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0

Not quite - you want the cold sink to be as cold as possible. The cold sink is the cold air (or water or coolant) flowing over the heatsink. The heatsink is designed to efficiently connect the chip to the cold sink, so it needs to be as low resistance as possible. In thermodynamic calculations consider the chip==heatsink and the airflow to be the cold ...


0

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). ...


0

The "$2$" in the "phase rule" conventionally written as $F=C-P+2$, where $F$ is the degrees of freedom, $C$ is the number of components and $P$ is the number of phases, refers to the temperature and pressure that are the usual intensive parameters in chemical equilibrium of several phases and components. But this equation is only a special case of a more ...


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, ...


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The fact that it is a cylinder has to be taken into account. Usually $\frac{dQ}{dt} = \lambda \frac{A}{d}\Delta T$, where A is the surface perpendicular to the heat flow and d is the thickness of the material between the to temperatures. As the temperatures are being maintained, the heatflow is a constant, but being a cylinder the outer surface is larger ...


1

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

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

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 ...


1

If I understand the question, you are wondering how to justify the statement that a (reverible) adiabatic process is isentropic from the point of view of statistical mechanics (the classical thermodynamics definition makes sense to you). Let us then start with the entropic fundamental relationship, S = S (U, V, N), where U stands for energy, V for volume, N ...


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 ...


0

I don't know exactly where to start either but one could think of the black body radiation coming from a source far away that was emitted by a source at temperature $T_1$ where $g_{00}$ had some value say $g_{00}^1$. Now,if we make use of the fact that $g_{00}$ influences the photon frequency via gravitational Doppler effect, then imagining balance of ...


0

From a physics point of view, you have some kind of mass-diffusion problem. It is true that a mathematician will name it "heat equation" because the classical problem with the operator $\partial_t - \Delta$ is the heat equation. Homogeneous Neumann BC is indeed appropriate to model a "no flux" condition at the boundaries, be it a flux of heat or of mass. A ...


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I think you missed the exchange of heat, as both gasses get in equilibrium. This means gas 1 either takes heat from gas 2 if T1 is lower than T2 or loses heat to gas 2 if T1 is higher than T2. This should give you a second equation.


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 ...


1

If you rearrange the ideal gas law to be expressed in terms of pressure $$ P=\frac{NRT}{V}\qquad\Rightarrow\qquad \frac{T_1}{V_1}=\frac{T_2}{V_2} $$ where the right hand equation assumes it is an isobaric process with no mass exchange. So, in an isobaric process temperature and volume vary inversely. If the volume decreases then the temperature must go up. ...


0

Yes, Temperature is halved. You are keeping the pressure constant. You are just imagining pressing a piston in which pressure is not constant. Volume is reducing. But we have to keep the pressure same. So we must reduce kinetic energy of our particles to keep the pressure same. The rest is obvious.


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I disagree with @Raskolnikov I think the principle of equivalence is described by the carnot engine: So basically $$Q_H =Q_C + W$$ This is equivalence as I understand, it doesn't have to be a law.


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In a system where multiple liquids and solids are mixed together with different specific heats at different initial temperatures, reaching an equilibrium temperature, how do all of these things relate to the masses? In general, you would need to consider Enthalpy of Dissolution and/or Enthalpy of Mixing. For example, if you mix water and sulphuric acid ...


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.


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A (3d) gas of particles with a gravitational interaction is an example of a system with long range interactions, where the energy is not additive and thus many basic results of classical statistical mechanics are not valid, including the equivalence of the microcanonical, canonical and grand-canonical ensembles. For a general introduction to the subject see ...


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 ...


0

$1$ mole of water = $18.015$ g so $1$ kg water = $1000/18.015 = 55.51$ moles. $1$ kg of water = $55.51 \times 6.023 \times 10^{23} = 3.34 \times 10^{25}$ molecules. Latent heat of water per kg = $2.257 \times 10^6$ J. Latent heat per molecule = $2.257 \times 10^6 / 3.34 \times 10^{25} = 6.76 \times 10^{-20}$J.


0

An important note to the other answers so far: the actual mass of a single particle (the rest mass as it used to be commonly referred to) does not change if the temperature rises (by temperature I mean that of the environment since temperature is a macroscopic quantity). The (rest) mass of a particle is the $m$ in $$E = \sqrt{|\mathbf{p}|^2c^2+m^2c^4}.$$ ...


-1

photon has very light mass almost negligible amount, so temperature does not effect mass of the object, but you take large object with high temperature it indicates measurable amount of mass change. example sun.


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There are two sides to this issue. If you imagine perfectly insulated room, with your fan and improvised air conditioner inside the room, and the room is fitted with a very sensitive thermometer. Issue 1. In the closed system you would have, with no fans switched on, the temperature remains constant."Nothing Changes" as our Prof used to say. Turn on fans, ...


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 ...


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 ...


0

Why do you say for N particles you have N! distinguishable state? Lets just make sure you got it right. Lets say you have N particles on a lattice having M sites. In the ideal case, no excluded volume therefore each particles can access M sites. The total number of states for distinguishable particles in M^N. Now, if you considere that the particles are ...


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 ...


0

In the real world, yes you can heat a Black Body. Ideally, any energy absorbed by the Black Body would then be radiated away. Since all the energy is radiated away, there is no energy remaining to raise the temperature of the Black Body.


0

I'm not sure whether I understood your actual question correctly. But in my opinion chaperones, in the context of statistical mechanics, can act in two ways. 1) They act as a catalyst, i.e. they merely assure that the minimum in free energy is found fast enough. 2) They can keep the polymer in a steady state. The polymer system without the chaperones is ...


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.


0

Light of any wavelength can cause heating. It depends on the material which absorbs it. Few materials absorb light present in microwave range while others simply reflect them . Consider a material which absorbs light at microwave frequencies. If the intensity of the light wave at microwave frequencies is increased then heating effect will also be observed. ...


0

A few things to ask yourself and keep in mind RE Aluminium: - Does the proportionality of the change in Volume matter in this context? - If you know the linear expansion (Remember this is the change in length/area/volume (all scaler units) VS the change in Temp), then you can work out what the final volume is by simply multiplying the initial volume by ...


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

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, ...


0

In fact, thermal energy is heat. Think about heat as atoms vibrating: The more vigorously they vibrate (i.e. the more kinetic energy they have), the hotter it gets. The relation between thermal and kinetic energy is summarized by the simple equation $k_B T=m\langle v_x^2\rangle$, where $k_B$ is Boltzmann's constant and $\langle v_x^2\rangle $ the expectation ...


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 ...


0

In other words, for, say, an elemental solid, should we expect a portion of its surface to be liquid at any given time, with this portion increasing steadily until the melting point when the whole thing becomes liquid? It is possible for many compounds be part solid and part liquid under the right conditions. As ice melts, you have this condition. ...


1

Yes that's possible. For a Carnot heat pump the coefficient of power is $COP= \frac {T_{hot}}{T_{hot}-T_{cold}}$ so about 2.4 in your case. Which means you would get 2.4 times more heat out than work in. In reality the COP is smaller than that due to non-reversibilities.


0

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. ...


0

I don't just mean reactions that require heat to proceed, storing surplus energy in chemical bonds. I wonder about strongly endothermic reactions that suck heat out of environment A reaction that requires heat to proceed, a reaction that sucks heat out of the environment, and an endothermic reaction are all the same. These are all just descriptions of ...


0

You'd be lucky to even generate 1 Watt. Compare yours to this much larger 10 Watt generator intended to operate off hundreds of degrees of temperature difference: http://www.devilwatt.com/products/17-10-watt-camping-stove-thermoelectric-generator


0

The details of how your particular device will perform is very much a function of how that device was constructed and how it will be used, but in general, the physical effect is called the Seebeck effect -- briefly, the conversion of a temperature difference across a device to a voltage. Efficiency for such devices is actually quite low, with $\eta = 0.02$ ...


0

Think of it as a resistors in parallel problem. You have a heat source inside the room and a constant temperature outside (if you assume that the outdoors is infinite and well connected to the walls. Then you have a thermal resistance/area for the wall, ceiling and floor material. Using the total area you have an overall thermal resistance and a ...


0

I found this one: Van der Waals molecules. This answers my question to some extent.


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, ...


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 ...


0

If you're looking to produce power from a temperature differential, go with a device optimized for the Seebeck effect. Peltier and Seebeck effects are essentially the same thing, or rather flip sides of the same thing, but thermoelectric generators (Seebeck) are optimized differently from thermoelectric coolers (Peltier).



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