurized gas flow through a well-designed nozzle (De Laval nozzle), the gas will accelerate to supersonic velocity (i.e., faster than the original thermal speed of the molecules) while the ...

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For commercial uses, usually these gases are in liquefied form .You can find more details, for LPG lets say , here: https://en.m.wikipedia.org/wiki/Liquefied_petroleum_gas In general you are right.

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The Ideal Gas Law PV=nRT explains what happens. It can be written as PV/T=nR. For our purposes nR can be considered a constant, PV&T are all expressed in absolute units. Thus if the container volume is constant, P/T is a constant.

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Apart from volume and temperature, there is one more parameter affecting pressure, that is number of molecules. At filling up the container you added more molecules in, consequntly you get higher pressure.

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Once the wax is heated with solar radiation during the day, it expands in every direction in the cup. When it cools down during the nights, the center wax returns back but the wax adjacent to the cup faces another attractive force called adhesion. Adhesive force is the force between different molecules in contact. So, my explanation for the concavity of the ...

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The mechanism for increasing the thermal conductivity is phonon assisted hopping. For a disordered system, one which do not preserve the long range order, the electronic wave function becomes localized. The wave function extent is typically much smaller than the system size and is characterized by the localization length $\xi$, a parameter in theory. In this ...

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The question operates under a false premise. It is not the case that fitting a blackbody spectrum to the Sun gives you its "surface temperature". The Sun does not have a black body spectrum, although sometimes that approximation is made. A star also does not have a single "surface temperature". What you can do is divide the luminosity of the Sun by $4\pi ... 0 If you have not just a uniform heat emitting cylinder but namely solid electrical conductor of certain non-zero diameter then consider that the current travels along its surface layer, not through inner layers. So happens because all electrons are charged '-' thus want to push away each other. Not too rough to say that in this system the heat is produced ... 4 There are some issues with the experimental setup you proposed (apart from the fact that when its temperature is lowered the gas would become a liquid and then a solid - if it's not$^4$He: in that case it will stay a liquid). Let's see why. In the picture above, I've sketched your experimental setup. The black box must be impermeable to matter in order ... 2 Your question is: assuming a rod of radius$R=1$cm with a fixed heat production per unit volume, totaling$Q=100$W/m, surface temperature$T_0$and thermal conductivity$\lambda$, what is the temperature$T(r)$as a function of$r$? First, observe that within a cylindrical shell of radius$r$, the total heat production is$q=Q(r/R)^2$. The temperature ... 4 It won't work because your perfect vacuum is permeated by the cosmic background radiation, which itself is only asymptotically reducing to zero with the expansion of the universe. Trying to exclude the cosmic background radiation backs you into the infinite steps that forms the basis of the third law again. Also, using a container results in quantum ... 10 You seem to make the implicit assumption that your vessel is placed in an environment that does not emit any thermal radiation, i.e. is already at 0 K temperature. The temperature of your container will asymptotically decrease to 0 K but will never actually reach it. Assuming black-body radiation, fixed heat capacity$c$, and sufficient thermal ... 2 For the hypothetical case of a thermally perfectly insulated system, I'm sure you can work out yourself from the specific heat and the enthalpy of fusion for water. Given that the enthalpy of fusion (330 kJ/kg) and the specific heat of ice (2 kJ/kg-K) have a ratio of 165 K, and you need the entire ice bucket to stay below the melting point, and your 20:1 ... 0 The speed of sound, which is effectively a measure of how fast particles are moving on average in a gas, is given by: $$C_{s}^{2} = \frac{\partial P}{\partial \rho} \tag{1}$$ where$P$is the pressure and$\rho$is the mass density. For an ideal gas, one can use an adiabatic equation of state such that Equation 1 goes to: $$C_{s}^{2} = \frac{ \gamma \ P ... 4 Looking around, the root mean square speed of air at 20 C is about 500 m/s, and given that you have \langle v^2 \rangle \propto \, T so that v_{rms}(T) = \sqrt{\langle v^2\rangle} varies with \sqrt{T} then have$$v_{rms}(15) = v_{rms}(20)\times \frac{\sqrt{15+273}}{\sqrt{20+273}} \approx 496 m/s$$and$$v_{rms}(25) = v_{rms}(20)\times ... 1 Your question seems to be about heat transfer through convection. The formula that describes this phenomenon is:$dQ/dt=h*A*(To-Tenv)$where$dQ/dt$is the heat transferred per unit time,$A$is the area of the object,$h$is the heat transfer coefficient,$Ta$is the object's surface temperature and$Tenv$is the fluid temperature (temperature of air ... 2 Your scenario is actually a classic transient conduction problem tackled in undergraduate engineering heat transfer, so we can handle this scenario easily. I took the figure below and adapted a derivation from a popular heat transfer textbook by Incropera and DeWitt: In this figure a warm object is placed in a tank filled with a known liquid (the ... 1 Whether something is solid or not, generally depends on two things: the attractive forces between molecules/atoms and the temperature. The attractive forces will try to lock the molecules together and make it harder for the material to change shape (or being solid), but this can only happen when the molecules aren't moving fast. If they are moving too fast, ... 1 Molecules of solid object are still able to move and can even move in high speed if the object in high temperature . Being solid only mean that those molecules is hard to separate or make room ( like liquids does ) ,it doesn't mean those molecules have to be standing still . That what I think . Sorry for my bad English 3 Solid and cold are two distinct concepts. As you can change the state (solid, liquid) with temperature (cold, warm), there must be a relation to it. For each degree of freedom you have a thermal energy of$k_\mathrm B T/2$. Atoms in a crystal have a certain binding energy. If the thermal energy is way larger than the binding energy, no crystal can form. If ... 3 The "buoyant force" does not arise from a principle, but is what remains of the gravity force when you subtract from it an average hydrostatic pressure. Let's assume that you have some reference density$\rho_0$. Then if$\rho=\rho_0$everywhere and the fluid is at rest,$p = \rho_0 g z$. Now let's call this baseline pressure$p_0(z)$, without loss of ... 6 It is entirely possible that the entire candle has melted in the sun. If this happens the molten wax has a significantly greater volume than the solid wax candle and so the whole level will rise up the glass ie liquid wax takes up more volume than solid wax. As it cools again the level will fall as it solidifies. Typically it will cool from the top and ... 3 We're talking about the Coanda Effect, right? Then I think this article could provide some useful insights. Quoting from the article: When the fluid flows over the heated curved surface in proximity of the curved surface as the temperature of the curved surface increases, dynamic viscosity of the fluid at the vicinity of the wall is increasing with ... 1 My hypothesis is this. When the Sun heats up the wax, the wax gets soften or melt. The surface tension between the glass wall and the wax then drag the wax up the wall. 12 Yeah, I would guess that the alternate heating/cooling of the wax in the sun pushes it up the side of the glass. Presumably the surface tension between the wax and the glass is quite strong and holds the wax up once it's been pushed up. Subsequent cycles cause wax to "backfill" the wax that's been pushed up. It would be interesting to design an experiment ... 41 Candle wax expands considerably when hot and molten. So while burning the candle the level in the glass rises. But when the candle is extinguished the outer region (nearest the glass) cools down quicker (candle wax doesn't conduct heat very well) and solidifies first, becoming immobile. The molten remainder then shrinks before solidifying. So it's the ... 0 The heat of the sun probably caused the wax to expand resulting in it crawling up the walls 7 The energy comes from the sun, that much is certain. One possible mechanism is capillary action resulting in a meniscus, assuming the sun heats the wax to a (near) liquid state. The other possible mechanism is a vaporization/redeposit cycle. During the day, heat from the sun creates wax vapor, which is heavier than air [citation needed] and therefore ... 3 Temperature can be thought of as the vibration or oscillation of individual particles. More the vibration, more the temperature. The frames velocity is just the velocity of its mean position, as the vibration is independent of the frame velocity, so is the temperature. 5 Concerning the meaning of "temperature coefficient of [measurement]". Generally, when it appears without additional adjectives, the "temperature coefficient" of anything refers to the fractional slope of a linear fit to that quantity as a function of temperature. In the introductory class we often introduce A linear coefficient of thermal expansion, in ... 1 While Kieran Hunt's answer is very good, a more blunt answer is that no, there is not such an algorithm. The easiest way to see that there is not consider not physics but economics. There are a large number of economically-important events which are crucially dependent on whether or not it rains -- as an example when wheat is harvested its moisture content ... 3 The quantity$\left(\frac{\partial S}{\partial T} \right)_{\{\alpha\}} = TC_v$is essentially proportional to the heat capacity of the thermodynamic system under study. As far as I know, there is no principle of thermodynamics that forbids such a quantity to be negative. Considerations such as "yes otherwise matter would not be stable" lie outside the ... 0 No. Under negative thermal temperatures the entropy of a system increases as T is lowered. For example, if the magnetic field is reversed quickly enough around a ferromagnetic material, then the system is in negative thermal temperature. 0 No, it doesn't. For example, consider a two-state system with ground state$| 0 \rangle$and excited state$| 1 \rangle$. At zero tempreature, the entropy is zero, since only state$| 0 \rangle$is occupied. As the temperature goes to infinity, the occupancies of$|0 \rangle$and$|1 \rangle$become closer and closer to equal. When they are equal, we have ... 0 You are asking about micro black holes. Some hypotheses involving additional space dimensions predict that micro black holes could be formed at energies as low as the TeV range, which are available in particle accelerators such as the LHC (Large Hadron Collider). Popular concerns have then been raised over end-of-the-world scenarios (see Safety of ... 0 We are talking about maximum amount of work, so you still consider an ideal (Carnot) cycle. But the efficiency is changing as the tank cools down, so there is an absolute maximum amount of work that can be extracted from this heat engine. Efficiency of Carnot engine is$\eta=1-\frac{T_{min}}{T_{max}}$, and is defined as work over heat transferred at the hot ... 1 If you are comfortable with electric circuits this method is useful: The temperature difference between 2 parts of a conducting body acts the same way as when you have a potential difference between 2 parts of a conducting wire. The heat conductivity is equivalent to the electric conductivity of the wire which is inverse of its resistance. And the last ... 1 Assuming the water can't cool, then after about 30 minutes, all the neutrons have decayed to protons and electrons, the resultant antineutrinos escape, but an average of around 0.5 MeV per decay will get thermalised in the water (the range of the electrons will be of order 1cm). As Floris points out, it takes 15.8kJ to heat a gallon of water by 1K, so this ... 2 One really, really energetic one might do it. But it would need to be very close to the speed of light, and I'm not sure you could capture its energy to boil the water. Assuming you want to convert the mass of a non-relativistic neutron to energy, it's about 940 MeV per neutron. Boiling a gallon of water (starting at room temperature, 20°C) requires raising ... 0 Read the "Formula" section of that Wikipedia article again. It gives a second set of coefficients which are good for humidities from 0-80%. 0 I think, the comment of lucas is the best answer; I'll elaborate a bit If I mix these two containers eventually both the Oxygen and Nitrogen will be at the same temperature. Why is that? By temperature you mean the mean energy per particle. We observe experimentally, that if you mix two heaps of particles with different mean energies, after some time ... 0 If you mix two gases, their atoms/molecules exchange with energy, momentum, etc. It is not surprising that a hot gaz heats a cold one. It (heating) can only stop when the mutual energy exchange becomes equal to both gases. So the energy distributions of "subsystems" are equal in the end, but the momentum distributions are still not. P.S. The subsystem ... 1 In a system of many particles, we essentially observe the most probable configuration, and relative fluctuations around it are negligible. Here I will prove that the most probable state of a 2-particle system is this with equal energies. The probability of a state is proportional to the volume of the corresponding part of phase space. If a particle has ... 0 (Thanks to Drebin J.) Let's suppose there's a building with many floors, labelled G, 1-10, and after that continues with A1, A2... A10, B1,B2 etc. A1 is floor 1 but actually the 11th, while A2 is 2nd but actually 12th. (Similar to Celsius in Kelvin) Supposing I jump from A2 to A1 (from the staircase), it would obviously hurt. The damage incurred ... 1 In an exam, Alice scored$50$, Bob scored$40$, Eve scored$10$. Now to raise the class mean, the teacher decided to add$20$points to every students. So Alice's score becomes$70$, Bob's score becomes$60$, and Eve's score becomes$30$. Now Alice complained to the teacher, "my score was higher than Bob's by$10$marks, and you see, according to the ... 0 This seems to me like the general "Problem", that differences of Observables behave in a different way, than the observables itself do: One other example (that has nothing to do with additivity or multiplicativity, as you stated) are differences of frequencies: You know that you can convert a frequency to a wavelength by dividing c by the frequency: ... 1 Celsius and Kelvin are two scales that differs only for an additive factor, but the single increment corresponds to the same temperature difference. In other words, an object become "hotter" in the same way if you rise its temperature by 1K or 1°C. You can use conversion formula in differences, just make sure you use it for both terms and keep in mind that ... 0 A temperature change of 1$^\circ$C is the same as a temperature change of 1K. So if you start at 30$^\circ$C (= 303.15K ) and increase the temperature by 5$^\circ$C (5K) the new temperature is 30 + 5 = 35$^\circ\$C (303.15 + 5 = 308.15K).

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