# Tag Info

1

Although this isn't obvious, the system doesn't return to its initial state. If you were to very slowly remove the weight from the piston, then the gas would do work on the piston as you removed it, which means that its internal energy would be reduced. If you remove the weight very quickly then the gas still does work on it, but it will do less work than it ...

1

If you want to separate hot and cold gas streams from a common stream, use a Ranque-Hilsch vortex tube,

1

From http://chemicals.etacude.com/z/zinc_chloride.php it would seem that the enthalpy of formation of zinc chloride is -415 kJ/mol. It has a melting point of 290C and a boiling point of 732C. Since it seems happy to boil (and make a smoke screen), you would need to go above (given the formation energy well above) 732C to get it to dissociate by itself on a ...

2

The key point I'm getting at is that when the pressurized liquid moves through the throttling valve, the auto-refrigeration effect is really a way of splitting the hot vapor "part" away from the cold liquid "part". I think this is the main misconception you have. Typically when a material boils, the gas that is released is at roughly the same ...

1

Instead: $$\frac{dQ}{dt} = hA(T -T_{amb})$$ and $$Q = c_pmT$$ where A is the surface area of the interface of the two temeperatures, h is the heat transfer coefficient and $c_p$ is the per-mass specific heat capacity.

1

Addressing parts of the question out of order: ...the heat capacity would smoothly approach zero around the transition. I have never seen anyone refer to these types of transition... The heat capacity of all substances smoothly approaches zero at absolute zero. $S(E)$ has a segment of zero first derivative no, the first derivative is a constant, ...

1

If I put the weight back on the piston, the system will again achieve its initial state. No, it won't. In the end, the pressure will be the same, but the temperature and therefore the volume will be higher. Firstly, in a real system there will be friction due to gas viscosity and piston/cylinder interaction. But even in an ideal system, after the ...

1

I am going to address the question as to why energy and information have time symmetric conservation properties whereas entropy does not. According to the Wikipedia entry on entropy - "The entropy of an isolated system never decreases, because isolated systems spontaneously evolve towards thermodynamic equilibrium, which is the state of maximum entropy." ...

0

The way I always thought about this was to pick one of the variables to be thought of as the dependent variable. Here I will pick $z$. Then we think of $z(x,y)$ to be a function which has partial derivatives $\partial_x z = \frac{\partial z}{\partial x}$ and $\partial_y z = \frac{\partial z}{\partial y}$. Now we must compute $$\left( \frac{\partial ... 1 x=x(y,z), y=y(x,z), z=(x,y)$$dx= (\frac{\partial x}{\partial y})_z dy + (\frac{\partial x}{\partial z})_y dzdy= (\frac{\partial y}{\partial x})_z dx + (\frac{\partial y}{\partial z})_x dz\therefore dx= (\frac{\partial x}{\partial y})_z [(\frac{\partial y}{\partial x})_z dx + (\frac{\partial y}{\partial z})_x dz] + (\frac{\partial ...

1

It depends on how you define your system or your control volume. If only the container is considered then indeed the entropy has decreased due to cooling. On the other hand if you account for the container plus the escaped vapour the entropy has increased, as the randomness of the molecules in the vapour state is larger than compared to in liquid state at ...

2

No, in fact you could even view the spontaneous evaporation as being driven by the fact that it increases entropy. Basically what's happening is the liquid particles have random speeds (with distribution characterized by temperature), and they bump into each other. Every once in a while, two particles near the interface will collide in just such a way that ...

1

Temperature is a macroscopic concept, so you're bound to run into some problems when you apply it on a molecular scale (what does temperature and equilibrium (or for that matter, friction) even mean on such a small level?). A thermal equilibrium does not mean all the molecules have the same energy. The distribution of their energies looks like this (normal ...

0

Is it only for an ideal gas that the work equals negative the heat in an isothermal process or is this general for all kinds of systems? In an isothermal process, $T$ is unchanged. In an ideal gas, $U$ only depends on temperature, so $\delta U=0$. Since $\delta U=\delta W+\delta Q$, you have $\delta W=-\delta Q$. This is not necessarily valid for ...

0

No it definitely depends on the Equation of state describing the system. You may check the third post in the link for a mathematical derivation below. Internal energy according to the van der Waals equation

5

The noise is either from the AC electricity, which would be a 60Hz buzzing, or from small bubbles forming on the heating element itself. When the electricity stops, both the buzzing and the bubble formation will stop as well. Bubbles create sound due to quickly expanding from a small nucleus. Here's a book I found with a section on noise from bubble ...

0

Well, those electrons tend to run into things, both other electrons and the atoms, a process called scattering. Scattering off the ions can generate phonons (lattice vibrations), and that transfers energy from the electrons to the lattice. Similarly, phonons can scatter off electrons. In thermal equilibrium the rates are the same forward and backward. If ...

0

in the end whater and milk and all containers will be at the same temperature, it's just 10 mins wasn't enough. the heat exchange is proportional to the temperature difference. so when milk is very hot the heat exchange from milk to water is fast, and it slows down as milk cools down and water heats up. they can meet somewhere in the middle at temp higher ...

1

As DavePhD has explained, the water would never be warmer than the metal. As you waited a long time, you can safely assume they are at the same temperature. But, why do they feel different? Your hands are warm, and when touching a thermal insulator, such as wood, they will quickly transfer energy and heat up the surface of what you are touching, so it feels ...

0

"I never understood is water continuously absorbing heat from the milk glass even when the water itself gone warmer? I thought they should ideally be at the same temp at the end.": Yes, the water will continuously absorb heat until there is no temperature difference. However, the rate of heat transfer will continously decrease. The rate of heat transfer ...

0

The definition of entropy is $$S = -k \log(\Omega),$$ where Omega is roughly the number of microstates (ways of ordering your particles) compatible with the macrostate (what you observe macroscopically). Intuitively, you can say that, if you have particles inside a box, and you increase the size of the box, you can arrange them in more ways; therefore, the ...

1

Exactly as you said: "(5) seems to say that the entropy change for an irreversible process is 0 for the case of the system itself, but is > 0 for the case of the surroundings." After the cycle is finished, by definition of the "cycle", the system returns to its original state irrespective of all the irreversibilities that may have taken place within it ...

1

I read somewhere that if the reservoir is infinite then any heat added is actually occurring infinitely slowly. Is this true? No, infinite reservoir does not imply infinitely slow transfer of heat. There is no direct relation. Equation (4) is incorrect; if the integral is over closed path in the state space of the reservoir, its value is 0. If the ...

0

There are many ways it's heated if you have not bathed hot. 1>Your body Heat. 2>Loss of Kinetic energy of falling water molecules 3>When a molecules drops into many smaller ones.. there is heat due to surface tension and surface energy 4>if you bath cold then you have lost enough heat to feel your surroundings warm. Amount of heat in the bathroom before ...

1

This means that you assume the energy of the initial state is approximately conserved and the initial state goes over smoothly into the state from which it originates when first adding the perturbation to your theory. Adiabatic here means "without changing the energy".

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In general it means varying (or turning off an interaction in your particular question) a parameter on a time scale that is much larger than the smallest energy separation of your Hamiltonian. More explicitly : Suppose you have a Hamiltonian $H$ with energy levels $E_n$ and suppose that $\left|E_a-E_b\right|$ ($a\neq b$) is your smallest level splitting ...

1

in welding a plasma is created, it's a mix of ions, electrons and atoms. alltogether they are a neutral mix. once you get plasma you get a ton of UV coming out of it, very dangerous to eyes not only on the direct contact, but also via reflection from other objects. in your case, I still don't know what exactly is the device you are creating. It sounds like ...

2

All materials emit thermal radiation (such as light). The hotter the material, the more the radiation is shifted to high frequencies (shorter wavelengths). The radiation comes from oscillating electrons (regardless of whether there is an electric current). Welding reaches temperatures high enough to cause significant emission of UV light. Oxyacetylene and ...

0

If the Reynolds number of laminar flow exceeds about 4000, it will go turbulent. Turbulent flow will choke mass transfer, and worse if it a supersonic shock. Rate of heat transfer through source and sink walls is another limit. And if you really juice it (accelerator)...the beta factor in special relativity. That last can be real world - ion engines in ...

0

There are basically three limitting factors in the system you set up. Both evaporation as well as condensation require in practice som sort of liquid films, which provide thermal resistance between the walls and the vapour phase and therefore temperature gradients are required to allow for heat transfer. This means that besides geometry and fluid dynamic ...

0

I've never heard of subcooled gas having a temperature below its saturation temperature. There are plenty of instances of subcooled gases. Cloud nucleation is a prominent example. Cloud chambers for radioactive particle tracking also rely on subcooled gases (or supersaturated solutions of a liquid dissolved in a gas, which is probably a slightly more ...

0

How does heat pass through glass? It is not heat that passes through the glass,but electromagnetic radiation in the infrared range, in the same way as the visible light passes. Glass is transparent to visible light and to most of the spectrum before and after visible light frequencies , depending on the optical properties of the specific glass. So ...

1

Be careful not to become confused between two distinct but somewhat similar concepts: thermal conductivity and specific heat. Conductivity is the ability of a material to transfer heat. That is, if you had a long tube made of a given material, conductivity explains the rate at which heat would flow from one end to the other. Specific heat, on the other ...

0

Yep, borrowed from thermodynamics. For a system that is constrained under fixed pressure, fixed composition, and fixed temperature, the Gibbs free energy is minimized. You're right to question the link between Gibbs free energy and potential energy (or kinetic energy or total energy ...). In fact we know that Gibbs free energy is defined by $$G = E - ST + ... 0 Natural circulation in tubes is driven by pressure gradient between warmer and colder parts of the fluid. The pressure gradient results from density gradient, which is induced thermally by heating/cooling the tube at different points. However, tube dimensions You've set, make it a very complex problem, probably impossible to solve. Normally models have to be ... 1 temperature is the measure of speed of molecules. when you compress the gas molecules start moving faster, which is the same as saying the temperature increases. why do molecules start moving faster? there are many ways of explaining this. here's one. when molecules are squeezed into a smaller volume their location is now more certain, it's locked in a ... 0 You did everything correctly. For the same compression ratio, the Brayton cycle efficiency is equal to Otto cycle efficiency. You're right that the cycles are different, however both start with isentropic compression, and efficiency of both in ideal case can be expressed as \eta_{th}=1-\frac{T_1}{T_2}, so only compression process matters. That's the case ... 0 I think it's not possible to relax the boundary condition where \frac{\partial T}{\partial r}|_{r=a} = 0 , because the temperature function is defined for a\leq r\leq b and 0\leq z\leq L. If you don't consider the boundary condition, you are eliminating the insulation on the inner surface of the cylinder, whereas the insulation on the inner side is ... 0 you have an open system, i.e. the heat is removed from gas through its walls. when the gas cools down, it shrinks, so the piston will squeeze into the chamber. since you still apply the force, gas will heat up, and heat will be removed through walls again. while gas shrinks more it'll be more and more difficult to remove heat from walls, because of finite ... 0 To increase the energy and temperature of the gas, you need a force and a displacement of the agent of the force, in this case the piston (\mathrm{d}E = - p\,\mathrm{d}V) A constant force with no displacement will not increase the energy and temperature of the gas. Any force, constant or not, with a displacement will increase the energy. The increase ... 0 There are physical and physiological aspects of this problem. Consider this: if your fingers are freezing in cold is it better to drink a hot tea or cold vodka? Definitely VODKA! Why? because it will immediately cause your blood vessels to widen, more blood will go to your fingers and you'll have a better chance to save them. Did I make this up? No. My ... 2 This article has some relevant results based on a study of bird plumage (it also happens to be cited in the abstract of the Nature paper mentioned in one of the other answers), and is summarized in simpler terms here. I'll attempt to summarize the summary. Black and fluffy/loose fitting clothing is best if it is hot out and there is any (>3 ... 1 Here's an semi-formal explanation. Define$$f(N,V,T)=\frac{p(N,V,T)}{kT}.$$While f(N,V,T) is a function of N,T and V, the variables N and V are partially redundant, and only the ratio \rho=\frac{N}{V} is needed, since pressure is an intensive quantity. Thus we can write$$f(\rho,T)=\frac{p(\rho,T)}{kT}.$$Every smooth multivariate function ... 1 In Landau's theory, the order parameter M should make G(M) minimal.$$\frac{\partial G}{\partial M} = 2 B(T) M + 4 C(T) M^3=0\frac{\partial^2 G}{\partial M^2} = 2 B(T) + 12 C(T) M^2 > 0$$Hence, M=0 or M = \pm M_0 = \pm \sqrt{ - \frac{B(T)}{2 C(T)}} When T is higher than the critical point T_c, the groud state of system satisfies M=0. ... 0 I'm not sure, what do You mean when You say "adiabatic system" - there are no adiabatic processes in real life, but in ideal case it is assumed that during adiabatic process, there is no heat exchange with the surroundings. If You mean that the heat can be added to the system, but there are no heat losses from the system and the heat source has a constant ... 0 Heat transfer by conduction or convection can only take place if there is a temperature difference between two bodies/air etc. If you have a stack of servers in a rack, each at the same temperature, no heat transfer will occur between them so you could consider them a single thermal mass, adding together their individual heat outputs. ... 0 Let me try to attempt to clarify in some direction that I feel comfortable. From usual mechanics, you know the definition of work is given by :$$ W = \int \bf{F}.\bf{ds} $$for simplicity considering the work done to be in the same direction as displacement$$ W = \int {F}{ds} $$Now this is just one version of defining work. In generalising, we redefine ... 0 IR emission and absorption for major Greenhouse Effect gases. Note that adding more absorbance to as saturated line (A greater than ~2, less than 1% tranmission) increases absorbance by log[concentration] not Beer's law [concentration]. ALWAYS look at the wavelength scale and the tranasmission or absorbance scale. The open transmisison windows are ... 0 It may just be the wording of your text that provides the confusion. Work is the same in thermodynamics as it is in mechanics. Work is an energy transfer process. In thermodynamics this often equates to energy in the forms of pressure, temperature, volume, etc. that is transformed into energy associated with position or movement (i.e. kinetic and potential ... 0 The following calculation gives the correct answer:$$Z\int_0^{\pi/2}\int_0^\infty 2\pi v \sin\theta\; v\; \mathrm{d}\theta\mathrm{d}v\; e^{-mv^2/2kT}\; v \cos\theta\; \frac{1}{2}mv^2,$$where Z is such that$$Z\int_0^{\pi}\int_0^\infty 2\pi v \sin\theta\; v\; \mathrm{d}\theta\mathrm{d}v\; e^{-mv^2/2kT} = n, where $n$ is the particle number density. The ...

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