New answers tagged

1

It does not violate any laws, the energy is not directly transformed into heat, which would be highly inefficient, but is used to remove heat from a hot source to a cold one (the inverse of a refrigerator) en.wikipedia.org/wiki/Coefficient_of_performance


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From the links I provided in the comments below your question it should become clear that entropy "meters" do not exist, you calculate it from other measured variables. If this does not satisfy you requisites for an experimental measurement, then your conclusion that the claims are only theoretical is justified. However, having said that, with your ...


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More of an extended comment, but here are two thoughts: 1) I'm not sure I completely agree with the statement Textbook discussions of the Second Law of Thermodynamics (SLT) often stress that this law applies only to "closed systems". Or, differently stated: if the system is not closed, its entropy can go down. Okay, this is correct, of course, ...


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To describe a chemical reaction first you have to define a model to describe the process, for example, in a reactive collision like A + BC -> AB + C you could use as a first approximation the transition state theory (TST) and perform the determination of the physical observables with some methodology like the quasi-classical trajectories (QCT), which solves ...


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there are lots of good books in thermal radiation, read Modest for example Author : M Modest Release Date: 15 Mar 2013Imprint:Academic PressPrint Book ISBN :9780123869449 eBook ISBN :9780123869906 Pages: 904Dimensions: 276 X 216


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When you learned this material as a beginning physics student, they taught you that $nC_p\Delta T=Q$, and the focus was typically on a solid or liquid (both of which are nearly incompressible). However, this was only introductory, and was not the correct and precise definition necessary to use in thermodynamics. In thermodynamics, it is recognized that ...


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The expressions are specifically for the cases "gas heated in a constant volume" ($C_V$) and "gas allowed to expand so that the pressure stays constant" ($C_p$). So you can't substitute one expression into the other because the values for $Q$ would be different. The $C_V$ case is easiest to understand: you add heat, increase the kinetic energy of the ...


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Other answers have handled the oxygen issue quite well. In regards to the temperature, space itself has no temperature because it's a vacuum. Objects in space, however, do have a temperature. If a human is exposed, unprotected, to space near the sun (or any other star), the temperature change in their body could very well be terminal. Even near our ...


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It is a bit of a hypothethical scenario where the body can breathe oxygen and does not rupture from the pressure difference (dissolved gases in the intestine and blood bubbling), but at the same time can evaporate and radiate freely. If we assume so, and also assume that there is no sunlight, then there are two major mechanisms for heat loss: radiative ...


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This is certainly an interesting question. As the question is currently put: The entropy argument says that the piston will eventually move. To know how fast it will move, we have to look at the rate of transfer of heat across the piston. If you have a perfectly insulating piston, this rate is zero. Therefore the first answer has to be qualified with: ...


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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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If an empty pan is put on the stove and a natural gas flame is placed under the pan, the pan will heat up to the flame temperature of 1083 C. This will probably destroy the pan. On the other hand, if pure water is placed in the pan, the pan will heat up to slightly more than 100 C at sea level, as the heat being put into the pan by the flame boils the ...


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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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You can't have strong enough electric fields to tear the proton away from the nucleus but it is really a very subtle thing and the inability is just "by a little bit". The strongest electric field that may exist is given by the Schwinger limit. In $\hbar=c=1$ units, the field is $m_e^2 /q_e$. Once you reach this value, electron-positron pairs start to be ...


1

All the particles that annihilate or decay as they become non-relativistic, heat the photon bath. A decaying particle, like the Higgs, also heats the photon bath by decaying to lighter particles, giving those particles alot of energy. In general, if the neutrinos had decoupled at a temperature $T_\text{dc}$, the temperature of the neutrinos would, at any ...


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Air molecules rush in all directions. There are a lot of molecules at the surface of the Earth. They continually bounce around, bumping into other air molecules or the walls of a container after traveling a microscopic distance. An open bottle has air in it. At the mouth, molecules fly in and out equally. Overall, they don't move much. If there was no ...


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The first suggestion that the piston will not move, is true, if the area of piston in contact with S1 = area of piston in contact with S2 (otherwise $F_1 = PA_1$would not equal $F_2 = PA_2$ and the difference in force will move the piston). At the time when there is restriction in the piston, the pressure exerted on the piston by the system 1 is equal to the ...


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I was going to make this a comment, but I thought of a phrase that I think intelligent non-physicists should grasp, so it might be worth keeping in your "teaching toolkit". At one level it's wholly a matter of taste and English usage that tells which phrase is correct, but the phrase "rushes in" is probably more evocative of the true physics than "sucked ...


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Yes, and eventually, the heat will be uniformly distributed to the copper. But since the formula of Q assumes a uniformly distributed temperature outcome, it is effectively used when there is none of fairly little amount of heat loss from the rod (due to air convection, radiation, conduction due to contact with other surfaces, etc.) will occur during the ...


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There are two components in this discussion. When we say two systems are in equilibrium, the two systems have the same temperature. When a process is a reversible process, the system is changing in a sequence of equilibrium states. Intuitively speaking, if we let the system change by 1 degree C from temperature A to temperature B and we want the change to ...


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It will try to distribute throughout the material but it takes a while therefore it is hottest for a while where it was introduced. Also depending on the shape of the bar much of the heat could be lost to the other sides before it is distributed. In the end the overall temperature of the material is influenced by all the surfaces.


1

Yes, the heat will eventually distribute itself uniformly through the material - this follows from the second law of thermodynamics. However, it can take a long time for uniformity to be achieved. The gradual diffusion of heat through a material can be modelled using the heat equation.


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This answer has been updated and the response has changed! You've discovered a famous problem in thermodynamics. In our case the piston will not move. The mechanical argument is right, while the maximum entropy argument is inconclusive. To see that $P_1=P_2$ is an equilibrium position you can also apply conservation of energy. Since there is no heat ...


2

' if we employ any other process that is not adiabatic, say an isochoric process, to take the system from one temperature to another, we shall need a series of reservoirs in the temperature range $T_1$ to $T_2$ to ensure that at each stage the process is quasi static ' This means if we want to have a reversible process between two temperatures, then we ...


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I will only make some qualitative and conceptual observations here that does not really deal with the specific problem, but shares some insight on why the two notions in the question are not conflicting. This may or may not answer the question. One way of treatment suggests that since $P_1=P_2$ and and since only mechanichal work (exchange) is allowed - ...


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If the phase change occurs at the temperature of interest, then the system can give off a lot of heat without cooling down very much. Thus, melting ice is a great way to maintain something at a temperature around 0°C, and melting paraffin-18-Carbons is good if you are trying to maintain temperature around 20 °C. With a melting point of 28°C, the material ...


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A thermodynamic process is called reversible if an infinitesimal change of the external conditions is capable of reversing the process. A thermodynamic process is called quasi-static if it is a dense succession of equilibrium states. Roughly speaking, a quasi-static process is one that can be represented by a continuous curve in the thermodynamic ...


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If it is pure carbon, it is quite difficult to form a flame. The possible molecular in the flame in air are, $$C+O_2=CO_2$$ $$CO_2 \leftrightharpoons CO + O_2$$


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I would assume that entropy would be a large factor. In any conversion of energy from one form to another, there will naturally be some degree of waste heat (entropy) generated. Therefore, heating is like rolling something downhill- it's the "natural" tendency anyway. Cooling something generally involves moving heat up a gradient, which requires more energy ...


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The liquids are either dissolved or in a colloidal form. When the liquid is boiled off, the solute or particulate matter comes together to make a solid. This is also achieved by the changing of the molecular structure. This procedure is also called denaturation. This is how liquids can be heated into solids.


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Simply Said: It doesn't attract itself enough to be solid. So much resistance to attraction that it is very hard to liquefy. The electrons prefer to repel because the electrons repel like the protons.


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This is a fuzzy answer to a fuzzy question. First, express the interaction between your system and the bath in terms of standard photon quantum operators $a^\dagger_k, a_k$. Use substitutions $a_k^\dagger+a_k\rightarrow q$ and $i(a^\dagger_k-a_k)\rightarrow p$ to shoehorn it in the form of the Hamiltonian that you want to see. (I'm omitting all the ...


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For the first statement, there is no problem with it as everyone talks here. For the second statement, it is wrong because the two system are separated by the system without exchange heat. There will be no change in entropy. One simple and quick way to reason is that entropy change is accompanied by heat exchange, $$dS=\frac {dQ}{T}$$


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Throttling the gas (Joule-Kelvin expansion) only lowers the temperature of the gas when the Joule–Thomson coefficient is positive. For Helium, that point (the "J–T inversion temperature") is reached at 43°K (source: Cryogenic Society of America; the wikipedia article gives an incorrect value of 51°K). Above that temperature, Joule-Kelvin expansion will ...


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For the first question, it is the low boiling temperature, 4.21K for Helium-4 and 3.19K for Helium-3, that makes helium difficult to be liquefied. Hydrogen's boiling temperature at 1 atm is 20.27K, or about 4-5 times higher. For pre-cooling, one can take a look of entropy $$\delta S = \frac {dQ}{T}$$ We can see that, because T is very small, a slight ...


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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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A significant factor in this is that helium is hard to produce. There is very little helium in the atmosphere because it is so light that individual molecules have a good chance of achieving escape velocity at atmospheric temperatures so it disappears into space fairly quickly. Also because it is very uncreative it doesn't get bound as chemical compounds, ...


0

Maybe the Cambridge Cosmology Lecture Notes can help to answer at least the first question. Besides I think you meant annhilation instead of scattering (?), because these equations are for the Dark Matter $\chi$ annihilation process \begin{align} \chi + \bar{\chi} \Leftrightarrow c + d~~~~. \end{align} However, a system is said to be in kinetic equilibrium ...


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The next approximation beyond the ideal gas is given by the Van der Waals fluid equation. It is a phenomenological law which takes into account the finite size of the molecules and their interactions with themselves. When you plot several Van der Vaals isotherms for a given substance, you observe that some of them show a phase transition from gas to liquid ...


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Getting from gas to liquid is a matter of interparticle interaction winning over thermal agitation. There are several reasons why interparticle interactions are very weak in the case of helium atoms. On one hand, it is a noble gas and thus cannot form covalent bonds. On the other hand, it is very light hence highly non-polarizable: its Van der Waals ...


2

It would be more scientific to have compared a covered with an identical non-covered bowl, with temperature measurements over an identical period of cooling time. But covering a bowl of hot soup with a plate does indeed reduce heat losses and slows down cooling of the soup, because: Radiative losses are usually small at these temperatures but the plate ...


1

There's an important difference between liquid water and water vapor. The difference is 539 calories per gram to be exact. That's why being in humid air makes you hotter - the water vapor has all this latent heat it can release upon you by condensing. Equivalently, it prevents your sweat from evaporating and absorbing this energy. Liquid water doesn't have ...


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When you feel hot, you perspire so as to benefit by evaporative cooling. As the relative humidity gets closer to 100%, the sweat cannot evaporate and evaporative cooling becomes less effective. Liquid water is a much better conductor of heat than air (even humid air) is, so if the water is even a few degrees cooler than your body, you feel cold because the ...


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You are describing two different mechanisms of cooling the human body: 1) When we sweat our body produces fluids that tranfer heat from us through their evaporation (fluids gets the heat from our body and evaporate). One can easily understand that the more the surrounding enviroment has a highly humidity value , the more this heat- transfer gets ...


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You feel cold when heat is flowing from you to the surroundings, your body tries to burn more energy to keep up your temperature, so you shiver. Water conducts heat much more effectively than air (more than 100x as well) so even with water at the same temperature as air you will lose a lot more heat and feel cold. When your body is too hot it losses energy ...


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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 ...


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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 ...


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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 ...


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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 ...



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