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From my heat transfer textbook by Cengel and Ghajar, they propose that you use the average thermal conductivity value as you specified in your last equation: $k_{avg} = k(T_{avg}) = k_0(1+\beta(T_a+T_b)/2)$ where $\beta$ and $k_0$ are material properties, assuming that the material thermal conductivity follows a linear function w.r.t temperature. You ...


1

The dimensionless Rayleigh number characterizes buoyancy driven convection. When the Rayleigh number is below a critical value, heat transfer is primarily by conduction (e.g. no Benard convection cells). When the Rayleigh number is above the critical value, heat transfer is primarily by convection (e.g. Benard convection cells spontaneously form and ...


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After studying this problem, I have found a method of solving it. First of all, since the ceiling is the heat source, we don't know what it needs to be to keep the room at 10 degrees C. Obviously it will have to be hotter than 10C. Secondly, the temperature will differ in basement. For example, it will be hotter near the ceiling and colder near the floor. ...


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As far as I remember, the RB convection system very far away from the onset has not been analytically studied to a great extent. There has been quite a bit of numerical work studying pattern formation and structure generation (finally leading to chaotic turbulence). Far from the onset, I suppose one might see ergodicity breaking and possibly chaotic ...


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A short answer for a specific system might be helpful here. Imagine a system of two containers linked by a valve. One container is full of gas - the other is empty - it contains no gas - it has a vacuum. Answer to question 1 If the valve is opened the gas will fill both containers - and the entropy will increase, but temperature remains constant as no ...


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I see your point, but I'm afraid your statement looks like a third conceptually different version. You're saying that the "other effects" regard only to a third subsystem. I can't see why. I might be wrong but I understand that the effects regard the system, that is, body A and body B.


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Imagine two systems. A and B. A is in a higher energy state (say is warmer) than B. Clausius says you cannot transfer energy from B to A without a corresponding change in another system which is neither A nor B (lets say 'C'). But you can transfer energy from A to B without affectng or invoking C. The 'other affect' bit implies another system/state/body. ...


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You will need more energy to convert water to steam in sealed one. The open one will be free to expand, while the sealed one have limited volume. When the sealed one evaporates, the molecules of steam pushes molecules of air in the container increasing the pressure. As the boiling point increases so as the enthalpy too, thus the difference in final and ...


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You find $N$ from the expression $$N = \frac{I_a}{e}$$ which you have stated yourself (equation (4) in the problem) - no need to find $N_0$ first. Plotting $\ln(N)$ as a function of $U_f$ will give you a straight line (according to equation (3) from the problem), with an intercept equal to $\ln(N_0)$ and a slope equal to $frac{e}{kT}$. Since you know $e$ ...


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Yes. Irrespective of the phase or reaction, chemical potential refers to the change in total free energy (Gibbs free energy) with the change in number of particles of the system. Fundamental definition is $\mu = \frac{\partial G}{\partial n}$ where n represents the number of particles in the system and G the total Gibb's free energy. For a pure substance ...


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If we want to examine gravitational collapse from a statistical mechanics point of view, we find that there's a tradeoff between the fact that a more spread-out collection of matter has more possible position states, whereas a more concentrated collection has more possible momentum states (because more of the system's potential energy has been converted to ...


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I would say no. The chemical potential is directly equal to the Fermi energy in some ideal ($T=0$) systems. This chemical potential is dependent on electron mass, number of electrons and volume.


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"a circuit that has different resistors at extremely different temperatures" -- Each resistor independently puts out its own noise related to its own temperature. "one long resistive element that has a temperature gradient across the whole thing" -- That's actually the same thing again. Treat it as a large number N of resistors in series, each with ...


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No, if $S$ is really the only thing you know about your system then there is no way to know its energy. There is no relationship between the energy and the entropy that doesn't involve some other quantity such as temperature. ...but surely you know something about your system, other than its entropy? I mean, you must know something about what it's made of, ...


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Actually one can flood the mines on an asteroid. The water will become boiling, but before it all boils out the water will freeze. As such, the scientists could say so only if they anticipated negativity in ice bulbs forming in the mines. But the process of freezing is very slow, so the water most likely would reach the bottom of the mines.


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When textbooks write $dS=dQ/T$, quasi-static process with no friction is considered. $S$ is entropy of the system, $dQ$ is heat accepted by the system from the environment and $T$ is temperature of the system. You can add indices to $dQ$, but they are not necessary since $dQ$ already means total heat accepted from the environment. There are no other kinds or ...


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Because the liquid would boil away. Boiling is what happens when the partial pressure of a liquid exceeds the ambient pressure. Liquids have higher partial pressure as they get hotter, so we usually associate boiling with high temperature. For example, water needs to be heated to 100°C to boil at 1 atmosphere ambient pressure. However, pressure is ...


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There are some nice ideas in other answers, but they are overseeing some basics. Let's do some thermodynamics. The efficiency of a thermal engine is bounded by the Carnot efficiency: $$\eta \le 1 - \frac{T_c}{T_h} $$ Where $T_c$ is the temperature of the cold end and $T_h$ the heat source. Assuming we are in a cool environment, $T_c=0 C$, $T_h=37 C$, so: ...


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Yes this setup violates the second law of thermodynamics .The law states that - "It is impossible for any device that operates on a cycle to receive heat from a single reservoir and produce a net amount of work." Now you know the answer


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It depends on your power supply. The power $P$ through a circuit element is always given by $$ P_\text{thing} = I_\text{thing}V_\text{thing} $$ where $I$ is the current through and $V$ the voltage drop across the thing that you're interested in. If your circuit element obeys Ohm's Law $$ V_\text{thing} = I_\text{thing} R_\text{thing} $$ then you can write ...


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There are actually two questions here. At phase equilibrium, yes independent water molecules do enter and leave the liquid vapour interface. The rate of crossing of individual molecules from one phase to another is characterized by an Arrhenius type rate equation of the form: $$\alpha \exp \left[ - \frac{\Delta E}{kT} \right]$$ where $\Delta E$ is the ...


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The definition does not require an "atmosphere". Your "environment" can be vacuum. A liquid can boil in vacuum.


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After some reflection on the answer of David Hammen, I think the following answer offers a slightly different perspective on spherical bubble formation where surface tension is the dominant force. If we assume that the bubble tries to minimize its potential energy then, for a given surface area, it will try to offer the maximum volume for the gas ...


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No, it is not possible to hide a person's heat signature indefinitely. Even with the best suit imaginable, you will eventually either begin leaking the heat, overheating the person, or both. One problem is that there are no perfect thermal insulators. This means that you must either use the best available and keep emissions below some threshold of ...


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Such technology is in its infancy, but it definitely exists. The images below are produced by several companies promoting their thermal/IR camouflage clothes. Obviously the applications are well-suited for the military, so who knows what more the military has developed. This last image is made by a company called Blucher Systems. The link provides much ...


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You could try to develop a material that acts like a fluorescent one, that is, transform infrared radiation into lower energy photons, such as microwave. So you will not glow on infrared but on some other wavelength of your choice. Now, if this is technologically feasible (using nano-engineered materials perhaps), I have no idea!


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I guess this is possible: for example, thermal radiation can be emitted within a very small solid angle, e.g., upwards.


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You couldn't make the heat "disappear". You can however make it go somewhere else. To talk about thermodynamics very loosely Anything (especially a vacuum) between the source and the detector will hide the source, as long as the medium remains at the surrounding temperature. This means that if you have a heat source behind a window, take a thermal imaging ...


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Let's go back of the envelope here: I'm going to assume that your "original" volume of water is 6 oz (I think that's fairly standard for a cup of coffee/tea), but you add an "extra" 2 oz of cold water. (I'll refer to the amounts of these waters by the quoted titles throughout the answer to keep the language precise). Whether or not this water will be added ...


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It forms a cone because it depends on a shock wave, and the region enclosed by the shock wave appears conical in shape. See, for example, the apparent cones here: They are also visible here: Wikipedia appears to be fairly clear on why vapor cones are related to shock waves. From the introduction to the article about vapor cones: Atmospheric water ...


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It would most likely be less warm if you add more water. It is because heat is measured in calories or temp. 1 calorie is how much energy that is required to heat up 1cm3 water up one degree Celsius. So theoretically if I added more mass or water it would take more energy to heat up because 25 cm3 of water would less energy to heat up because it has less ...


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It will come out cooler if you add the extra water at the end. It could be a lot-if you literally mean it came out boiling hot at the end of 60 seconds, you could have used the last 15 seconds to boil water. If you had added the extra water at the beginning, it would come out at 100C. If you add it at the end the mix will be cooler. The extra heat in the ...


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I propose this : As long as the detailed balance is respected, the system can be studied with equilibrium thermodynamic. Therefore we can write the free energy of the solution, for exemple $ F=A \log(A) + B \log(B) + S \log(S) + \chi_{A S} A S + \chi_{B S} B S + \chi_{A B} A B $ ($S$ the solvent, $\chi$ the interaction parameters), plug the equilibrium ...


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based on casual observation, believes that coffee will cool faster than ordinary hot tap water Let's revisit the "casual observation" part. Are the initial conditions the same? Water from a kettle will usually be hotter than from a coffee machine. And we tend to do things to coffee that we don't do to water, like add 10% of another liquid (milk) at low ...


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It is because of the second law of thermodynamics. There are many irreversible processes that can be used to heat something. It is the natural flow of things because entropy will increase in isolated systems, and much of the internal energy of objects can be dissipated as heat (and this heat used to heat something that is colder). However, to cool something ...


1

Heat is simply released to the surroundings during freezing. Freezing occurs because at a temperature below the freezing point, solid has a lower Gibbs free energy than liquid. The plots of G vs T is mathematically derived. At the freezing point, $G_{solid} = G_{liquid}$ There is no change in Gibbs free energy in a reversible freezing process. Therefore, ...


2

What makes water boil/evaporate is the thermodynamic concept derived from the first and second law of thermodynamics. You can read this article to find out the derivation from entropy to the Clapyeron equation. http://en.wikipedia.org/wiki/Clausius%E2%80%93Clapeyron_relation#Derivation_from_state_postulate $$\frac{\mathrm{d} P}{\mathrm{d} T} = \frac ...


0

The only temperature that would make sense in your question is the temperature of the thermal distribution of the photons (if the photon gas is thermalized). So let us assume we have a photon gas at equilibrium. Its chemical potential is $\mu=0$ because photons have no ground state. Consider now the atom. One of the characteristics of an atom is that the ...


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For stars (which have huge amount of mass and density), gravity is taken to be responsible for the heat increase. because heat and volume (thus density) thus gravitation of a (massive) star, are related. This is exactly one of the factors that make nuclear fusion (in stars) possible. The two effects thermodynamics (and kinetic energy) and gravity are ...


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Consider a satellite in orbit about the Earth and moving at some velocity $v$. The orbital velocity is related to the distance from the centre of the Earth, $r$, by: $$ v = \sqrt{\frac{GM}{r}} $$ If we take energy away from the satellite then it descends into a lower orbit, so $r$ decreases and therefore it's orbital velocity $v$ increases. Likewise if we ...


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So a while ago I did a little project where I grabbed a "standard solar model" from this paper, which gives me some information that's useful for actually making an estimate. (Unsurprisingly the link given to download the data has changed in the last ten years; I haven't sleuthed to see whether the data is still publicly available.) Only about 1.5% of the ...


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To add to Whelp's Answer. Even though the $$\bar{d}Q=T\,dS$$ does not hold in an irreversible process, it still gives us something general. Consider a thermodynamic system linked to a system of reservoirs and which can only exchange heat and nothing else with the reservoirs. Let's call the thermodynamic system an engine for our convenience. Then the engine ...


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At the end of your question you ask if there are other ways to express the temperature of the Sun in terms of energy. This is probably not exactly what you're looking for, but from Wien's law $\lambda_{\rm max}T=b$ ($\lambda_{\rm max}$ is the peak wavelength of the Sun's $\sim$blackbody spectrum, $b$ is Wien's displacement constant) and ...


1

Fun, So you are asking about the thermal energy content of the sun? If we assume that all the hydrogen is dissociated. (single atoms) Then each atom has three degrees of freedom and carriers 3/2 kT of energy. So count up the number of atoms at each temperature.... That will work until the atoms ionize. Then there will be equal energy in all the ...


0

The sun is more than a cloud of hot gas that is radiating energy. The sun also has energy stored in hydrogen "fuel" that will be "burned" through nuclear fusion into helium, releasing a lot of energy. The sun is $2 \times 10^{30}$ kg, and about 70% hydrogen, so around $1.4 \times 10^{30}$ kg of hydrogen or $8 \times 10^{56}$ protons. Let's estimate that ...


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Convective heat transfer is an engineering topic and is subject to much uncertainty. Heat transfer coefficients are normally obtained from correlations based on experimental data and backed up with theoretical analysis. If you have no other source of information, I suggest you start by taking a look at Thermopedia : http://www.thermopedia.com/content/652/


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This relation is not true for general processes. For a closed system, the general relation is $\delta Q \leq TdS$, as is illustrated by the Clausius Theorem (http://en.wikipedia.org/wiki/Clausius_theorem). Another way of writing it is $dS=\delta Q/T + dS_{irr}$ where $ dS_{irr}$ is the entropy change due to irreversibilitiy in the closed system ...


2

Ever heard of the cosmic microwave background? The CMB is a relic from when the universe became "opaque" - when, as Wikipedia says, protons and electrons combined to form neutral atoms. These atoms could no longer absorb the thermal radiation, and so the universe became transparent instead of being an opaque fog. So photons decoupled and the CMB was ...


1

Interesting question. I would have thought that if you were aware of the exact number of energy states and the populations thereof, you could apply boltzmann statistics to each of the levels in order to fit an appropriate temperature to the population in each state. This temperature, if comparable amongst the included levels, would therefore require that the ...


0

Since the question is not constrained in any way, the answer is yes, it is possible. Is it "realistically" possible, no. To make it possible, all you have to do is concentrate the moonlight (optically,or electrically) until you have the energy required to reach the ignition point of the material. what makes it not realistic is the size, cost, and/or time ...



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