New answers tagged

1

Temperature is defined for a system in (at least local) thermal equilibrium. The electromagnetic field is present everywhere and, when in thermal equilibrium, has a blackbody spectrum. Thus, a body made of charged particles "vibrating" at temperature $T$, in the presence of an electromagnetic field at the same temperature $T$, will on average gain ...


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One geometric way to look at it is as follows: Imagine that the solid sphere is in fact a collection of concentric shells (like a spherical Russian doll). The hollow sphere on the other hand obviously is exactly the same thing, except that it has only one hell, namely the outermost one. Let’s ignore the contribution of the inner shells in a first approach....


3

There is only one equation, but fairly straightforward and I have included an explanation at the atomic/molecular level also. A uniform mass distribution when heated to a certain temperature will expand to a certain volume according to $$\Delta V =V_0 \alpha \Delta T$$ where $\Delta V$ is the change in volume of the mass, $\alpha$ is the coefficient of ...


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There are two main errors: Single particles don't have a temperature. Temperature is a statistical feature of bulk matter. Single particles don't emit EM radiation when they move. Instead their energy is quantised. Under Classical theories all atoms would quickly collapse as their electrons radiate all their energy away.


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Near absolute zero, the quantum treatment is necessary. Consider the Einstein solid, Heat capacity plot look like: The simple explanation for $C\rightarrow 0$ as $T\rightarrow 0$, because changes in the temperature has no effect on the internal energy when the temperature is so low that only the lower level is occupied and even a small change in ...


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Short answer: elementary particles, no. Composite particles, yes. A charged particle just moving without acceleration does not lose energy, but if it accelerates then the Larmor formula applies (in the non-relativistic case): $$P=\frac{\mu_0q^2}{6\pi c}a^2.$$ So the relevant question is, will a hot particle accelerate? If the particle is elementary even ...


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They are cold and lonely, and other particles of the same charge find them repulsive ;) More seriously - cold/hot is the concept based on temperature, which is the measure of the average kinetic energy of the particles. As such, the temperature is not directly related to the charge of particles, although in some situations one may affect the other. One ...


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In reality M. Plank advanced first (constructed) a simple interpolation formula bridging two asymptotics - infrared and ultraviolet, both being well established experimentally. Then it turned out that his semi-empirical interpolation formula worked so well in different conditions that it seemed to be an exact formula. After that M. Plank tried to derive it &...


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Let's assume that the volume V of the vessel is constant (neglecting expansion of the vessel itself). Let $\rho_0$ be the density of the liquid at T2 and P2, and, at arbitrary P and T, the density is $$\rho=\rho_0\exp{\left[\frac{(P-P_2)}{B}-\alpha(T-T_2)\right]}$$So the mass of liquid 2 in the tank is always $M_2=\rho_0V$. After the pressure rises to P1, ...


0

An ideal gas is subset of ideal fluids. The specific heat of an ideal fluid depends only on temperature. The temperature of concern is in a static reference frame. One common form of the temperature dependence for specific heat capacity is the Shomate polynomial as on the NIST Chemistry Webbook site $C_p = A + BT^2 + CT^3 + \ldots$. A subsequent assumption ...


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This is indeed an error. Emissivity is the emission relative to that of a black body of the same temperature.


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If we now increase the water temperature in the region, the evaporation rate will briefly exceed the condensation rate, filling the air with more vapor until a new equilibrium is reached. This isn't quite right. If you were studying the liquid/vapor equilibrium in a small, closed container, this would be the right idea. However, the atmosphere is somewhat ...


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Maybe worth to derive it from the differential definition $$ dQ = PdV+ dU \tag{1}$$ Recalling that $\frac{dQ}{ndT}=C$, $$ C= \frac{1}{n} ( P \frac{dV}{dT} + \frac{dU}{dT}) \tag{2}$$ From ideal gas law and polytropic equation we can state $$ (PV^{\gamma} )V^{1- \gamma} = nRT \tag{3}$$ Considering differentials while noting that $PV^{\gamma}$ is constant: $$ ...


0

How can we determine the heat absorbed by measuring the fall in a weight? Heat is not absorbed due to the fall of a weight. The purpose of the falling weight experiment is to demonstrate the equivalency between energy transfer by work and energy transfer by heat. The fall of the weight in the diagram given by @Cross does paddle work (stirrer work) on the ...


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For the situation you are looking at $$\Delta H=Q+W_{elec}$$where $W_{elec}$ is the electrical work done by the surroundings on the system. From the 2nd law, if follows that $$Q=T\Delta S-T\sigma$$ where $\sigma$ is the entropy generated within the system by irreversibility of the process (e.g., oversupplying electrical work). So, combining these we have $$...


3

Can one go more fundamental than "heat flow is proportional to temperature difference, and this then leads to exponentially slowing approach to equilibrium"? Of course one can. Newton's law of cooling $$Q_{flow} = h\Delta T = h(T_{body}-T_{environment})$$ is of the same form as many other equations of conduction, where there is a transport ...


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Very fundamentally, the reason is that the heat flow is proportional to the temperature difference, which in turn depends linearly on the amount of heat flown. This (the rate of change is (inversely) proportional to the overall change) describes an exponential relationship: The steepness of the exponential curve is proportional to the y value at that point. ...


1

The principle that the fall in weights can be used to measure the energy absorbed by heat follows directly from the conservation of energy for an isolated system: $$\Delta U_g + \Delta E_{int} = 0$$ The system under consideration comprises of the falling masses, the Earth and the engine. The change in gravitational potential energy $\Delta U_g$ is negative ...


0

In Thermodynamics and an Introduction to Thermostatistics, Callen writes We [follow] the standard convention of restricting attention to systems that are macroscopically stationary, in which case the momentum and angular momentum arbitrarily are required to be zero and do not appear in the analysis. The familiar material properties frequently used in ...


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We can derive this exponential behavior from the laws of thermodynamics as follows: on the one hand, heat is a quantity that is subject to a conservation law (first law of thermodynamics): locally, the amount of heat $H$ stored in a body can only change due to a flow of heat $J$ out of it. On the other hand, the second law of thermodynamics tells us that ...


0

Energy balance in living organisms can be studied on all levels of life: viruses, cells, multicellular organisms, and whole ecological communities. There are two key elements in applying the thermodynamic thinking to live organisms, associated with the first and second laws of thermodynamics: Energy conservation (1st law of thermodynamics) The organisms can ...


1

Slow or fast, it is a matter of the relevant time scale used for comparison. It is not the human time scale that should be used, but the microscopic time scale associated with every thermodynamic system's dynamics. On the human time scale, waiting for water boiling may seem slow, while the freezing of a bottle of supercooled water, or an explosion, may look ...


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I think you would get better appreciation of your question on Chemistry SE because chemists usually have a lot of first hand experience with slow thermodynamic processes (although there are also many reactions that are fast, of course). As far as I understand you, the processes you mean are all somehow related to diffusion/heat conduction. Why is diffusion ...


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This behavior is basically described by Newton cooling with heat generation, using the equation: $$MC\frac{dT}{dt}=G-k(T-T_{\infty})$$where T is the temperature, t is time, M is the mass, C is the heat capacity, G is the heating rate, k is the Newton cooling coefficient (convective heat transfer coefficient times surface area), and $T_{\infty}$ is the ...


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Yes there is. Provided that the temperature of the (single) heat sink can be considered constant during cooling, and the heat capacity of the components between the device and the heat sink are negligible, the cooling part is described by the differential equation $$\frac{dT}{dt}=-A\cdot (T-T_{sink})$$ no matter what geometry connects the device and the sink....


2

[...] it looks like m is taken to be equal to $\frac{1}{N}\sum_i s_i$ which seems a bit odd since I believe that m is supposed to be a thermal average not an average over "sites". Yes, that's right. The $m$ which appears at that section of Tong's notes is not the thermal average magnetization, but rather the average magnetization of a particular ...


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@Jon Custer comment is correct. One of the reasons for cooking bagged frozen vegetable in the bag in a microwave oven is that the steam cannot escape. This cooks the vegetables faster and avoids drying them out. Unfortunately, not all foods that you would like to steam come in a microwavable bag. Thus the possible purpose of a microwave steamer. Though I ...


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Why do we need to compress the gas adiabatically after the isothermal compression of gas? What if we just skipped it? The short answer is it is necessary in order to increase the temperature and internal energy of the system to its original state and complete the cycle. The work done in the compression equals the increase in internal energy (since $Q=0$). ...


3

The variables are related by the ideal gas equation $pV=nRT$. We can write $n = m/M$ where $m$ is the mass of the gas and $M$ is its molar mass. Therefore we have $$pV=nRT \\ pV=\frac{m}{M}RT \\ p = \frac{m}{V} \frac{RT}{M} = \frac{\rho RT}{M}$$ The equation for speed can also be written as $$v = \sqrt{\frac{\gamma RT}{M}}$$ Your statements depend on what ...


1

You're told that both substances are being given heat (energy) at a rate of $6 \text{ cal s}^{-1}$. This means that every second, your substance gains $6 \text{ cal}$ of energy. All you have to do is work out how many seconds each substance is in their "fusion" phases and multiply rate by time. The "known result" you're looking for is ...


2

The ideal gas law (which is an approximation to real life, but good enough for a back of the envelope calculation) tells us that $$PV \propto T$$ or, in an alternative formulation $$P \propto \rho T$$ So if pressure $P$ stays the same but temperature $T$ decreases then density $\rho$ must increase. In other words, a given mass of gas takes up less space as ...


0

It turns out that terms representing the kinds of energy in a system generally are products of two factors. One factor is an extensive variable (depends on how much of the system you consider), and the other is intensive (does not). For example, $PV$: $P$ is intensive, $V$ extensive, or $\mu N$: $\mu$ (the chemical potential of a species) is intensive, $N$...


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Because that what experiments at the time indicated. For instance, the first (!) line of the 1957 BCS paper states: The main facts which a theory of superconductivity must explain are (1) a second-order phase transition at the critical temperature, $T_c$ (...) This is the whole answer. There is actually a really interesting story here. At the time of ...


0

What is the differentiating factor between work and heat?' Both heat and work are means of energy transfer. The main differentiating factor is heat is energy transfer driven solely by temperature difference whereas work is energy transfer driven by force acting through a distance. For example, The way in which entropy is defined is a very good way to ...


1

You are correct: heat transfer can be thought of as work done at a microscopic level. Consider the simplest example of two molecules colliding. There will be some electrostatic repulsive force for the very short duration of time when they are colliding, and this force will act over a very small distance. As a result, one molecule will gain energy and the ...


1

If we take the water molecules (both those in the liquid and vapour phase) as a system, does this system do work on the surroundings when evaporation occurs? That is, when the most energetic of liquid phase molecules break free into the gaseous phase, do they push against the atmosphere and as a result do work on it? Indeed there is work done on the ...


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$\newcommand{\ket}[1]{\left|#1\right>}$ Here is a not very rigorous explanation. In equilibrium, any macroscopic quantity A should have a constant value. This means the corresponding operator A should have a constant mean. For convenience, I use the same symbol here for both the quantity and the operator. Suppose that the Hamiltonian operator H is ...


2

Does evaporation do work on the surroundings? Actually, apparently it does, at least according to the Hyperphysics website: http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/phase.html#hvap. To quote from the site: "In the process of vaporization of water, a large amount of energy must be added to overcome the remaining cohesive forces between the ...


0

The formula $$ U_{thermal}=nf\cdot\frac{1}{2}kT $$ can be derived from the equipartition theorem and holds for systems of molecules such that their Hamiltonian is only made by $f$ quadratic terms. In practice, it is valid for a very dilute gas (in the limit of the perfect gas, i.e. negligible inter-molecular interaction) of rigid molecules or non-rigid ...


0

Yes, the water will freeze, but not into the normal hexagonal crystalline ice form that we are familiar with. Assuming that the container is capable of withstanding an internal pressure of at least 300 Mega Pascals (about 43,500 pounds of force per square inch), the water will freeze into its Ice II rhombohedral crystalline form. This also assumes that the ...


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This is not an opinion based exercise. Thermal energy is generated through conversion. It is not inherent. The natural state of empty dark space is absolute zero. Stars and other celestial bodies generate thermal energy which is distributed through space and absorbed and retained by other objects. It is not possible to add cold. It is only possible to ...


0

The first documented steam engine was built by hero of Alexandria in the first century AD. It worked much as you describe, except that it was a rotary engine and the whole boiler rotated. In the last century, ships and power stations increasingly came to employ continuous-flow steam turbines. Many, including all coal and nuclear fuelled power stations, still ...


2

The key word in the description is "quasistatic", meaning the process is happening so slowly that it appears to be static (not moving). Then the gas pressure is always equal to the pressure of the surroundings so there is no acceleration of the piston and its velocity approaches zero. Hope this helps.


1

We can make a steam engine just by putting huge amount of water in large tank and heat it and then use the steam to run the wheel. We just have to put huge amount of water and heat then engine will work for days. But in my books, all engines are based on cyclic process. Why? This is essentially a question of definition. You can indeed extract work from a ...


2

Entropy, as the conjugate thermodynamic variable to temperature, is the "stuff" that's transferred or shifted when a temperature difference drives energy flow. (You can think of this as analogous to a pressure difference driving a shift in volume, or a surface tension difference driving a shift in area, or a stress difference driving a shift in ...


0

No, this is not implied that velocity of piston surface is constant. What needs to be implied here is that there is no friction involved in piston movement, hence making work done a path independent function. In this regard, the velocity of the piston or its history of movement may not be taken into account as it does not affect the final state of ...


0

We can make a steam engine just by putting huge amount of water in large tank and heat it and then use the steam to run the wheel. We just have to put huge amount of water and heat then engine will work for days. But in my books, all engines are based on cyclic process. Why? Yes, early steam engines were like that. https://en.wikipedia.org/wiki/Steam_engine#...


0

Question does not make any sense. An engine has to operate on a cycle, otherwise you will only get a small amount of work in the beginning as your source and your sink thermalize, and that is it. The moment you reheat stuff, you are re-doing the cycle, making an engine.


0

All heat engines operate between a hot temperature source and a low temperature exhaust, where useful work gets extracted from the heat source and whatever remains is dumped out as exhaust. The operating principle of a particular heat engine can be plotted on a certain kind of chart called a P-V Diagram, and when you go through each step of the heat engine ...


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