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What the book is saying is that your system has to be in thermal equilibrium at every point in time during this process. A quasi-static process is slow with regards to the relaxation-time of your system. If your system is in thermal equilibrium (constant N), that means it can be described by 2 variables - i.e. pressure P and volume V. A system that changes ...


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Often this reaction can occur simply due to the release of pressure allowing the chemical change to occur. I used to work at a fast food restaurant, and the frozen cola dispenser was always over-pressurised, meaning that new staff would fill the cup with frozen cola, and then panic as it continued to "grow" in the cup. The liquid was also supercooled, and ...


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In fact, $\delta Q/T$ is the entropy of $\delta Q$, and if $\delta Q$ is considered as the heat energy in transfer, it follows that $\delta Q/T$ is the entropy in transfer. For the inequality \begin{align}\oint dS \ge \oint \frac{\delta Q}{T}.\end{align} Whether reversible or not does not need to be considered for $\delta Q/T$ in that the entropy production ...


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Some safe distance could probably be computed for the long (indefinite) stay. However human body (mostly water) has a huge thermal capacity and lots of heat should be absorbed just to raise its temperature by at least six degrees (36°C to 42°C) to dangerous level. Before this happens, it is possible to stay in the zone where the equilibrium conditions are ...


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I got close enough to slowly flowing lava to stick a rock hammer in it, but you had to pull back quickly -- it felt like a bonfire. It was tolerable 8 feet away. The lava was about 6 inches thick, oozed less than an inch per second and showed orange-red on an advancing toe that was only about six inches in diameter. The rest of the flow was silvery black ...


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As to the data, of course it varies from one lava flow to another, let alone explosive eruptions, but at http://www.youtube.com/watch?v=4b6n8riJaFo you see a man walk onto the air-cooled crust of a current lava flow and draw a still molten sample from it. Related news videos show lava flowing across vegetation and houses slowly enough that you could easily ...


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More broadly, the free energy functions for phases of a number of elements, including carbon, are gathered in 'SGTE Data for Pure Elements', A. T. Dinsdale, CALPHAD 15(4), 317-425 (1991). This provides a self-consistent set of free energy functions to use in phase diagram calculations. For carbon, the source listed is P. Gustafson, Carbon 24, 169-176 ...


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It's tempting to think that heat will flow more readily across a cavity from a high emissivity surface to a low emissivity surface than in the opposite direction when the surface temperatures are interchanged, but this is fallacious. The reason is that repeated inter-reflection between the surfaces restores symmetry. Call the radiant flux across the cavity ...


2

If we assume that both the aluminum foil and the paint have scalar heat conductivities then their composite will also be scalar, and thus symmetrical. Being both material either poly-crystalline or amorphous this is probably reasonable assumption. @Floris suggested that I expand on this, but not being my area I can only summarize a few ideas from ...


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The factors that most matter when you are near lava: The fractional solid angle of lava as subtended at the observer ("how much lava do you see") The temperature of the lava The reflectivity of the clothing you are wearing Any effect of air flow (wind blowing towards lava or away from it) Toxic fumes... In essence, if we treat lava as a black body ...


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One should never write $d S_S= \delta Q_{E\to S}/T_S,$ unless it is a reversible process in the system, as well, because the temperature $T$ is whatever temperature the heat is supplied and that is the temperature of the environment (heat reservoir), presumably behaving independently of the amount of heat it supplies. The reservoir is always assumed to ...


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Heat is defined as a transfer from the environment to the system. No subscripts are needed for that purpose. I think you would have to be very clear about the specific situation for the other processes that you mention before there can be any discussion about them. Friction where? Turbulence implies a non-equilibrium condition...


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The second law of thermodynamics forbids materials that conduct better in one (forward) direction than the reverse direction - such a material placed between two containers at thermal equilibrium would drive the temperature away from equilibrium, decreasing the entropy of the whole system and paving the way for a perpetuum mobile... Reflectance and ...


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I'd look at it as an energy storage vs loss situation. Take a patch of earth (square slab) and neglect rotation of the earth around its axis (days) so that the patch always faces the sun. At any time it's receiving an incident solar flux (assume constant) and emitting due to its own temperature. The slab also has some thermal mass (capturing the ground, ...


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The main problem with your approach is that you are using the wrong area for the radiation "window". The area over which the exchange of radiative energy can take place is just the window in the door - the walls "see each other" and that part of the radiation has no net effect. So you want to use just $H\cdot W$ for the area. Secondly the window is ...


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Experimental data is given in http://journals.aps.org/pr/abstract/10.1103/PhysRev.98.889 - unfortunately I only have access to the abstract. It may be worth taking a look. The shape of the cathode does not matter. The material does. Key to solving this problem is knowing the work function of the material - that is the minimum energy that an electron needs ...


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The definition of a physical concept can be a differential form but can’t be the difference of functions. $\Delta S=S_{final}-S_{initial}$ is an equation but not the definition of entropy. Thermodynamics itself now can hardly explain “what is the entropy really" , the reason please see bellow. 1.Clausius’ definition \begin{align}dS=\left(\frac{\delta ...


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The Stirling cycle as you describe it is not reversible. The transfer of heat from thermal reservoirs along paths 4->1 and 2->3 is not a reversible process, because heat is being transferred between two objects at different temperatures. To reverse the process, you would need to spontaneously transfer heat from a colder to hotter reservoir, which violates ...


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There's actually not one simple answer to your question, which is why you are a bit confused. To specify your problem fully, you must specify exactly how and whether the gas swaps heat with its surroundings and how or even whether it is compressed. You should always refer to the full gas law $P\,V=n\,R\,T$ when reasoning. Common situations that are ...


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Thermodynamics only deals with homogenous systems? That is, we suppose always that pressure, temperature and all those quantities are the same on the entire system under study? Of course not. Thermodynamics would be a rather useless field of study if it only addressed homogeneous systems. That thermodynamics does far more than that is what makes it ...


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This topic makes some sense, “Thermodynamics quantities like pressure, temperature and entropy are associated with overall states of a macroscopic system” or associated with overall states of a local, this is a fact. In the equation \begin{align}dU=TdS-pdV+Ydx+\sum_j\mu_jdN_j\end{align} $T, S, p, V $ “are associated with overall states of a macroscopic ...


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Even assuming all of the energy it gains is turned into thermal energy (heat) it's just too small to worry about. $$\Delta KE = mg \Delta h$$ Let's assume it falls $2m$. $$\Delta KE = mg (2m)$$ Given that energy, we can compute the heating of the water. $$ Q = m C_p \Delta T$$ $$ \Delta T = \frac{Q}{mC_p}$$ $C_p$ for water is $4.18 \frac{J}{gK}$ $$\Delta T ...


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Temperature doesn't rise as an object falls. Temperature is just the average kinetic energy of the individual particles within an object, but their individual velocities is random. A water droplet falling is an ordered falling that you can consider as a form of work, and you feel it as the splattering on your feet.


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Dielectric heating is the priciple of a microwave oven. Water $H_2O$ has a strong dipole moment. Since the water molecule is not linear and the oxygen atom has a higher electronegativity than hydrogen atoms, the oxygen atom carries a slight negative charge, whereas the hydrogen atoms are slightly positive. As a result, water is a polar molecule with an ...


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Thermodynamics does deal with inhomogeneities and non-equilibrium conditions... ...but indeed it requires some "macroscopic smearing" of quantities. Problems like diffusion of particles, or heat are dealt with and very well explained by thermodynamics, specifically with Maxwell's Thermodynamic Potentials formalism. But the thing is, the thermodynamic ...


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From a heat transfer perspective, steam is a lot more like air that it is like liquid water. The chemical makeup doesn't matter as much as the state. As a gas, steam is a bunch of $H_2O$ molecules flying around at random, bashing into each other occasionally. The mechanism of conduction in that case is that the hot molecules will gradually bounce past the ...


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I assume that when you say 'microwave' you are talking about a microwave oven. As you say sand unlike water dense foods cannot conduct electricity. But in a microwave the foods are not heated up using electricity but by bombarding it with microwaves. The microwave use electricity to generate microwaves of frequency 2.4 GHz. In your question you have already ...


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You can break the system down into smaller subsystems. If we have a room with an inlet of air at one corner and an outlet at the opposite corner, we can make a grid of cells of whatever spacing we want. We can then assume the pressure and temperature are uniform within the cell and compute the flows between the cells. This is what the weather people do. ...


1

I can't think of any source that would claim you need homogeneity to do thermodynamics. Textbooks might usually assume systems have a single constant pressure and temperature because it's easier, but it's not a requirement. Intensive variables can be functions of position. Any discussion of the buoyant force needs position-dependent pressure. Any discussion ...


0

when heat is provided to a metal (which is a conductor) it increases the vibrational motion of electrons inside the metal surface. as heat goes on increasing like u said to 100 degrees the vibrational motion(energy) of electrons become so fast dat it over comes the attractive forces between the electrons and the nuclei to which they r bounded.so they leave ...


0

OK, I think I've got it, thanks to your comments above as well as this link, which shows how to calculate the temperature of a solar oven. (My situation is very similar to a solar oven, except that the power dumped inside the craft is electrical -- but watts are watts, right?) So, I believe that what I need to do is: Calculate the steady-state ...


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According to Wolfram|Alpha, for Copper it would increase by 39.2% while for Aluminum it would increase by 41.6%.


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Typically, for a conductor, the coefficient that relates the resistivity of the material with temperature is positive. This means that when the temperature increases, the resistivity increases as well. Here you can find a table with various coefficients, and an online calculator. The relationship can be described by $$ \Delta\rho = \alpha\cdot \Delta T + ...


0

This page gives the temperature coefficient of resistance of copper (near room temp) as $0.004041 K^{-1}$, so an increase of $100K$ would increase the resistance by a factor $1.4041$


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Internal energy is due to motion of particles in a system. As internal energy depends on temperature. As we know temperature in isothermal process is constant so the internal energy will also be constant thus the change in internal energy will be zero.


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To see that you are correct, look to radiative heat transfer amongst black bodies. Consider two black bodies, call them bodies A and B, arranged as flat plates facing one another. Suppose body A has a temperature $T_A$ and body B has a temperature $T_B$. Both plates are black bodies, so each radiates energy at a rate given by the Stefan-Boltzmann law: $dE/dt ...


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As @DavidVercauteren hinted in his comment and @user31748 pointed out in his answer (albeit with other words), the problem here is that the process undergone by the system is in fact not irreversible: the irreversibilities occur elsewhere. There is a missing piece to the puzzle, which, after some thought, I present below: The key here is that the ...


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Regarding the first viewpoint, it has also been long emphasized that thermodynamics is about studying phenomena related to a property of equilibrium systems called temperature. The equilibrium is always important in all thermodynamics --- although you can make up some measure of temperature for a nonequilibrium system, none of the familiar thermodynamic ...


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If two systems $X$ and $Y$ are in contact such that they can exchange energy, then the statement $X$ and $Y$ are in equilibrium. is equivalent to the statement $X$ and $Y$ are at the same temperature. Here is a proof. The total entropy of the combined system is $$S = S_X + S_Y \qquad (1)$$ where $S_X$ is the entropy of system $X$, and ...


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Here's a point of view from thermodynamics that might be useful. Typically, the intensive quantities (in the form they're usually defined) arise as derivatives of the total (internal) energy $U$ by some particular extensive quantity. Thus: Temperature $T=\frac{\partial U}{\partial S}$, the derivative with respect to the entropy Pressure $P=-\frac{\partial ...


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This is a very interesting question. Velocity can be considered as either an intensive or an extensive property, depending on whether we are inquiring about the parts of a single system, or considering relations among separate systems. Velocity must be an intensive property, for consider: If I and my passenger and my books are traveling in my car, and if ...


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The $Q=H$ refers to the system itself, the $2.3MJ$ latent heat rejected by the $1kg$ steam as it condenses to liquid at $100C$ is the same heat that would be measured if the surroundings were at $99.9999999999C$ that process being effectively reversible, and therefore then we would have $\Delta S=0$. But since $Q$ is rejected to the environment that is at ...


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If there's hardly any ice at all, the water will cool as the ice warms to 0°C, then cool some more as the ice melts. The cooling stops when the ice melts. If there's hardly any water at all, the water will cool to 0°C. If the ice is cooler than 0°C, that tiny bit of water will freeze. If there's an intermediate amount of water, the water will cool to 0°C ...


1

Yes, your analysis is correct. Water in equilibrium with ice is at a temperature of 0 degrees C. The reason that the water doesn't spontaneously turn to ice has to do with the latent heat of fusion of water: in order for water to turn to ice at zero degrees C, you need to remove quite a lot of heat from it. In the case of water / ice, the latent heat is ...


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Correct me if I'm wrong, but your line of thinking goes like this... Since quantum fields do not commute in general one can have finite variances for, e.g., particle number. Since the vacuum states defines a probability distribution we can find the corresponding entropy. However, here we are dealing with quantum physics. The entropy is in general $S ...


1

Essentially from the definition of thermodynamic equilibrium yes. The Zeroth Law of Thermodynamics states that if a body $A$ is in thermodynamic equilibrium with $B$ and $B$ is in thermodynamic equilibrium with $C$ then $A$ is in thermodynamic equilibrium with $C$ and we say that $A$, $B$ and $C$ have the same temperature. This is precisely the setup you ...


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You can combat that the same way we avoid getting too cold: apply insulation. The outer surface of the spacecraft may be very cold, but that doesn't mean the internal temperature is that cold. That is what the greenhouse effect does-it insulates the surface from space. Power dissipated inside the body all becomes heat. If you have solar arrays making ...


1

You may start from the density matrix $\rho$ of the grand canonical ensemble (assuming $k_BT=1$ as the unit of energy), $$\rho=Z^{-1}e^{-H+\mu n},$$ where $Z=\text{Tr}\,e^{-H+\mu n}$ is the partition function. Then $\langle n\rangle$ and $\langle n^2\rangle$ are defined by tracing with the density matrix $$\langle n\rangle=\text{Tr}\,n\rho\text{, and ...


1

It is more complicated than the physics of heat transfers. Our tactile sensations are pretty weird. One example would be that humans can taste "cold" and "cold" affects other tastes. There is not enough research done regarding the processes. In the multitude of cutaneous receptors, you have several that relate to temperature. One type of nocireceptors, ...


3

Interesting and complicated question. The things to consider: "Black body radiation" assumes perfect absorption / radiation at all wavelengths. The greenhouse effect comes about from having absorption in the IR: the hot (short wavelength) radiation from the sun can penetrate the atmosphere, but the cooler earth radiates at a lower temperature - longer ...



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