Tag Info

New answers tagged

3

The fact that the apparatus is using outer space as a heat sink or just using the atmosphere should not have any significant impact on radiative cooling properties. Is this reasoning correct? No, it is not. The cooling properties of the apparatus depend on the heat sink to which the heat is aimed, as the final result will be an equilibrium of ...


1

This is the general form in which a conservation law can be expressed, when the quantity being conserved can't be converted in other forms, or disappear and reappear anywhere else. The flux density $\mathbf h$ is defined as that vector field that satisfies the equation you wrote, for every closed surface. More precisely, $\mathbf h \cdot \mathbf n$ ...


0

we feel warm after the bath in the bathroom because as the temperature of the water is cool when compare to temperature of the body.So,that the body cools & the surrounding temperature become hot or warm(humidity).


0

What people don't understand is that the laws of thermodynamics are not exact in the same way as, for example, energy conservation is. They are only quite probable, meaning that for a finite system there always exists a non-zero propability to break them. So, even though it's quite improbable to reach $T= 0 \ \textrm{K}$, in principle it is possible. With ...


-1

article quote: "Hot minus temperatures: At a negative absolute temperature the energy distribution of particles inverts in comparison to a positive temperature. Many particles then have a high energy and few a low one. This corresponds to a temperature which is hotter than one that is infinitely high, where the particles are distributed equally over all ...


0

when u spray through nozzle the water beam get converted into tiny droplet thus increasing the surface area due to increased surface area the heat dissipates to the surrounding air molecules thus lowering the water temperature also if the surrounding air is at higher temperature than of water then the water will get heat up :D i hope it would help you ...


0

It always will help you to go back to the derivation of the heat equation. The heat in a region $D$ at a time $t$ is given by the integral, $$ H(t)=\int_Dc(t)\rho T(x,t)\,dx $$ where we've added your assumption of $c$ being time-dependent. Thus, the change in the heat would be $$ \frac{dH(t)}{dt}=\frac{d}{dt}\int_Dc(t)\rho ...


0

Try to move away from "equations" and think about what you are trying to achieve. You take a certain amount of heat from your mass of steam - some to turn it into liquid, then more to cool it to 30C. This heat should be equal to the heat needed to heat your ice from -10 to 0C, melt it, then heat it to 40C. And since this all happens in a 50 gram copper ...


3

The question of AdS (in)stability is indeed a hot topic in current research of the AdS/CFT correspondence. It is a field that ties together many interesting subjects: Gravity in AdS (i.e a confining box), thermalization in QFTs, the theory of non-linear differential equations and their perturbative treatment, turbulence etc. This explains the explosion of ...


6

How opaque is that -- would we be able to see a couple of meters, some kilometers, or nothing at all? The photosphere of our sun is somewhere on the order of 500 km thick. For a quick ballpark, you can imagine an exponential decrease in the transmission of light which about this characteristic thickness. It might be a little less, but it's still ...


1

The key question is "Per mole or per gram?" Because the both values can be found tabulated as "specific heat" in various sources. Perhaps it would be useful to distinguish "molar specific heat" from "specific heat per unit mass". You seem to be using the intuition for the molar quantity, so if the table is by mass, the answer is simply that you need more ...


2

Well, I can give you a definitive answer to Q1, but my answer to Q2 would only be educated speculation. Perhaps some of the astrophysicists on here can be more help with that one. However, before I tackle Q1, a very important disclaimer: Temperature is a measure of the average kinetic energy of the particles of an object, and cannot be used all by itself ...


0

NOT THE ANSWER BUT MAY COME OF SOME HELP TO YOU : The Big Pot of Soup As part of his summer job at a resturant, Jim learned to cook up a big pot of soup late at night, just before closing time, so that there would be plenty of soup to feed customers the next day. He also found out that, while refrigeration was essential to preserve the soup overnight, the ...


0

You should keep it at atmospheric pressure because of the thermal conductivity. A vacuum is an under pressure or a space that is devoid of matter, therefore it is a very good isolator. The second best isolator is stationary air, so if you want to cool the can as fast as possible you should have a decent airflow in the fridge to abstract the heat from the ...


0

So, in a irreversible process , to calculate the variation of entropy you must subdivide all the variation is some little parts, that could be considered some isotherms and reversible(in this case, in fact, the entropy change is right zero).Obviously this passage isn't so simple, but I want to give you a simple explanation: These little parts tend to zero. ...


0

Fundamentally there's a simple difference: When you are working with a perfect gas at constant volume you can take the variation of inside energy equal to the heat absorbed in the transformation. In this case you must use Cv, obviously. In the other side, where the pressure is constant you can't consider the equality defined previously. In fact you must ...


1

For the record, it might be a partially blown head gasket letting water flow from the coolant channels, into the combustion chambers, and out the exhaust. A lot of water flowing out of the exhaust isn't normal.


0

Entropy is a state function which means that the value of entropy change between 2 states depends only on the initial and final states and not the path/process taken from the initial state to reach the final state. I'm not sure about this part but I think entropy calculations are more easier for reversible processes than irreversible ones.


2

To answer your question first we need to know why do we need quantum mechanics in thermodynamics: In Quantum mechanics you can attribute a wave function(to be precise wave-packet) to a particle. . At high temperatures particles can be pictured as billiard balls because their size is much smaller compared to interparticle distance. But as the gas cools down ...


0

From the engine we recharge the battery using the alternator, Consider the range of an electric vehicle with a fully charged battery. Now, to this vehicle, add a tank of water, an electrolysis system, and a combustion engine that burns the resulting H2 and O2 and drives an alternator. The range of the vehicle is necessarily less than before. Why? ...


0

Conservation of energy screws you every time. No, you cannot get more energy out of a system that you put in. However, there is a huge caveat to that statement, which is when the system already contains energy to be extracted by some minor input. For example, a nuclear power plant. However, we know that with chemical processes it is a zero sum game, which is ...


1

Following is computation of $E[(\sigma_a - E[\sigma_a])(\sigma_b-E[\sigma_b])]$: $E[(\sigma_a - E[\sigma_a])(\sigma_b-E[\sigma_b])]=\\ E[\sigma_a\sigma_b-\sigma_aE[\sigma_b]-E[\sigma_a]\sigma_b+E[\sigma_a]E[\sigma_b]]=\\ E[\sigma_a\sigma_b]-E[\sigma_a]E[\sigma_b]-E[\sigma_a]E[\sigma_b]+E[\sigma_a]E[\sigma_b]=\\ E[\sigma_a\sigma_b]-E[\sigma_a]E[\sigma_b]$ ...


1

The general answer seems to be that everything you know is correct and we cannot have a material with a negative bulk modulus. But we can have materials with a negative incremental bulk modulus. In other words, I can't make a material that will always expand when you try to compress it. But I can make one that will compress for a little while, then expand ...


1

I think this should be roughly correct. If you want to also roughly estimate the difference that internal heat, tides and atmosphere (the Earth's) would have had, you can look at the moon, which is the same distance from Titan but has an average surface temperature of $268K$. So your error is about $20K$ for estimating earth's temperature from Titan's. ...


0

It's good to consider both radiation along with conduction and convection. Evaporation is also in a way convection only as it happens through the medium of heat transfer. In your case , please give maximum attention to convective heat transfer for all practical purposes. If you are working on a practical design problem , then also consider radiation and ...


0

Beryllium Oxide - about 75% the conductivity of Copper Diamond dust More on diamond thermal paste ... and yes, people do sell diamond loaded thermal paste Now, does anyone sell diamond loaded Beryllia paste (ignoring it's toxic properties)?


2

Your teacher is wrong... in a way. What he probably meant was same mass, different temperture.


6

if the pressure is the same, there in no net force on the piston, so it will remain at rest, so neither gas will expand, and this will stay this way because there is no heat exchange that could change the pressure on either side. Ask your teacher to return his diploma.


1

The best vacuum we can make is about 10^-12 Pa. Atmospheric pressure is about 10^5 Pa, so it's 17 orders of magnitude lower pressure. The best vacuum recorded is the intergalactic void, at about 10^-17 Pa. Even if you managed to remove all matter, there would still be energy from any light or electric fields.


1

It doesn't. They'd have to exchange heat through the piston.


1

You can consider "average kinetic energy" equivalent to "temperature". The question doesn't directly mention if the contents of the flask are all at the same temperature, but if they are at the same temperature then they all have the same average kinetic energy. Considering that water has a triple point at 0 degrees Celsius and a certain pressure, I think ...


2

The best performance (from a theoretical standpoint) would be to use diamond dust as a filler with something like silicone as a binder. In principle you can get about 5 times better thermal performance than you can with silver. Of course, you need the particle size as small as possible, which is harder to do with diamond than with silver (you can't ...


0

I agree with the last part of David Hammen's answer that the discrimination between classical thermodynamics and non-equilibrium thermodynamics is sometimes a bit arbitrary and essentially a cultural trend. There is in fact nothing conceptually new in the quantities introduced: temperatures are still temperatures and so are pressures, densities and so on and ...


0

Thanks for the comments. I understand it is an ugly problem, however I wanted to find a way to do it computationally somehow. I solved it by using the Newton-Rhapson method of finding the roots, basically what Mathematica already did on the background. I have now an iterative relation to calculate $p_1$ and $p_2$: $p^{n+1}_1 = f(p^n_1, p^n_2)$ $p^{n+1}_2 ...


-1

human being is heat engine as chemical reactions take place and some of energy is utilized in working of body rest is released in atmosphere. when temperature of sourrounding is higher then human body still it releases heat in form of sweting.


1

The formula you cite refers to the work done when the change of volume is made at constant pressure, which is not the case here. Let's say I start with the balloon in the air with volume $V_0$ at pressure $p_0$. The state I want to end up with is the balloon underwater at a depth with pressure $p_1$ and the balloon having volume $V_1$. I cannot directly ...


0

I think I will go with the version from Landau and Lifshitz with quantisation of phase space: $$ N = \int f \ln \frac{e}{f} \, \frac{dr^3 \,dp^3}{(2 \pi \hbar)^3}, \quad N = \int f \, \frac{dr^3 \,dp^3}{(2 \pi \hbar)^3}. $$ With the equilibrium function of $$ f_{\text{eq}} = (2 \pi \hbar)^3 \, \frac{N}{V} \Bigl(\frac{1}{2 \pi m k_B ...


0

It is impossible to transfer heat from a colder body to a hotter body without any other effect... in the surrounding bodies. Every reservoir is a body too.


2

The convergence of the iteration you are using becomes very slow below 93F. I might be inclined to switch to a binary chop instead. Something like: Const pc = 33.5 Const tc = 126.2 Function n2z(p as Double, t as Double) as Double Dim zstart As Double Dim zend As Double Dim z As Double zstart = 1 zend = 2 Do z = (zstart + zend) / 2 If ...


6

Heat will always travel from the hot place to the colder region. So if the air outside is colder than the air inside the box, even insulated walls will let the heat travel outwards and lower the temperature of the box. You will have to keep supplying heat into the box if you want to keep the air and wall temperature constant. There is no such thing as ...


-3

While the above is true, the atomic nuclei of a composite gas do not line up neatly in nature such that O2 and N2 are neutral in terms of "trapping" heat energy. In fact the physics of light refraction or bending, scattering and altering wavelength in that process still takes place in our atmosphere the GHG value of Strictly N2 and O2 combines for ...


1

Assuming you have a liquid water reservoir available and the liquid is heated along with the gas, water will evaporate to keep the partial pressure of water equal to the vapor pressure of water at the current temperature. As the total gas mix is maintained at a given pressure, the rest of the gases must expand more than Charles' law would indicate, reducing ...


1

Under atmospheric conditions , the most important heat loss mechanism is convection as this involves the physical medium , which is air. That is the most abundant element in a room atmosphere. Conduction needs to be considered if the physical solid material area is significant. Radiation in most cases is not very relevant in local conditions. It is however ...


0

Since it is being heated by an immersion heater,so convection process is must along with radiation which increases with the increase of temp.but since surface area is not so large,evaporation process can be neglected, i think. Let me know what you think.


0

JonT provided a nice answer, but I'd also like to add that heat is always generated whenever wind is. When turbulence is generated (and there is always some level of turbulence generated when one generates wind), the energy in each eddy cascades down to smaller eddies (with very little energy lost) and eventually the eddies reach a size where viscous forces ...


5

I will address this part of the question: Why exactly does a cavity with a hole behave like a black body? Why can I make such a contraption and simulate and skip considerations of a special chemical makeup that has to occur in a blackbody? One starts with a black body, absorbing all radiation and emitting it in thermodynamic equilibrium with the ...


1

As you know from thermodynamics, heat cannot flow one one body to a hotter body. Imagine a body with a cavity that is not a perfect black body emitter. Let's put it face to face with a second cavity - but this one is a perfect black body. Initially they are the same temperature. Radiation will flow from the perfect black body to the imperfect one - but a ...


3

To answer a question that the OP put to Irish Physics's answer: What about the fact that radiation isn't really absorbed in the box, and instead is really just bouncing all over the interior? Why doesn't this matter? Actually you need the radiation to be absorbed by the box and re-emitted, or at least it must interact with the box material in some ...


3

The whole point of a black body is to make a "box" that shows, when you keep its temperature as constant as possible at some value, say 300 K, how much the intensity of the radiation emitted through a tiny little hole in the box changes with respect to frequency. So the chemical composition of the box does not really matter as long as it does nothing to ...


0

This is simply the boiling of water, with the high rate of volumetric expansion producing a vibration in the air, thus producing sound. You can see that the sound is most evident when the nickel ball is in direct contact with the water, as the thermal conduction is greatest and thus water boils fastest when in direct contact. There is also a period of time ...



Top 50 recent answers are included