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6

I suspect this will be closed as "opinion based". I don't believe there is a canonical answer. Usually microscopic scale relates to phenomena that occur on a level much smaller than the system under consideration (atoms in a crystal when you are thinking about the crystal, for example). There is an analogy with micro- and macro-economics. Micro-economics ...


4

The black hole initially lost the gravitational energy that was needed to create the pair. The pair-creation model is a bad description of Hawking radiation, which for macroscopic black holes is really photons. The second particle that gets created above the event horizon doesn't have nearly enough energy to escape. It does, however, produce photons above ...


4

It is a dynamic equilibrium. To see this notice that the Boltzmann distribution law is statistical in nature, for instance, the molecules that have energies between $E_1$ and $E_1+\Delta E_1$ at time $t_1$ are not necessarily the same than the ones in between those same energies at time $t_2$. Particles collide (and thus exchange kinetic energy) all the ...


4

The equal probabilities are meant for states of an isolated system with constant total energy. Each state with this energy is then equally probable. The Boltzmann probabilities are meant for systems in thermal contact and equilibrium with reservoir of definite temperature - in that case the energy of the system may change due to interaction with the ...


4

In thermodynamics the quantities we conventionally use, like Gibbs free energy or entropy, are statistical averages. Remember that our systems are ensembles of around $10^{23}$ particles all buzzing around at random. If you could measure, for exammple, the Gibbs free energy with sufficient accuracy you'd find that it varied randomly with time and only the ...


3

No, you cannot prove that mathematical law. All you can say is that your experiments are consistent with your premise. I recommend you take the time to read some introductory books on physics, statistics, and the scientific method (which is to say, don't just take my word that my initial statement is valid).


3

It is a matter of definition. In the case of the heat engine, you are seeing how much of the heat can be converted into useful work - since something will be lost in the conversion process, a number between 0 and 1 makes sense. When you have a heat pump, you are comparing the amount of work needed to move heat from one reservoir to another with the amount ...


3

The usual way to achieve minimal radiative losses (and thus a low thermal signature) is using MLI - multilayer insulation. The use of multiple layers of reflectors creates a "thermal series" - each layer gets a temperature that is closer to the radiative temperature of the environment, and it allows you to achieve a low thermal emission. This is very ...


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Roughly speaking solid matter is on a lattice form, A three-dimensional lattice filled with two molecules A and B, here shown as black and white spheres. The molecules fit like LEGO , the forces tying them together are mainly the spill over electric field forces , attractive and repulsive forming the patterns of the lattice. In a single crystal one ...


2

Physics is not mathematics. Physical laws are not axioms. A physical theory is not the derivation of all possible hypotheses from a set of axioms. Please repeat this like a mantra a hundred times a day for the next three weeks. Instead a physical theory is patterned as a set of naive ontological assumptions about the approximate usefulness of a set of ...


2

In short, ice-cream is a solid, but it is filled with air bubbles. All this air that was incorporated in the ice-cream as it froze (through various industrial processes) means that the dairy product doesn't freeze as a solid, but rather a foamy delicious treat. There are many websites explaining what it actually is, and how it is made, such as this one. ...


2

Perhaps you are looking for the notion of Legendre transform. Here is the general idea: suppose you have a function of some sets of variables, say $\mathbf x$ and $\mathbf v$, and so $f = f(\mathbf x,\mathbf v)$. You usually get a new set of variables, conjugate to $\mathbf v$ by taking the Legendre transform of $f$ w.r.t. $\mathbf v$. So what you do is to ...


2

Actually, there is a reversible process that will allow you to end up with a larger temperature, and this does not violate the second law. To see this, you can couple your system to Carnot engine, which will extract and "store" work until both the reservoir and the case are at the same temperature. Now use the stored energy to heat the case. Actually your ...


2

The type of blankets is not defined, different can mean many things. If one makes two classes, more insulating ones and less insulating ones the answer of Photonicboom applies. The more insulating one next to the skin will give a comfort value sooner than the less insulating one because the body will have to heat up a smaller amount of matter (air and ...


2

In a still air environment, the previous answers are correct and sufficient. In an environment with unheated air moving over you, however, the answer greatly depends on what blankets you have, exactly. Humans lose a lot of body heat to evaporation of sweat, so the amount of air passing across you makes a big difference in how warm you stay. Let's play a ...


1

Whilst I'm not sure that I disagree with what has already been written, I think that there is another variable - the temperature of the ground. I do a lot of cold weather camping, and it doesn't matter how thick your sleeping bag is, if you are not properly insulated from the ground with a thick sleeping mat, you get cold. Presumably this is because heat ...


1

Work is force acting over a distance. That means there can only be work done on the gas if the piston moves, i.e. if the gas changes volume. In that case, $W = \int F \text{d} x = \int P\, \text{d}V$, where P is the pressure difference between the inside and outside of the calorimeter. Note, however, that in your example the gas is doing work on the ...


1

According to p. 303-304 of the book Gravity from the Ground up by physicist Bernard Schutz, viewable on google books here, it's because in terms of the pair-production explanation for Hawking radiation, one member of the pair actually has negative energy and thus causes the black hole to lose mass (negative mass/energy falling into a black hole can also ...


1

Let's break up the process into two steps, which necessarily must happen in sequence--the first step is one in which the reservoir heats the body up from a temperature of 273 K to its own temperature of 363 K, and the second step is one in which the reservoir heats the body from 363 K to 373 K. The first step involves a positive entropy change, so it can ...


1

On the long run both scenarios will be equally as comfortable. However, when you initially cover yourself, you would want to cover urself with the thicker blanket first, so that you restrict the movement of air as much as you can, and thus warm up a few seconds faster. This is key because your body will warm up the air directly above your skin and the thick ...


1

Thanks to David Hammen for pointing me in the right direction. I was able to find some information on modern methods to measure gas densities. The most accurate modern method is the Magnetic Levitation Densimeter, which works by suspending a weight from a balance plate using magnetic forces. The gas whose density is to be measured is then introduced to ...


1

To clarify, the full assumption is that all states are equally probable, in the absence of any knowledge about the state of the system. With the Boltzmann distribution (AKA canonical ensemble) this assumption doesn't apply since we have knowledge about the system. In particular we know that if the system is put into contact with a thermodynamic heat bath of ...


1

First of all I will address your last concern, which translates into: equilibrium doesn't necessarily mean that nothing is moving. As an example, particle in a solution at equilibrium can move from one side to the other as long as almost the same number of particle move the opposite direction (this usually happens, say, because of thermal agitation/Brownian ...


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Now, if the system is isolated, how has it changed to bring the equilibrium?? The system is isolated from its surroundings, but different parts of it can exchange matter and energy within in. Take, for example, a long metal ruler. Put one extremity on hot water and the other on ice and wait a certain time. A thermal gradient will be established along ...


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(This is not an answer. I cannot comment because I don't have enough reputation). Have you tried googling heat release chemical bond? See this result, for example.


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The short answer is that when energy is released in any form, it will always find a way to become heat - in other words, spreading itself around all the atoms nearby. In a gas this process takes a little longer, but still occurs. The stable state of oxygen is of course $O_2$, so it's not the formation of ozone itself that introduces energy into the ...


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I’m not sure I fully understand the question but I’ll try. The “curious one” says that if the initial condition is picked at random then it should be a state of thermal equilibrium (maximum entropy) since the overwhelming majority of states look equilibrium-like. Now there has to be something wrong with that in the cosmological context. States of thermal ...


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Thermal expansion from an atomistic perspective: The energetic potential between two atoms can be approximated by two exponential functions, one for the attractive force between the atoms, one for the repulsive force. The superposition of these two force fields has a minimum at a certain distance. Examples for such empirical potentials are Stillinger-Weber, ...


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You can always say that $\Delta S \ge \frac {Q}{T}$ irrespective of whether $Q$ is transferred reversibly or not, but for equality to hold you need a reversible process. If the process is irreversible then $\Delta S \gt \frac {Q}{T}$. In fact, the difference $\Delta S - \frac {Q}{T}$ is characteristic of the dissipation taking place in the system. These ...



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