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3

The showers I have used that have two knobs have something akin to a globe valve for each line As you unscrew the handle, there is more opening between the plug and the body, letting more water through. The resistance of valves like this is quite variable when the plug is near the body. Once the plug is withdrawn a certain amount the resistance is small. ...


1

For (1): the energy required to vaporise a liquid is known as the enthalpy of vaporisation, or alternatively as the latent heat of vaporisation. For water at room temperature this is about 44kJ per mole, and a mole of water is 18g. For (2): the density of air at room temperature and pressure is about 1.2 kg per cubic metre, so 1kg of air is 1/1.2 or about ...


1

Christoph gives a good answer from the point of view of statistical physics, but the first context we actually encounter entropy is in classical thermodynamics. It is useful even without the interpretation of being "the measure of disorder". This got slightly out of hand and is now quite technical, but I think it is still digestible and provides a complete ...


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What is entropy, more than disorder? Don't look at entropy as disorder. Thinking of it as disorder has long been a source of confusion. Many texts are moving away from using the disorder description. Macroscopically, it's better to think of entropy as a measure of energy dispersion rather than as a measure of disorder. Microscopically, it's better to think ...


3

What is entropy, more than disorder. Mathematically, entropy is just a measure of spread of a probability distribution: The lower the entropy, the more spiked the distribution. In statistical mechanics, a state is generally only partially defined via some macroscopic constraints, and entropy is a measure of microscopic indeterminacy. Maximizing entropy ...


0

To solve for the steady-state temperature everywhere, you need to solve the heat equation: $$\frac{\partial T}{\partial t} - \chi \nabla^2 T = \frac{S}{c_p \rho}$$ Here, $\chi$ is the thermal diffusivity of the sphere in question, which is defined as the ratio $\kappa / c_p \rho$ (ratio of thermal conductivity to product of specific heat capacity and ...


1

When studying dynamical systems you consider a low-dimensional phase space that only includes macroscopic variables. In this phase space dissipation does indeed cause volume contraction - but only because the energy that has been lost has been transferred to the far bigger phase space including all the microscopic variables. In this space energy must be ...


1

For $N$ particles that can occupy $M$ different states $\left(M\geq N\right)$, it seems to me that the answer is $$M\cdot M-1 \cdot \dots \cdot M-(N-1)=M!/(M-N)!\ $$ if one imposes that each must occupy a different state. That is since, to the first particle, $M$ states are available. For the second, $M-1$. We continue thus until we run out of particles, ...


0

The characteristic modes can be excited by photons whose energy matches these modes. However, heating mechanisms are dependent on several other parameters, not the least of which is the surface shape. Take a look at the literature on black-body cavities and on ultra-dark roughened surfaces for some info on how to maximize radiative absorption.


0

The Stefan-Boltzmann law for net power radiated pertains to the object. That is, we're simply asking, how much radiation leaves this object (this depends on the object's emissivity), and how much radiation is absorbed by this object (this depends on the objects absorptivity). The emissivity and absorptivity in the equation you present thus pertain to the ...


1

It is because of convection effect. Convection is fluid (in our scenario) movement driven by temperature difference. In this case grater temperature difference at the beginning - grater speed of fluid at the end. Fluid is cooling down near dish walls. This causes cooled water to sink faster in this part of dish. Picture shows water movement in dish in ...


3

Actually, in your rotating torus your are mimicking gravity, which is point outward. I.e. the outside of the torus acts as a floor. The centrifugal acceleration would be $g\approx\omega^2 R$. This $g$ plays the same role as the gravitational acceleration on liquid or gas pressures under normal gravity conditions, so you can say $\Delta p = \rho g h$, or in ...


0

Convection is very hard to model. It most likely changes with increasing temperature difference. (With very a low temperature differences there maybe no convection.) The best link I've found is here, http://www.thermopedia.com/content/660/?tid=110&sn=7 (or go to Thermopedia and look under convection.)


2

The equation you state is a very general expression related to heat transfer, and basically everything goes into that constant. Convection of course is one thing, but what about radiative cooling (often important), diffusive cooling (might be important), and heat resistance, since the temperature of your object is not uniform. All these contributions can be ...


1

For simple conductive heat transfer, h is $\kappa$, the thermal conductivity, divided by the length over which the temperature gradient exists. You can look this up for a given material. For convective heat transfer, this constant will depend on the details of your problem, including the dynamics of the liquid in question (can't simply look it up, you'd ...


2

PhotonicBoom is correct in saying that the airflow created by blowing across the top of the coffee will replace the coffee-heated air with cooler air that will absorb more heat from the coffee. It also allows more of the coffee to evaporate (which might seem like a bad thing, but evaporation is simply the hottest molecules becoming gaseous and leaving, so it ...


-2

I have the similar doubt but I think like this. See consider a cylinder with piston fixed, (i.e. it doesn't move) and the system is provided with a source with temperature $T$. Now as the piston does not move the volume is constant, so no work is done and internal energy is also constant and no heat is added since system and source are at the same ...


0

No, because $F$, in fact, has a statistical meaning, $F$, or rather $e^{- \frac{F}{kT}}$, is a sum along different possible energies configurations : $Z = e^{- \frac{F}{kT}} = \sum_i e^{- \frac{E_i}{kT}}$ So you cannot compare with one-particle (non statistical) physical quantities (for instance the energy of the particle would have only one value). ...


1

In principle, the gravitational potential energy should be included into total internal energy, but in practice, most often it is not. I know of two reasons. because for systems that are discussed in thermodynamics, it is believed that gravitational energy is negligible compared to electromagnetic potential energy of the constituting particles; because it ...


1

Your separation into potential energy of the system as a whole due to external force fields and energy contained within the system known as internal energy seems a bit arbitrary. Still, if you want to split the PE up this way gravitational interactions within the system would have to go into internal energy. Take the Solar System as an example. Everything ...


1

The gravitational potential for particle 1 is $V_1(r_{12}) = -Gm_2/r_{12}$ and for particle 2 it is $V_2(r_{12}) = -Gm_1/r_{12}$. $m_2$ is in $V_1$ and vice versa because the gravitational potential energy is the potential energy in a gravitational field per unit mass, and therefore only depends on the mass that is generating that gravitational field. When ...


0

This is a good question, but the answer is that the energy equations of thermodynamics and dynamics now cannot correspond to one by one, completely, in that the theoretical structures of the two theories are different. Since the internal energy \begin{align}U=TS+Yx+\sum_j\mu_jN_j-pV.\end{align} Helmholtz free energy ...


1

This is a very difficult problem. I try to explain why. If we want to obtain the mathematical proof of the second law, we must consider the mathematical proof of the entropy first, as a state function. Clausius’ definition $dS=δQ/T$ cannot be proven in mathematics, as an exact differential, so the definition $dS=δQ/T$ must depend on imaginary reversible ...


4

When a cup of coffee is hot, the air molecules directly above it get hot as well. After some time, they reach equilibrium and no heat transfer (or maybe very little transfer) occurs. By blowing, you disturb that equilibrium and replace the hot air molecules directly above the cup with colder air and therefore create once again a steeper temperature gradient. ...


0

Thermodynamic free energy is a subset of potential energy. There are at least two types of thermodynamic free energy - Gibbs free energy and Helmholtz free energy. Both describe the potential energy of a system under certain conditions. For Gibbs free energy, the conditions are constant temperature and pressure - useful for biochemists - because the two are ...


0

There is a new statement on the second law: "irreversibility root in a fundamental principle: the gradients of the four thermodynamic forces spontaneously tend to zero". Please see http://arxiv.org/abs/1201.4284


2

No. You wouldn't say that pair of beams has a temperature. Temperature is defined by the zeroth law of thermodynamics, which states that if $A$ is in thermal equilibrium with $B$ and $B$ is in thermal equilibrium with $C$ then $A$ is in thermal equilibrium with $C$ and $A$, $B$ and $C$ are said to have the same temperature. Temperature is fundamentally a ...


0

You can use a Boltzmann distribution at any point in time to describe the density of electrons as a function of position (or potential) only if there are e-e collisions. These collisions thermalize the distribution of electrons and allow you to define a temperature. The distribution you start off with (I assume this is what you mean by "initialization") ...


4

To start with, "water freezes faster when it starts out hot" is not terribly precise. There are lots of different experiments you could try, over a huge range of initial conditions, that could all give different results. Wikipedia quotes an article Hot Water Can Freeze Faster Than Cold by Jeng which reviews approaches to the problem up to 2006 and proposes a ...


4

This happens due to cooling affect of evapourisation. As you must be knowing, the temperature of the lquid is a factor of evapourisation. So as the temperature of hot water is more, the rate of evapourisation is also more. Now this is where thwe cooling effect of evapourisation takes place. As the water evapourates, it takes away some heat thus cooling ...


0

It is a good question and your confusion is genuine, especially in view of the fact that you are trying to understand the meaning of the statements given in Wikipedia, that need a lot of elaboration. But simple tings first. If you try to apply the concepts/equations of the first and second law of thermodynamics to closed system containing gas governed by a ...


3

I believe it was Boltzmann who first made the connection between entropy and micro states. chapter 12 of "Classical and Statistical Thermodynamics" by Ashley H. Carter discusses Boltzmann's arguments. To summarize from that book: Entropy ($S$) corresponds to a particular configuration of an ensemble of particles called a macro state. A macro state can be ...


1

I think part of the perceptual difference has to do with humidity. The human body doesn't really feel the temperature we read off of our thermometers. The thermodynamics of the human body are complicated, but people have designed various scales that are supposed to measure "apparent temperature". One of these is the Canadian humidex, which is a ...


4

As everyone else is saying, if you assume Newton's law of cooling: $$ \dot Q = m c_p \dot T = h A \Delta T $$ The equation for how you heat or cool is an exponential $$ T(t) = T_\infty + \Delta T e^{ -\frac{hA}{mc_p} t } $$ The rate constant for growth (or dying) of temperature is the same (assuming other details of the material don't change much), so ...


0

A simple approach is to use Newton's Law of heat transfer: $Q = k\Delta T$ for the rate of heat transfer. Assuming the conduction, convection, and radiation aren't all that much different (same $k$) between sitting in the fridge and sitting on the counter, in one minute, the food will absorb or release an amount of heat proportional to the difference in ...


1

Not always. The rate of heat transfer from one body to another depends on the difference in temperature between the two bodies and many other factors. Higher the temperature difference, faster is the rate of heat flow. So, when the object is brought out of the refrigerator, it will depend on the temperature difference between the object and air and when kept ...


2

I would use a very simple model, and assume that the item has only one temperature. Also i don't think that this will change the result, the real thing is more complicated. As far as i know, the thermal energy flux is dependent on the thermal energy difference. So the time-dependent solution is something like an exponential function. That means that the ...


0

I'm not sure what our question is exactly, so I'll try to answer what I guess your question is. In general, the entropy of a fluid is a function of both $V$ and $T$. During an isentropic compression, the decrease in entropy from the reduction of volume is compensated by an increase due to the temperature rise. The net effect is zero. For an ideal gas we ...


1

By definition, $dS =\frac{dQ}{T}$, so an adiabatic process doesn't change entropy. But you can find more details at http://en.wikipedia.org/wiki/Adiabatic


0

If you think it again, it is not that difficult to understand. First, assume that S and the reservoir are an isolated system. In any such system, it is an intuitive assumption (axiom, due by symmetry) that the system will spend the same amount of time on each of these states (the total energy is a constant, E). So if you know that S is in a specific ...


2

The relation between entropy and information is well established; indeed, Shannon entropy is the seminal measure of the information in a system. The other question, about determinacy and information, is more complex, and even more complex yet when extended to the entire universe. Let us set aside, for now, the fact that quantum mechanics would seem to ...


2

You'll want a much bigger heatsink!! (and maybe just one TEC) If it's being cooled only by convection then maybe a heat sink area* that is 10 times that of the TEC. (maybe bigger) The classic mistake with a TEC is to make the heat sink too small. With too small a heatsink the hot side of the TEC gets hotter, more thermal leakage through the TEC, it has ...


2

If efficiency is the issue, then definitely parallel TECs (or use a single unit rated for twice the power, same thing). The only reason for stacking TECs is to get a lower temperature. However that comes at great expense to efficiency and overall power consumption. Another point is that paralleling TECs is actually more efficient overall. The reason is ...


1

It is best to always stick to the rigorous formulation of the second law, which comes in two parts. Quote from the book "Fundamentals of Statistical and Thermal Physics "by F. Reif: 1) In any process in which a thermally isolated system goes from one macrostate to another, the entropy tends to increase, i.e., $$\Delta S\geq 0$$ 2) If the system is not ...


0

The material contained in the given link suffers from mistakes such as: For a process carried out at constant temperature and pressure, the Gibbs free energy change is equal to the maximum amount of work (wmax) that can be done by the process. $G° = –nF$. Coming to answer your questions, the second equality in your equations: δQ−δW = δQ−(pdV + other ...


0

Start with an ideal gas, $$ V=\frac{nRT}p $$ Then take the natural logarithm of this: $$ \ln V=\ln T+\ln\frac{nR}{p} $$ The derivative of both sides with respect to $T$ gives $$ \frac{d\ln V}{dT}=\frac{d\ln T}{dT}+\frac{d\ln\frac{nR}{p}}{dT}=\frac{d\ln T}{dT}=\frac{1}{T} $$ where we assume an isobaric situation so that $p$ contributes nothing. The left hand ...


1

A question like this appeared recently: Is it possible to have a perfectly black material? Which I ANSWERED: We can make it absorb a lot of energy but if you read about black-body radiation effect you will notice that as energy is introduced into the object it will similarly radiate a small amount of the energy back, at room temperature appears ...


1

This material absorbs visible light, but ideal black body absorbs the whole spectrum regardless of frequency (including gamma rays). There are better approximations of black body than this material, like the stars. Black holes are probably the closest black body in the universe. In laboratories we are using things like this to approximate black bodies.


0

As I am new to Stack Exchange, I happened to see the question only now and mine is a belated answer. If Mark is still interested in an answer, then: It is a good question gaining added strength with the EDIT. For thermodynamic analysys, a process must connect two equilibrium states A, B of a system. Take a state C of the system on the way from A to B. We ...


0

Thermodynamics has simple answers to offer regarding reversibility or otherwise of a given process. For a process to be reversible, it must be reversible at every point along the path ie, the system must be in equilibrium, both internally having well defined values for its properties such as temperature, pressure , internal energy etc., and externally with ...



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