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## Hot answers tagged thermodynamics

8

Dimensionless equations have the advantage that they work for any value of the parameters. They are scale invariant. So the solution in terms of a single dimensionless variable applies to all values of $D$ and $t$. It also allows the definition of characteristic values for the dynamic variables. In your example, one could say $u_0$ = ...

6

Let's say your goal is to describe the shape of some object, such as a box. You could create a completely arbitrary ruler and measure the three axes of the box, coming out for example with lengths of 11.72, 23.44, and 35.16 of your arbitrary ruler units. Or you might look at your results more closely and think hmm, something is going on here, since the ...

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It's the water itself that forms the lens. Lenses work via refraction. The refractive index of water is about 1.333, which is different from the refractive index of air (about 1.0), so rays of light bend at the junction of the air and the water.

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Here's an intentionally more conceptual answer: Entropy is the smoothness of the energy distribution over some given region of space. To make that more precise, you must define the region, the type of energy (or mass-energy) considered sufficiently fluid within that region to be relevant, and the Fourier spectrum and phases of those energy types over that ...

5

In terms of the temperature, the entropy can be defined as $$\Delta S=\int \frac{dQ}{T}\tag{1}$$ which, as you note, is really a change of entropy and not the entropy itself. Thus, we can write (1) as $$S(x,T)-S(x,T_0)=\int\frac{dQ(x,T)}{T}\tag{2}$$ But, we are free to set the zero-point of the entropy to anything we want (so as to make it convenient)1, ...

3

The entropy of a system is the amount of information needed to specify the exact physical state of a system given its incomplete macroscopic specification. So, if a system can be in $\Omega$ possible states with equal probability then the number of bits needed to specify in exactly which one of these $\Omega$ states the system really is in would be ...

3

You can set the entropy of your system under zero temperature to zero in compliance with the statistical definition $S=k_B\ln\Omega$. Then the S under other temperature should be $S=\int_0^T{\frac{dQ}{T}}$.

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Have a look at this chapter THE THERMODYNAMICS OF THE HUMAN BODY AND THE BIOPHYSICAL FEATURES OF THE THERMAL ENERGY In effect the main cooling method for the human body once the outside temperature gets larger than the body temperature is evaporation of sweat which is increased by fanning oneself with make do fans, as the humid air is pushed away and less ...

3

Much of the vapor you see at that stage is unburned material, not a true "smoke" which would be ashes or non-burnable material. A true solid is very difficult to burn. Most fuels instead volatilize as the temperature rises, increasing the surface area. This material coming from the heated fuel appears similar to smoke. A visible flame is the burning of ...

2

This problem has certainly been solved in practice in a number of applications - cryosurgery, cryogenic treatment of metals, and cryogenic cooling of x-ray crystallography samples, but the approach will depend on the application. The flow of liquid nitrogen is always always unstable and fluctuating, because it is usually impossible to prevent small amounts ...

2

First, you have to understand that Rudolf Clausius put together his ideas on entropy in order to account for the losses of energy that was apparent in the practical application of the steam engine. At the time he had no real ability to explain or calculate entropy other than to show how it changed. This is why we are stuck with a lot of theory where we ...

2

In classical thermodynamics only the change of entropy matters, $\Delta S = \int \frac{dQ}{T}$. At what temperature it is put zero is arbitrary. You have the similar situation with potential energy. One has to arbitrarily fix some point where the potential energy is put zero. This is because only differences of potential energy matters in mechanical ...

2

The current entropy in the Universe is all stored in photons. The first reference by Qmechanic gives you the precise value. Since the photons of the CMBR do not at present interact with anything, the entropy of the Universe is very close to being a constant. What evolution there is, is all due to non-reversible processes in baryonic matter, but it amounts to ...

2

One cannot "derive" any non-equilibrium rate law from thermodynamics, simply because they are beyond the scope of the theory. Thermodynamics simply does not deal with such phenomena and hence cannot tell you how such processes occur (in this case heat conduction). All that thermodynamics does is relate mean values of certain properties of systems amongst ...

2

Evaporative cooling mentioned by Anna is the way to go. You could consider making your clothes wet, that saves you from having to sweat. Every liter of water lost in the form of sweat requires you to drink 1.5 liters of water to rehydrate yourself, so you save water this way.

2

As a general rule, physics gets easier when the mathematics gets harder. For example, algebra-based physics comprises a bunch of seemingly unrelated formulae, each and every one of which needs to be memorized separately. Add calculus and wow! Many of those supposedly disparate topics collapse into one. Add mathematics beyond the introductory calculus level ...

2

John Baez has a nice article on his website that goes into some detail on this question. Broadly, your intuition is right that at face value, it looks like structured systems are born out of nearly featureless initial conditions. However, as (for instance) a gas cloud collapses into a galaxy, it heats up (see the virial theorem, a theorem any ...

1

I think the assumption that radiation is required for a collapse in general is mistaken. Think about a cloud of gas. If it is going to gravitationally collapse it must have a negative total energy; if it doesn't parts of the gas will fly off. If it has a negative total energy then there is some finite maximum size for the gas cloud, where it only has ...

1

A higher entropy equilibrium state can be reached from the lower entropy state by an irreversible but purely adiabatic process. The reverse is not true, a lower entropy state can never be reached adiabatically from a higher entropy state. On a purely phenomenological level the entropy difference between two equilibrium states, therefore, tells you how "far" ...

1

Strictly speaking, you're using the relation $a\ln b = \ln(b^a)$ outside of its domain of validity. When $a$ is a quantity with units (and $b$ is dimensionless), it's perfectly valid to write $a\ln b$, but it is not equal to $\ln(b^a)$, because $b^a$ is undefined and so is its logarithm. If you want, for notational convenience you could specify that the ...

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If we assume our universe as an isolated system, then its entropy can only increase. It cannot decrease because of the second law of thermodynamics. It cannot stay unchanged because the universe is undergoing all kinds of irreversible processes.

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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 ...

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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

For a potential dark matter cloud around the Sun to affect the corona you would have to propose a mechanism. The most obvious one would be for dark matter particles to transfer energy to corona particles, but dark matter interacts with ordinary matter far too weakly for this to happen.

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"The gas would slowly radiate its heat through the glass to the ambient container housing the vacuum, and solar panels lining this surface could feasibly collect this energy." No. If we assume the gas inside and the cells outside are both at temperature $T$, then no (thermal) energy can be extracted. They will be in thermal equilibrium. Whatever ...

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

Because it is easier for dimensionless quantities to be combined in arbitrary polynomial terms (or other terms e.g exponential) with no loss (or extra) factors. Think like "characteristic times" used in exponential factors. Especially quantities appearing in solutions of the form $e^{a} = \sum_0^\infty \frac{a^n}{n!}$, one can see why a dimensionless ...

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A free dark matter cloud (without the presence of ordinary matter) will simply not "collapse" the same way a radiating gas cloud does. In both cases total momentum, angular momentum and energy are conserved, but in the case of a gas cloud the photons can carry away some of the angular momentum and most of the energy, in case of a dark matter cloud they ...

1

These solutions are preferred because they directly embody the scale invariance of the equation. In general, when a physical problem has some sort of symmetry - like the parabolic dilation invariance of the heat equation - then this establishes a corresponding action of the symmetry group on the solutions. The canonical forms based on dimensionless ...

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I will not derive this exact formula for you, but instead show you the derivation for a single solid (following Schroeder's 'Introduction to thermal physics', where I learned it). The generalization to two solids is then very simple (mostly a conceptual difference; the math hardly changes). Consider an Einstein solid with $N$ oscillators and $q$ units of ...

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