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Got it! as $H=U+PV$ and $PV = n_{tot}RT=const$ throughout the process we get $\Delta(PV)=0 => \Delta U= \Delta H$

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When a gas expands, it has do pressure-volume work against the piston, atmosphere etc. to make its room to occupy. Now, as the system spends energy to work, part of its kinetic energy decreases. Since, temperature is directly proportional to kinetic energy, the system's temperature decreases. An example: Much of chemistry takes place in vessels that are ...

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Intuitively, For a gas,if you apply heat to the container of gas the kinetic energies of the molecules or atoms increase,means heat added is used in increasing the kinetic energies of the molecules. As we know ,temperature of a gas depends on how fast the molecules of gas moving or vibrating ,so on heating temperature of the gas increases. Now these ...

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It means that the heat supplied to the gas is used in its expansion. From the text you quoted, it seems that part of the heat energy supplied to the gas is used for its expansion. This is what you usually see around you: when you heat a gas, that gas can expand a little but this does not mean that you don't manage to heat it.

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This is a difficult question for many reasons. One reason is likely because most of the introductory thermodynamics textbook problems that we are familiar with from childhood do not involve gravity. To illustrate this difficulty with gravity consider, for example, this snippet from an article in the New York Times Review of Books by physicist/mathematician ...

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If we consider temperature to be due to translational motion of the molecules and we assume the system has reached equilibrium, then the velocity distribution of the molecules is given by the Maxwell distribution: $$f(v) = \sqrt{\left(\frac{m}{2\pi k T}\right)^3} 4 \pi v^2 \exp\left(\frac{m v^2}{2 k T}\right)$$ which will give you the velocity ...

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Your error is in assuming that the efficiency of a non-reversible heat pump is the reciprocal of the efficiency of the corresponding heat engine. A reversible engine can be run backwards to act as a reversible heat pump, and the efficiency when acting as a heat pump, $\eta_p$, is related to the efficiency of the heat engine, $\eta_e$, by: $$\eta_p = ... 1 There is a degree to which this is just terminology, but in cosmology a distinction is somtimes made between the Heat Death and the Big Freeze. The Big Freeze is the point at which the universe reaches absolute zero, while the Heat Death is the point at which the entire universe has a constant temperature. These are not necessarily the same thing, because a ... 0 Feeling cold to the touch is slightly complicated, as are all of our senses. In as much as what we feel as temperature is related to actual temperature at all, it is the temperature of our skin or perhaps the amount of heat our body must supply to keep that approximately where the body wants it to be. Heat is the energy in unordered motion of atoms. Hence ... 2 Your breath is the same temperature either way. The difference is how much ambient air is brought along with the breath by the time it reaches the object. Emitting a thin and fast stream of air will cause a lot of other air to follow along with it. When you are blowing on the soup to cool it, what you're really doing is using your breath to move a lot of ... 0 The Molar Specific Heat at constant volume C_v is:$$\frac{fR}{2}$$and the Molar Specific Heat at constant pressure C_p is:$$\frac{(f+2)R}{2}$$(f is the degrees of freedom of a molecule of the gas in concern, and R has its usual meaning.) If you verify, the difference (i.e. C_p-C_v) is always R regardless of the value of f. 12 You're not getting enough Make-Up Air. I strongly encourage you to install a CO detector in you home and reassess your HVAC situation (especially the 'V' part, ventilation). Lack of sufficient makeup-air in a house with gas fired equipment is dangerous. This is usually more of a problem with new-construction homes that are built to be nearly hermetic. ... 1 This is a typical problem when starting up a fireplace, wood stove, or what-have-you. The air pressure inside your house at 75 degrees is lower than the outside air pressure at 40 degrees because warm air is less dense than cold air. Result? The higher-pressure air outside the house wants to flow down the chimney and into the lower-pressure area inside your ... 3 There seems to be a slightly lowered pressure inside the room the fireplace is in, which leads to air actually streaming down the chimney to equalize it, and this blows the flames into the room. When you open the door, you allow air draught to stream into the room and up the chimney, as it is supposed to, and that sucks the flames into the chimney. ... 14 There's two possibilities that are immediately obvious. The first is that the pressure inside the house is slightly lower than the pressure outside the house before you light the fireplace. This would cause air to flow down the chimney into the house which would keep push the flame into the house instead of up the chimney. This would be easily testable if ... 4 You need a "draft". When you light the fireplace, initially the heated combustion products want to rise directly up the chimney, but after that occurs for a few seconds it results in a partial vacuum in the house, attempting to suck air back down the chimney. This cause the fire to be blown/pulled outwards into the room. (With a conventional wood-burning ... 0 Hot always goes to the lesser degree it is faster depening on mass Convection on speed of air has to be considered also 0 In general, the idea looks like this: you keep track of a "stuff" (in this case both air and energy) as it flows into and out-of some "perfectly mixed reactors" (in this case "reactor" means a tank where things which react are often stored). The "stuff" can be anything which obeys a conservation law in that context, so momentum and energy and volume (at ... 0 If you have a volume of air V at temperature T_B, then you replace a part of that air with air of volume \Delta V and temperature T_A, then the new average temperature is a weighted average of the temperatures of the room's air and the new air.$$T_B(t+\Delta t) = \frac{(V-\Delta V) T_B(t) + \Delta V T_A(t)}{V}$$We get a symmetrical expression for ... 1 You're correct an adiabatic expansion results in work being done by the gas, thus has a positive sign. This is seen in part b of the 'an example of work done' section. The section you are referring to - 'adiabatic processes' - states that the change in the internal energy of a system, U, is equal to the negative of the work done (U will be negative if the ... 3 Generations of physics students, including me, have got mixed up about the sign of work done. That's because the phrase work done can mean work done on the gas or work done by the gas, and these are equal but with opposite signs. I don't think there is any perfect way to deal with this except by using your common sense. If an expanding gas does work then ... 0 Some comments on the premises may be useful and answer the question: [1] Unitarity of quantum mechanics prohibits information destruction. What this really means is the von Neumann evolution equation for density matrix D(t) of an isolated system prohibits change in the von Neumann entropy:$$ N[D]=\sum_{k,l} -D_{kl} \ln D_{lk}. $$Very similarly, ... 0 Simply because the Second Law of Thermodynamics states: Entropy > 0 (always) (think of taking a basketball and squeezing it into a golf ball) so that the density gets larger as the mass gets smaller. therefore, Entropy will always increase in a black hole, gaining Entropy [heat] as it gets smaller.The stars reaching critical mass(limiting mass) ~ ... 0 Now we do -ve work on the system while stopping it so the work done by the system is +ve. Is this part right? Not always; you've stumbled upon a subtle point in the definition of work in mechanics. Generally, if body A has done work W on body B, it does not follow that the body B has done work -W on the body A. This is true only if the two ... 2 The energy of any infalling mass is absorbed by the black hole. Classically, the temperature of a black hole is absolute zero, since it is a perfect absorber. If you include quantum mechanical effects, as Stephen Hawking did, you can show that black hole horizons will emit radiation in such a way that is consistent with the horizon being a hot body with ... 4 If I recall it correctly, the information sunk into black hole can be considered encoded in the ripples on black hole surface, much like egg impact parameters which could in principle be deciphered (at least partially; even quantum theories give us certain confidence intervals) from shattered egg fragments. Falling objects will necessarily have mass, and ... 1 "They can't erase it, hence they won't heat" : If they do nothing indeed they won't heat. Maybe you don't understand when erasing would be needed for a demon to work : imagine you have a box full of gas : you could put a wall in the middle, with a door controlled by a demon : the demon would let only in the high momentum particle, so that you can have 2 ... 5 I'm not too qualified to speak about the Black Hole Information Paradox, but I think I can say a thing or two about Maxwell Daemons. I think the essential "flaw" in your description is the assumption that Maxwell Daemons destroy information. The do not, and I describe what they actually do in my answer to the Physics SE Question "How can the microstates be ... 7 If you drop an egg then it looks as though the process is irreversible and that the information about the original state of the egg is lost. However this isn't the case. The equations describing how the egg shatters are all time reversible so in principle, if not in practice, we could take the shattered egg and evolve time backwards to reconstruct it. ... 0 The information interpretation of entropy, also known as Shannon entropy, interprets entropy as an increase over time in the amount of information contained in a closed system. This interpretation is appealing to information theorists, including many computer programmers, but it does, on the surface, appear to contradict the quantum mechanics conservation ... 2 No, because theres still the same amount of particles vibrating around. How fast the particles are moving around that actually matters not the number of particles. Ans will be yes. The system has done some work on the sorrounding (or work is done on the system) but no heat is added to the system(as adiabatic process) so the internal energy will now change ... 1 Ballpark (based on the iron starting out at 0 degrees Kelvin and melting at 1538 and the earth's radius of about 6000000 meters and the mass of a fly about 12 milligrams and velocity of a fly about 2 meters per second) (EDIT: Also based on the assumption of no radiative cooling of the sphere, i.e., perfect transfer of fly-bumping into heating the sphere and ... -1 Watching the video frame frame one sees that an incandescent ball is falling and a tail of gasses follows. The ball must have been ejected from the tube by the rest of the explosion, like a rocket, and falling, generates a tail of incandescent gasses. My argument against a reflection is that the ball ( falling in my view) and its cone/fan tail are too ... 4 It's internal reflection in the camera. It's an inverted image of the actual exploding material. 2 Yes, this works. It's called radiative cooling. This phenomena has been known for a long time, considering the ancient Egyptians used to make ice this way. Ideally, something open only to a clear sky would "see" the temperature of space, which is the microwave background radiation. In practise there is enough stuff in our atmosphere that radiates so that ... 0 With nothing but clear dark sky above, I assume there is nothing to radiate appreciable heat back into the box and maintain the object's temperature. In this case there wouldn't really be an "ambient temperature" though. To elaborate: The inside of the box would cool down until thermal equilibrium is reached between the inside and the outside. At this ... 15 I would suggest that the bottom penny does not have the same composition. A zinc penny is mentioned in the comments (BTW, what a tedious video) and zinc has a much lower melting point than copper (420 C vs 1085C). Copper plated zinc pennies were introduced in 1982. Given that it started to melt almost straight away, I think that is your answer. At about 10 ... -2 Entropy argument might not work here, because amount of information is going to be influenced by number of moles, not by arrangement of individual molecules. In the same fashion as 1TB hard-drive is 1TB no matter what is written onto it. However, more precise crystal structures will tell you exact picture stored in those bits. So measuring, essentially, ... 11 Quite simply, no. Water memory doesn't appear to violate any physical laws, and the claims made about it are not well-defined or specific enough to be falsified (e.g. with an entropic argument). It's revealing that while a scientist could be convinced that he's wrong, there's nothing that could change the mind of a homeopath. The best we can do is test the ... 0 Equilibrium is dynamic microscopically but macroscopically it is static because net flow is zero. 3 That is nothing but a mechanical relation valid for every type of isotropic fluid (even viscous, in the quasi-static regime), so gases in particular, no matter any further state equation. 1 Because conduction is the transfer of heat through a substances as a result of neighbouring vibrating particles, The particles in air are far apart. Heat does not travel easily by conduction through air. 0 The window will not protect you. Due to the Maxwellian distribution of thermal energies of the weapon plasma, a significant fluence of visible light can be transported through the atmosphere, the window, and your eye prior to cataract generation. So while the window can block the infrared, it won't block the visible. Nuclear weapons generate thermal ... 3 As you said, the relation you have is$$ dE = TdS + FdL $$So one maxwell relation is$$ \left(\frac{\partial T}{\partial L}\right)_S = \left(\frac{\partial F}{\partial S}\right)_L $$Which is the one you have obtained. For get the other do the following transformation$$ dE = TdS + FdL = d(ST) - SdT + FdL  \implies d(E-ST) = -SdT + FdL $$Since -SdT + ... 2 Specifically the result you have is$$\left(\frac{\partial F}{\partial S}\right)_L=\left(\frac{\partial T}{\partial L}\right)_S$$So using the cyclic rule for partial derivatives we can write that$$ \left(\frac{\partial T}{\partial L}\right)_S=-\left(\frac{\partial T}{\partial S}\right)_L \left(\frac{\partial S}{\partial L}\right)_T So if we sub this ...

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Okay, the "cylinder" thing means that we can think of this as a 2D system; we can assume that the temperature of the cylinder has reached a steady state. Then there are three constraints at play here: Circular symmetry. The temperature of the disk is uniform as the disk is rotated about its center. Equilibrium. When you draw a circle around the center, the ...

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Yes, all of them. You have reduced amount due to aeration (bubbles), cooling via convection and cooling via conduction into the side of the cup/bowl.

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So, this is an old post that I came across when I had a similar question. Here's a paper where they dissolve different amounts of ions in the water and found that the ability for the microwave oven to heat the water actually reduces as more ions are introduced.

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There's several things involved in your question. The first is what would realistically happen if you set up this system and you managed to measure its temperature. Here there's no easy answer: you have to consider in detail how you set up the system and how it will interact with its surroundings. Simple idealizations like "isolated" aren't adequate here ...

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Even if light from stars cannot enter the box, eventually the box and the surrounding space will come to thermodynamic equilibrium and the box will emit blackbody (or, more likely, graybody) radiation both inside and out and the equilibrium temperature attained. If you take the box far from any starlight there is still the sea of photons that make up the ...

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