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17

If the particles are not point-like, they will take up some volume. As the gas is compressed, the collision frequency will rise more quickly, which will make the pressure-volume curve change. The corrections in the Van der Waals model of a real gas accounts for the volume of the particles. Also if they have internal structure, that structure can have ...


5

Every body in thermal equilibrium radiates the same amount of energy that it receives, otherwise its temperature would change until it attained equilibrium. This is not unique to black bodies. Suppose an object, not necessarily a black body, is at a temperature $T_1$ and its surroundings are at a temperature $T_2$, then the rate of radiation by the object ...


3

Because if those particles aren't point objects, you must also take into consideration that they take some space in the system and have properties like density and size which have to be taken into consideration when formulating the laws. This makes everything extremely complicated. A huge part of classical mechanics is only true for point objects for the ...


3

You've overlooked gravitational entropy. The entropy of a black hole horizon is given by: $$ S = \frac{kA}{4 \ell_p^2} $$ where $A$ is the area of the horizon, $k$ is Boltzmann's constant and $\ell_p$ is Planck's constant. This entropy is absolutely huge, so if you take a uniformly distributed gas in thermal equilibrium and concentrate it into a black ...


3

Without a doubt, it is the zeroth law of thermodynamics, as it defines an equivalence relation. It states that If two systems are in thermal equilibrium with a third system, then they are in thermal equilibrium with each other.


3

That $C$ is the specific heat for the given cycle, i.e. $$dQ=nCdT$$ This is for $n$ moles of gas.(not the $n$ you stated in question) I will assume $$PV^z=\text{constant}$$ $$nCdT=dU+PdV$$ $$\int nCdT=\int nC_vdT+\int PdV$$ We will integrate it using Pranjal's method : $$nC\Delta T=nC_v \Delta T+\int \frac{PV^z}{V^z}dV$$ As numerator is a constant, ...


3

As Jan Lalinsky has already pointed out (and indeed the OP has implied), the concept of macrostate is defined by the macroscopic quantities you have available experimentally. Suppose you know the expectation values of some set of extensive observables $X_i$. Typically these observables will be energy, particle number, average magnetisation, etc. This set ...


2

The simplest reaction deuterium and tritium. Tritium is common in big labs (like NIF, JET, Omega) [1]. Tritium sucks - practically speaking. It is expensive, radioactive, and hard to stockpile. Omega spent millions and years on a tritium facility. It may even never be used in fusion power [2]. The next easiest reaction is deuterium with itself. This ...


2

The Wikipedia article answers most of your questions. What are the requirements for hydrogen atoms to go through fusion? Two atoms must overcome the coulomb barrier, which can be done by forcing two atoms very close together, or by leaving them moderately close for long periods of time, which allows them to tunnel through the barrier. Is it a ...


2

This is a classic conundrum and it is called the "problem of the adiabatic piston". You can find it discussed in books on thermodynamics by Landau & Lifhsitz and by Callen. Another very thorough analysis is by Gruber "Thermodynamics of systems with internal adiabatic constraints: time evolution of the adiabatic piston". (You can find Gruber's article ...


2

The paper suggested by Trimok seems to answer your question. The paper gives an entropy for the observable universe of: \begin{equation} S_{obs U}= 3.1×10^{104} k \approx 10^{104}bits \end{equation} where $k$ is the Boltzmann constant and $S_{obs}$ is the entropy. However I would like to answer your two questions with a back of the envelope calculation. ...


2

Your description of the disturbance wrought on the system by the thermometer is sound. You may be able lessen the effect with a thermal diffusion model of the thermometer and by calculating what the system's temperature was before it brought the thermometer into equilibrium with itself, but for that approach to work, one must know the system's heat capacity ...


2

The equation you give: $$ Q = m C_p \Delta T $$ just tells us the total amount of heat transferred, and does not tell us anything about the rate at which the heat transfer occurs. To calculate heat flow we have to solve the heat equation. If you do a physics degree this apparently innocent equation will cause you many hours of frustrated head scratching, ...


2

It may actually work, as evaporating liquids need heat to evaporate, and water will somewhat evaporate even in the fridge. I am not sure it works in practice, because the paper also causes an adverse effect, it provides insulation, Hard to tell which effect is dominant. I'm pretty sure that the balance of both effects depends in a very large part on the ...


2

John Rennie's answer is correct for a DC series connected motor and, almost certainly, this is the kind of motor you (the OP) are talking about. An interesting way of writing John's answer "backwards" is that you have just observed the reason why the most powerful traction motors are exactly this kind of motor - almost all DC train and tram motors are ...


2

When a motor is turning it acts as a generator and produces a back EMF that opposes the applied EMF. See my answer to Top angular speed of electric motor for more on this. A frictionless motor would draw no current when not under load, though obviously real motors do draw some current because of frictional losses. If you load the motor you reduce the back ...


2

The simple answer: Satellites do feel this force, but obviously don't get ripped apart. The tidal forces are simply too small (for the satellites' materials) to actually rip them apart. The Why: Tidal forces happen because one side of an object feels such a larger huge difference in force than the other side. The magnitude of the force not only has to deal ...


2

By definition a reversible adiabatic system has $dQ = 0$. We also know the following from the Clausius Theorem : $dS = \frac{dQ}{T}$ Then it is easy to see that there can be no change in entropy. Note that irreversible adiabatic systems CAN see a change in entropy because in that case the above equation is no longer an equality but an inequality : $dS ...


2

I don't think I quite understand your question, but I'll do my best. In Thermodynamics, pressure is defined in a bevy of ways. If we look at the Thermodynamic Identity: $$ dU = TdS - PdV + \mu dN$$ (where $U$ is the Energy, $T$ is the Temperature, $S$ is the Entropy, $P$ is the Pressure, $V$ is the Volume, $\mu$ is the Chemical Potential, and $N$ is the ...


2

I believe the author is thinking here of things like electrochemical work that don't involve a change of volume (but rather, in this case, moving a charge through a potential difference). In this case a full thermodynamic description involves the chemical potential $\mu$, as a later equation shows. (I agree that the description of “different free energies ...


2

There is a universe where the second law doesn't hold. In fact, the opposite holds: entropy always decreases. Things are a bit strange there, though. The people there can easily predict the future many years in advance (although they don't call it "predicting" but rather "gnirebmemer"), but even their best scientists can't remember what the weather was like ...


2

As far as I understand your problem, you ask for the energy to vaporize $\Delta m = 1\,kg$ water at an atmospheric pressure of $200\,kPa$. ($9\,kg$ are already vaporized.) You have to increase the internal energy $\Delta U$ of the water to vaporize it and expand the vapor against the air pressure $p\Delta V$. The change in energy is: $$\Delta E = \Delta U + ...


1

I believe the 'urban legend' you are referring to is about cooling a bottle when you do not have a refrigerator. On a hot and windy day you could store your bottle in the sunlight, but it would be better in the shade, but if you really wanted to cool the bottle by a few more degrees, the 'myth' says wrap it in wet paper or cloth. During the time when the ...


1

Yes, infrared radiation which is invisible to human eyes but still radiates heat. This is how they make thermal cameras, they are using your body's infrared radiation which is detected by the camera and forms an image based on visible light.


1

Since this appears to be homework I can only give you a hint. Start with $dU = TdS -pdV$ and $C_v = \left( \frac{\delta Q}{\delta T} \right)_v = T \left(\frac {\partial S}{\partial T}\right)_v$ Now differentiate both sides with respect to $V$ and use the Maxwell relation $\left(\frac {\partial p}{\partial T}\right)_v = \left( \frac {\partial S}{\partial ...


1

Just because the entropy of the subsystems increases, that doesn't mean that the entropy of the whole system increases. This is possible here because of entanglement: an entangled pure state has zero overall entropy, but the subsystems have non zero entropy. A simple example is the state \begin{equation} \frac{1}{\sqrt{2}}(|00\rangle+|11\rangle). ...


1

Most, if not all, scientific analysis of real situations involves approximations. If some of the kinetic energy gained from falling was converted to kinetic energy of downstream flow (like a more sliding board shaped waterfall) it could affect the calculation. The environment could affect the temperature of the pool of water at the bottom of the falls, ...


1

The term you are looking for is premelting or "surface melting." It is an observed phenomenon (which could explain how ice skating works) with some thermodynamic descriptions. Basically what happens is the system is separated into two distinct phases, a solid (ice) and a vapor (air). There is a surface energy associated with this interface. If it happens ...


1

The exact times to equilibrium are difficult to calculate. It depends on the size of the domain, the method of pouring, surroundings, etc. You are right the with a higher temperature difference, the heat transfer will be higher. However, as you are getting closer to the equilibrium temperature, the heat transfer will also decrease. So it is only the initial ...


1

I think that you can state that it can not measure the temperature of a substance before it have been test but it can measure the temperature after it have come in equilibrium with the substance. For example, in an experiment that need to push a liquid to T temperature, what you do is heat up the liquid with the thermometer already inside. With this you ...



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