# Tag Info

1

Ovens in the kitchen are not suppose to heat above 450°F. Any type of glass can handle that temperature, if the change in temperature is not too fast. For safety, I would think that tempered glass is used on the ovens.

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"High" and "low" are relative terms that usually also carry an anthropocentric connotation. What "high" means depends on what humans think of as a large quantity but to thermodynamics the absolute scale does not matter! What matters is only that there is a change from one entropic state to another. As long as there is such a change, no matter how slow, there ...

1

In the early Universe, entropy is preserved (dS=0). This comes out of the equations of general relativity, but it can be also understood by thinking in terms of classical dynamics: the Universe is a closed system, no heat is exchanged when expanding, so its entropy must not variate.

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The low-entropy initial state of the universe is an open problem without a satisfactory answer. Your question is the first time I've heard the suggestion that the initial state should have been a crystal; you remind me that the quark-gluon plasma, which was the state of the universe while it was too hot for nucleons to be stable, has been shown to be a ...

3

Joule's law, and thermodynamics in general, is a model of the classical world. Here, classical should be interpreted as non-quantum-mechanical. Thermodynamics is the study of large collections of particles and their collective behavior. No microscopic model is assumed, and one tries to extract as many (non-trivial) features as possible based on purely ...

2

That's a very hard question to answer with the appropriate level of detail! Very broadly speaking in an ideal metal all atoms are forming a perfectly regular crystal lattice. Conduction band electrons can move freely around these atoms, which makes it easy to pass a current trough the metal. In a (theoretical) metal with perfect crystal lattice the ...

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The answer lies in the fact that, in graphene, there is an effective long range interaction mediated by the inverse biharmonic operator (which in 2D goes as $x^2\ln(x)$ and is extremely long-ranged) coupling the gaussian curvature at any two points on the sheet. Due to this, any static ripples or thermally produced dynamic ripples interact at arbitrary ...

-1

The heat generated by combustion of fossil fuels is slowly building in the environment. Contrary to atmospheric degradation theories. We have gone from a half a million camp fires to burning billions of gallons of oil.

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You have come to the right place, and your question will be answered. But, be warned, you may not like the answer. You can not talk about the heat of anything. No object, system, or molecule can be said to contain this much heat, at all. A physicist states this as the following: Heat is not a state variable. What you can talk about is called internal ...

0

The heat as we deal with in everyday life is due to the MOTION of the ingredient particles of that matter which had gained energy. when you give energy the particles motions get faster which means the particles are more energetic and it get's hotter through our sensing. (then heat as we (possibly) mean here = motion of the particles)

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It's due to the temperature difference in the vapor, caused by turning off the fire as user3823992 said. And u can feel the temperature difference by putting your hand ((or your face if you want to feel it better ;)-don't do it because it may burn your face)) on the flask where the vapor is coming out and feel the enormous difference in the temperature. ...

1

I think you have two questions here, namely What is heat? Is it a fundamental property of stuff? I'll answer both at once. In modern physics we view heat as an average property of a system containing many particles. Roughly speaking, the heat of box is a measure of microscopic energy transfer between the box and its surroundings. More precisely, we use ...

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You might get an order of magnitude estimate as follows. We make the rough assumption that everything ends up in its vessel as a monoatomic ideal gas - actually it will be a plasma, with a thermal energy per mole of $\frac{3}{2}\,R\,T_{final}$, where $T_{final}$ is the thermodynamic temperature of the plasma. Neglecting heats of vaporisation (we assume ...

1

When opening the bottle in space, all the air that was initially in it will flow out due to the pressure difference. The inside of the bottle will then become approximatelly vacuum, so when you open it on Earth air will flow in it again. (Unless it's not sturdy enough (for example a plastic bottle), in which case it will be compressed/crumpelt before you ...

3

First, I want to say that different people use different notation and I welcome any comments. I also feel as if I am about to enter a minefield. Here the answer is made up with examples of use of $d$, $\partial$ and $\delta$. I would say for $d$ that $dV \over dx$ would be the total derivative in one dimension for $V(x)$ where the potential $V$ is a ...

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Typically: $\rm d$ denotes the total derivative (sometimes called the exact differential):$$\frac{{\rm d}}{{\rm d}t}f(x,t)=\frac{\partial f}{\partial t}+\frac{\partial f}{\partial x}\frac{{\rm d}x}{{\rm d}t}$$This is also sometimes denoted via $$\frac{Df}{Dt},\,D_tf$$ $\partial$ represents the partial derivative (derivative of $f(x,y)$ with respect to $x$ ...

-1

Okay, so first: Light can be [is] broken down into a spectrum ranging from shortwave lengths to long wave-lenghs. Visible light (the part of the spectrum we humans can see) contains the colours which are all around us. An objects colour directly relates to which wavelength(s) of light it reflects. White cars are best at refracting light as it reflects ...

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The reason is the same as why a metal pipe feels colder than wooden plank at the same temperature: thermal conduction. The heat from your tongue (including the moisture) is absorbed faster than your body can replenish it. This has the effect of freezing your saliva in the tongue's pores to the metal surface (which itself isn't too smooth at small scales). ...

1

It's a sum over states. The partition function can also be written as \begin{align} Z = \sum_\alpha g_\alpha e^{-\epsilon_\alpha/kT} \end{align} where $\alpha$ is an index which labels levels, $\epsilon_\alpha$ is the energy of level $\alpha$, and $g_\alpha$ is the degeneracy of level $\alpha$. This follows from the sum over states by noting that in that ...

2

The formula is actually better written $$\Delta S = \frac{Q}{T}.$$ That is, the change in entropy associated with the flow of heat is inversely proportional to the temperature at which the heat flow occurs. Note that $Q$ is already a change itself: it is not a state variable, but rather something more like $\Delta W$. Physically, this is because adding ...

1

A formula such as this (being a analog to parallel electrical resistors, as pointed out in the comments) can get a little more complicated when different areas are involved. However, I think that the author of your formula restricted the discussion to one single heat transfer area in order to avoid dealing with that. The heat transfer rate originally starts ...

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The answer depends on many factors, but here are the basic bits of physics that play: The power and wavelength of the laser The reflectivity of the surface (function of wavelength of the laser) The size of the focal spot The thickness of the sheet The thermal conductivity of the sheet The reflectivity of the copper is a particularly important one. If you ...

1

Well it all depends on what assumptions you are able and willing to make. For a start you can work out how much of the laser light is reflected. The reflected power will be something like (for normal incidence) $$\frac{P_{r}}{P_i} = \frac{(\eta_m - \eta_{vac})^{2}}{(\eta_{vac} + \eta_m)^2},$$ where $\eta_{vac}=377$ Ohms and $\eta_m$ is the impedance of the ...

0

There are so many variables here i think its nearly impossible to give an answer like how thick is the lazer, how far away is the lazer how reflective is the surface of the metal ......... the more variables there are the harder something is to predict if you wanted an answer it would be much more practical to get one by doing the experinment, probably not ...

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The specific heat of a molecule depends on the number of degrees of freedom the molecule has. There are several degrees of freedom available: translation (3), rotation (3), vibration (depends on the number of bonds in a molecule) and electronic modes. Now, for something that is monatomic, you have 3 translational modes (x,y,z directions), zero rotation ...

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Since $n=\left(n_1,\,n_2,\ldots\right)$ with each $n_i\in\{0,\,1,\,2,\,\ldots\}$, then you can write the sum as $$\sum_{n_\alpha}=\sum_{n_1\in\{0,\,1,\ldots\}}\sum_{n_2\in\{0,\,1,\ldots\}}\cdots\sum_{n_m\in\{0,\,1,\ldots\}}$$ while your product is, $$\prod_\alpha e^{n_\alpha \beta}=e^{n_1\beta}e^{n_2\beta}\cdots e^{n_m\beta}$$ Thus, we have $$... 1 yes, the collisions between molecules in the gas and or walls in the container can change both the kinetic energy and the direction of motion, like millions of billiard balls moving in a huge pool table. The analogy with the billiard ball is not that good because the billiard balls will eventually end up at rest, as many collisions are inelastic and there ... 2 1-How much energy(in calories) do we need to change the temperature of 1 kilogram of metal A from 0 to 40 Celsius? You simply integrate your function (find the area under the graph). I guess, what you have on the y-axis is c=\frac{Q}{m \Delta T}. Integrating gives you the amount of energy per mass, Q/m. 2-How can we choose a single value (as an ... 4 1 You could find a relation between C and T from the assumptions(quarter of an ellipse) and just substitute that in the integral$$ m\int C dT $$2 We usually say specific heat capacity of metal is some value at particular temperature and pressure say ( Standard Temperature and Pressure ). When we know that the specific heat changes with temperature ... 0 In the case of the atmosphere, we usually assume that the stationary atmosphere is adiabatic, that is, there is very little transferral of heat between regions of different height . Combining hydrostatic equilibrium with the adiabatic law of an ideal gas gives you a variation of temperature (and pressure) in the direction of acceleration. Check out Richard ... 0 A good, semi-technical discussion of the general problem (how the post-Big-Bang evolution of the universe, including the formation of galaxies, stars, etc., can be reconciled with the 2nd Law) can be found here: http://arxiv.org/abs/0907.0659 It's important to realize that while the ensemble of atoms in the gas cloud does indeed, as your intuition suggest, ... 0 The efficiency of engine is defined as ratio of the useful work done to the heat provided: \epsilon=\frac{work done}{heat \space absorbed}=\frac{Q_2-Q_1}{Q_1} Notice that the work done relates to the power delivered at the clutch or at the driveshaft. This means the friction and other losses are subtracted from the work done by thermodynamic expansion. ... 2 I guess you refer to the free expansion of a gas, which is an irreversible process. During free expansion, no work is done by the gas. The gas goes through states of no thermodynamic equilibrium before reaching its final state, which implies that one cannot define thermodynamic parameters as values of the gas as a whole. For example, the pressure changes ... 0 Yes, when we heat an object its mass increases. The complete equation is$$E^2=(pc)^2 + (mc^2)^2 And from this equation if a system has zero momentum (p=0), then it has energy $E=mc^2$ When you heat an object the molecules or atoms begin to vibrate, rotate with more kinetic energy. But this doesnt increase the momentum of the system (the object is a ...

2

All internal energy such as thermal, rotational, and internal potential energy contributes to the rest mass of an object. In fact the vast majority of the mass of an atom is due the internal energy between quarks that make up the nucleus rather than the rest mass of the quarks themselves. So yes, a hot objected has greater rest mass and would weigh more ...

2

Several things are happening here that may make the sensations of touching metal and touching water similar when they are at room temperature (~ 25 C), although the thermal conductivities are a couple of orders of magnitude different. The sensation of coldness comes from the loss of heat from the part of your body contacting the material. The rate of heat ...

1

The parts of your body that generate heat and that can sense temperature and the loss of heat are insulated from the environment by a layer of dead skin cells. The total thermal conductivity to the environment is the thermal conductivity of the materials that you touch in series with the thermal conductivity of this layer of skin. Since this layer has a ...

3

However, my hands seem to feel more comfortable being exposed to winter air than to snow Air (still air) is a better insulator than snow. In fact the insulating properties of snow are due to the fact is has a lot of trapped air within it. The other component of snow, water, is not noted for being a good insulator. The trouble with using air as an ...

2

Firstly, the OP is forgetting that the classic microwave polariser experiment is done with EM radiation in a pure state, not a mixture. We simply have polarised light from, say, a Gunn diode and this pure quantum superposition is forced into a polarisation eigenstate by the polariser. So we begin with near to zero entropy light, absorb some of it (adding ...

0

The anwesr is independent of the surface to volume area ration, unless some conditions mentioned at the end are present. First, unless the syatem is covered, the preasure of teh gas will be always the atmospheric preasure. Second, if it is closed, the preasure will depend on the amount of liquid evaporated and the volume available. The second is fixed by ...

0

Yes, but probably not much. Snow is great as far as natural insulators go, but your hot tub cover is probably a much better insulator. Adding extra insulator on top won't hurt, but the top of the hot tub cover is already probably quite close to ambient air temperature so heat isn't being lost too quickly by convection. If much heat is being lost through ...

4

As always the answer is a simple thing. You calculated the change in entropy using the definition of entropy $$\mathrm{d}S = \frac{\mathrm{d}Q_\mathrm{rev}}{T}$$ Note that this applies to heat transferred reversibly. More generally we must use Clausius theorem $$\mathrm{d}S \ge \frac{\mathrm{d}Q}{T}$$ ...

2

I think the factor you are ignoring is that the polariser will emit thermal radiation. If we continue with the ideal polariser, then it should only emit the polarisation which it absorbs (ideal components are weird). This means that there will still be a component of the absorbed polarisation in the beam after the polariser and so there will always be some ...

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I believe the error is in assuming that the polarized beam is a pure state of zero entropy. If you characterize it in terms of polarization only, then the characterization is not complete. You need a complete set of commuting observables to charactherize a pure state. The macroscopic polarized beam is still compatible with many different quantum microstates ...

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Absolute temperature relates only to translational degrees of freedom (connection to pressure via momentum exchange with a supposed exterior membrane). Since energy is constantly being randomly reshuffled between translational and non-translational degrees of freedom, the molar heat capacity is greater.

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No relation. The second law of thermodynamics can be proven on general statistical grounds. It is independent of what particles are in our universe or what equations describe their motion. That is why physicists are very confident about the second law of thermodynamics, even though nobody knows for sure what particles are in our universe or what equations ...

0

A really helpful discussion of this question is given by Daniel F. Styler, "Insight into entropy," Am. J. Phys. 68 (2000), pp. 1090 - 1096. A (macro)state with high entropy is one that corresponds to many microstates, i.e. the system has many ways that it could configure itself on the microstate level to achieve the given macrostate. Styler gives some nice ...

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Other small effects that you may be ignoring are heat conduction to the air and temperature drop along the pipe. If the fluid is a gas, its temperature may drop by some small amount due to pressure drop along the pipe. The fluid temperature will also drop due to the heat transfer that you calculate. An old reference for these effects is Chapter 8 of ...

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It all boils down to energy and heat capacity. Water has a specific heat of 4.186 J/g degreesC, versus air, which has a specific heat of 1.005 J/g degreesC. To keep a radiator at a temperature designed to heat a room, 70C or more, it would take a multiple amount of air blown through, as not only the specific heat per gram but also the density of air ...

2

Simple answer: When you blow harder, more surrounding air gets mixed in with the stream of air from your mouth. The faster air moves, the lower pressure it has (Bernoulli's principle). So when you blow faster, your stream of air is lower pressure than the surrounding air. Thus the surrounding air fills in the stream. The surrounding air is obviously cooler ...

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