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Normally temperature probes measure the internal temperature of food being cooked. The surface of food can heat quickly, while the interior is still cool. Most food being cooked will have a known cooking time at a certain temperature setting (from previous experience).


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The Maxwell speed distribution curve is based on non-relativistic mechanics where the kinetic energy of a particle of mass $m$ with speed $v$ is $\frac{1}{2} m v^2$. In non-relativistic mechanics there is no limit on how fast a particle can move.


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Sure, the distribution goes off to $v\to\infty$, but the tail of the distribution drops off almost exponentially. Therefore, the probability of observing a molecule with higher and higher speeds becomes extremely unlikely. I think it is still (classically) reasonable to not have a cut-off speed. Perhaps the molecules will collide in such a way so that one ...


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It's because Maxwell developed it before the Einstein had worked out special relativity, when there was no maximum speed known. If you want to rework it for special relativity you need to rework it in terms of special relativistic kinetic energy. Specifically, the special relativistic version is: $$ f(p,x) = \frac{4\pi p^2}{V h^3}\exp\left(-\frac{1}{kT} \...


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This is a more precise and detailed assessment of the issue based on fundamental Newtonian fluid mechanics. For a compressible gas like air contained within a cylinder with a piston in which irreversible (or reversible) expansion of compression is occurring, Newton's law of viscosity in 3D tells us that the local normal compressive stress $\Pi$ exerted by ...


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This can be obtained from the Maxwell Distribution for the speeds of a molecule in an ideal gas $$P(v)=\left(\frac{m}{2\pi kT}\right)^{3/2}4\pi v^2e^{-mv^2/2kT}$$ where $m$ is the mass of the molecules, $k$ is Boltzmann's constant, $T$ is the temperature, and $v$ is the speed. $P(v)\,\text dv$ tells us the probability of observing a molecule with a speed ...


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This has to do with the probability distribution of the velocities of the molecules. Without knowing the distribution, one can only calculate the rms of the velocity (see e.g. here). Which leads to a different equation without $\pi$: $v_{rms} = \sqrt{3RT/M}$ If you know the distribution, you can integrate the probability distribution of velocities over the ...


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The reason for the appearance of Pi in the Maxwell-Boltzmann distribution can be rationalized by the following steps: 1) The probability of a given state in a thermal ensemble is proportional to the Boltzmann factor $\exp{(-E/kT)}$. 2) The energy of the point gas particle is expressed by its momentum $p$, the Boltzmann factor is then $\exp{(-p^2/2mkT)}$. ...


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The Fermi-Dirac distribution is a statistical expression for the fact that only a single Fermion can occupy a specific state. Then, the probability of a state to be occupied is given by the expression. Now let's think of a situation where we have a system with $N$ states of energy, ordered from the lowest to the highest as $0 < \epsilon_1 < \epsilon_2 &...


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1) Yes. These curves (for example) were computed with Mathematica with $T$ very small, approaching zero but still finite. Discussion of the limit: $$ \lim_{T\rightarrow 0} \frac{1}{e^{\delta/T}+1},$$ where $\delta = (E-\mu)/k_B$. If $E>\mu$, then $\delta >0$ and $\delta/T \rightarrow +\infty$, hence $e^{\delta/T} \rightarrow \infty$ and $f=0$. If $...


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If you heat a sphere of water from the center in the absence of gravity, I see no reason for convective instability (in the usual sense of buoyancy of hot fluid). The heat transport should purely conductive, and the idealized heat conduction equation could be solved by the familiar Green function method. Hot liquid expands, so the sphere will inevitably ...


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When an electron is bounded by the electric charge of a nucleus, it doesn't have enough energy to "run away" to infinity. Meaning you actually need to give energy to the electron to tear it apart from the atom, creating an ion (that is the meaning of ionizing energy). Looking at it the other way around, if you have a free electron and you want it to bind ...


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In absence of gravity water start to evaporate due to kinetic motion of water molecules.The making of water vapor absorbing latent heat from either near heat source or from water itself .So in this way water become solid ice which was locate far from heat source and other part become water vapor.


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The heat flow will be radial. A bubble will form at the center of the sphere and increase in volume with time. Because of the much greater specific volume of the vapor compared to the liquid, the vapor will push the surrounding fluid outward radially, just as if a non-condensible gas were being released at the center of the sphere. So, within the ...


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The first frame of reference is moving with the gas particles. The gas particles have ZERO velocity in their frame. The next step is to "put them in a box" and have them "collide with the walls". The step does work on the particles. It causes them to move from one reference frame to another. The work that is done is the integration of the force over ...


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When you have a distribution of particles in a single energy state and compare that to a distribution with particles arranged in multiple energy states the entropy values will be significantly different. You can work that out for a even if you do not view it from the perspective of microstates and instead view the system by understanding it from the ...


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Foundations The first law for a differential change in a closed system by the IUPAC form is below. $$ dU = \delta q + \delta w $$ Allow that the system has only mechanical work to obtain this expression. $$ dU = \delta q - p_{ext} dV $$ In this, $p_{ext}$ is the external pressure and $dV$ is positive when the system expands. Problem Statement Consider ...


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The important thing to remember is that U is a function of state (a property of the material) independent of any process. Between two thermodynamics equilibrium states, the change in U is independent of the specific process path that was used to transition the system between these two states. There are an infinite number of paths that can go between the ...


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As @Asron Stevens has already pointed out the answers to your questions really depend on the thermodynamic process involved as well as the type of gas (ideal or real). For example he has shown you that, for an ideal gas, you can have an increase in pressure with no change in temperature or internal energy (isothermal compression). When you say you read “...


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This Because of the Decrease Potential Energy (of the attractive force) between the Nucleus and the electron (Which Dominates over the the increase in potential due to electron -electron interaction) You can think of it as the electron "Falling down" the "Potential well" and thus releasing energy (like a ball which falls down hits the ground releasing ...


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For an ideal gas the internal energy only depends on the temperature of the gas. How the temperature relates to pressure is easily seen in the ideal gas law $$PV=NkT$$ So I suppose one could make the argument that the internal energy for the ideal gas depends on the quantity $$\frac{PV}{Nk}$$ and it's up to you how you want to explain the dependency. The ...


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Before Clausius there was Sadi Carnot, who many consider the father of macro (as opposed to micro or statistical) thermodynamics. Although I'm not aware of Carnot presenting the concept of entropy per se, his understanding of the maximum theoretical efficiency of a heat engine operating in a cycle laid the ground work of it. Then of course after ...


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Clausius invented the term entropy in 1865, which he understood as 'loss of movement of activity' i.e. energy (heat) losses. This is how engineers used (and continued to use) the concept, e.g. steam engine design, in the 19th century, even after Boltzmann introduced statistical thermodynamics in 1877.


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At the heart of a thermos jug lies a Scottish invention, the vacuum flask: A vacuum flask (also known as a Dewar flask, Dewar bottle or thermos) is an insulating storage vessel that greatly lengthens the time over which its contents remain hotter or cooler than the flask's surroundings. Invented by Sir James Dewar in 1892, the vacuum flask ...


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It happens because the kinetics of boiling have a finite time scale. This is because to begin boiling, a nucleus must be furnished to trigger the phase change. That nucleus usually takes the form of an air bubble in a crack or crevice in the water container's walls. If the bubble exists when the boiling point is reached, boiling begins without delay. If no ...


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An extensive property depends on the size of the object. Absolute entropy (J) is an extensive property. Molar entropy (J/mol) is an intensive property. We can double internal energy U (J) by methods other than doubling the size of the object. For example, double the absolute temperature of an ideal gas at constant volume. Since $U = C_V \Delta T$, the ...


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I just wanted to come back and update this, as I realized after-the-fact there was a much simpler solution to the above problem rather than making complex calculations: Maintain the same steam “power”, belt velocity, etc.; confirm all other factors remain the same as control sample (using the detained export product as your referenced control data). ...


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My best guess would be that it's due to convection flow passing through the pasta's cylindrical gap in a stable manner when upright. Convection when the bottom is heated causes hot water (and possibly an amount of steam formation if very hot) at the bottom to flow upwards. Pasta's that are lying down will be pushed around all the time and never really find ...


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I will only comment on your assumption of the efficiency of the cycle. The efficiency equation you are using assumes a reversible cycle where all the heat added $Q_H$ occurs at a single temperature $T_H$ and all the heat rejected $Q_C$ occurs at a single temperature $T_C$. In your cycle the heat added during the isochoric process and rejected during the ...


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The QH you are asking for is the net heat given to the system which is nC v( Tf - Ti ) here as heat in adiabatic process is 0 and in the isobaric process, heat is taken out from the system here ( since Tf < Ti for the isobaric process) Therefore, now you have net heat and net work ( both work positive and negative work would be considered ) I hope you ...


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You incorrectly applied the Clausius inequality to this problem. The temperatures that must be used in the denominators are the values at the interfaces between 1 and 2 and between 2 and 3 (see Moran et al, Fundamentals of Engineering Thermodynamics). So the correct application of the Clausius inequality should look like this: $$\Delta S_1\geq\int{\frac{...


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I don't know if this helps you "complete your argument", consider the following involving only heat transfer between the substances: For an isolated system, system energy must be conserved, so the heat exiting $S_1$ equals the heat entering $S_3$. Consequently $Q_{12}$ = $Q_{23}$ = Q Assume $S_1$, $S_2$, and $S_3$ are thermal reservoirs, so that the heat ...


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Depends on how you are defining "weight". If you mean weight as in the vector or scalar of the gravity acting on an object, Marco's answer is what you want. There is a third meaning of weight. This is essentially "what a scale measures". This definition also fits with how people typically experience the effects of weight. If we use that definition, then ...


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Okay today I'm less tired and I look at all that values again and I find a more "closer to be real" answer: - 16 cubic feet minute = 0.45 cubic meter minute. - 0.45 cubic meter minute = 27 cubic meter in a hour. - 27 cubic meter of dry air = 27cm * 1.29density = 34.83Kg. - Temperature difference stable after a hour = 50C from outlet - 10C from inlet = 40C....


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For all practical purposes the weight will remain 1000 pounds when you freeze it. In theory the mass of the water will reduce somewhat through cooling because it will contain less energy when frozen, but the effect would be utterly negligible. Although the weight will be the same, water expands in volume by about a tenth as it freezes (which is why, for ...


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Here's a formula you will need. I'm sure you know it already: $$P = \frac{\Delta E}{\Delta t}$$ Here's the major problem you have. In your post, I see lots of effort to calculate $\Delta E$, with temperature changes and such. However, calculating how much time the air spends inside your heater is rather difficult because the air goes in and churns ...


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In isotherm process and with no change of phase we can't use those formulas and there's no formula for defintion of Q (correct me if I'm wrong). You are correct that you can't use those two formulas for an isothermal process with no phase change, but you can determine $Q$ from the first law. Usually we use $\Delta U=Q-L$ but so we are defining Q by ...


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The work done is the change in KE of the body. You have caused the contradiction by assuming that the change in KE is exactly counterbalanced by the generation of heat- that is a spurious term.


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Work done by a force is always given by the line integral in your question. The work energy theorem, as you have written it, is basically a statement of a conservation of energy. Work is a mechanism of moving energy from one place or form to another. In a system where the only places for the energy to go is into kinetic energy or heat, then the total ...


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I believe the issue is that you are including heat in your work calculation when you shouldn't. The formula $W_F = \Delta KE + H$ is not correct. The change in internal energy ($\Delta U$) according to the first law of thermodynamics is given as $$ \Delta U = Q - W$$ where $Q$ is the same as your $H$. Work should just be $W = \Delta KE$, the heating does ...


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Frankly, I don't see any significance to the blue vertical lines either, other than perhaps to label specific materials. But then, why vary the height? I agree with you the horizontal lines probably indicated ranges due to material variations. That would make sense for wood being a natural material with lots of variations. Interestingly, there is a note in ...


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There are two different definitions of heat capacity, heat capacity at constant volume and heat capacity at constant pressure. The reversible expansion of an ideal gas cannot be done at constant volume. It cannot be done at constant pressure without adding heat.


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Actually there were many debates on it, and some paper also established by Cambridge and Oxford University but actually Nobody really know why the entropy is so big. My view is that black holes manage to excite the quantum-gravitational degrees of freedom, so to really understand them you should work in quantum-gravity. In a such a theory it should be ...


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If black holes are to satisfy the second law of thermodynamics, then since black holes do not remember what went into them (other than the total mass, angular momentum, and charge), their entropy has to be larger than the maximum entropy of all possible matter distributions that could have formed the black hole. Some fiddling around on the back of an ...


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You don't give the context, but this idea tends to be used in scattering experiments where the particles are either massless (e.g. photons) or highly relativistic so their rest mass is negligible compared to their energy. In that case the de Broglie wavelength is related to the energy by: $$ \lambda = \frac{hc}{E} \tag{1} $$ And as you say, $kT$ has the ...


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$\frac{hc}{k_BT}$ is a length, where $h$ is Planck’s constant and $c$ is the speed of light. Apart from a numerical factor of order 1, this is the thermal de Broglie wavelength for a photon gas at temperature $T$, the critical wavelength at which quantum effects begin to dominate. Also, apart from another numerical factor of order 1, this is the ...


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It's not really clear what you're asking, since you can no more "convert" energy into length as you could convert mass to length -- they're fundamentally distinct. However, I'll assume that what you want is something like this: As you note, $k_BT$ has dimensions of energy, equivalent to $M L^2 / T^2$, where $M$, $L$, and $T$ are mass, length, and time, ...


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If you are asking whether you can determine the final thermodynamic state of the system, the answer is yes, provided you know the details of how the external pressure is applied as a function of the volume. The process is going to be isentropic if the added external pressure is applied gradually and in tiny pressure steps. If it is applied rapidly, then ...


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Kuma spotted my mistake by pointing out that \begin{equation} \frac{\partial (\beta E)}{\partial E} \neq \beta. \end{equation} I wanted to provide the answer in a slightly shorter way. Also setting $k_B = 1$, \begin{equation} \begin{aligned} \frac{\partial S}{\partial E} &= \beta + E\frac{\partial \beta}{\partial E} + \frac{\partial \ln Z}{\partial E} \\ ...


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