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## New answers tagged thermodynamics

0

As as been pointed out by @engineer, this one indeed is a straight-forward homework question. The answer would be 0 J. The explanation for the answer lies in a fact that the internal energy of an ideal gas depends only on its temperature. This page, will give you some angle of explanation provided, you also know calculus. Also look up Free expansion of a ...

1

you're still converting the same amount of energy to heat, correct? Yes. In other words, brakes made from the same material on the same wheel should experience equivalent wear if each one converts X Joules of kinetic energy to heat energy, no matter if they are applied on the front or back wheel? I don't know. Why do you expect that "wear" is ...

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I don't have enough "points" to leave this as a comment: it appears the question asks how much energy is required to brake, not how much kinetic energy is converted by braking.

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Actually, if you apply the front brakes, there is a tendency to topple, while if you apply the rear brakes, there is a tendency to skid. Now, as the linear momentum is converted to angular momentum in this case, by an impulsive brake force, there is some loss of energy. Also, the time of contact of the brakes is less. That is why applying the front brakes ...

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The answer depends on whether the wheels skid. When you brake with just the rear wheel, it's quite possible to skid; if you apply the front brake, the increase in normal force on that wheel tends to prevent skidding (although in extreme cases it could make you fly over the handlebars). Applying the rear brakes hard enough to block the wheel would generate ...

1

It doesn't matter if the brakes are the same. The bike+rider has a certain kinetic energy. If you stop it, you need to dissipate that much energy. The problem with braking the rear wheel is that braking reduces the load on the rear wheel making it prone to skid. If you brake gently enough, the wheel will not skid. You will then dissipate the kinetic ...

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Intuitively, the moment of inertia of a single atom is far smaller than a diatomic molecule because the nucleus is at the origin, while in a diatomic molecule the nuclei are half the bond length from the origin. The minimum excitation energy for rotation is then much higher, well above room temperature, so it doesn't contribute, because $E=\frac ... 1 As you calculated, but seem to dismiss, the temperature change in water pumped up from deep, high-pressure zones to the surface depends very little on the change in pressure. Other factors would have far more effect: Pumping energy loss into the water Conduction from the pipe and its surroundings as the water rises Friction as the water moves through the ... 0 Ideal gas equation of state - $$PV = Nk_bT$$ For ideal gas (can be calculated directly from entropy (Sakur-Tetrode) or via equipartition theorem) - $$E = \frac{3}{2}Nk_bT=\frac{3}{2}PV$$ thus - $$H = E+PV=\frac{3}{2}PV+PV=\frac{5}{2}PV=\frac{5}{2} \frac{2}{3}E=\frac{5}{3}E$$ 0 In slow expansion the gas is always in equilibrium state. Rapid expansions are basically rapid so there is no time for energy transfer so they are adiabatic. 1 The internal energy of a system is directly proportional to its temperature. Formally, $$E_{sys}=\frac{3}{2}RT.$$ You could then note that $$PV=nRT=H_{sys}-E_{sys},$$ or $$H_{sys}=RT\bigg(\frac{3}{2}+n\bigg)$$ or, identically, $$H_{sys}=\frac{3}{2}RT +PV.$$ Your method should work, however, this is in my opinion a more "elegant" solution. 2 I think you have a fundamental misunderstanding of what the heat death really is. Any observer, whether they are a time traveler, observer from another universe, or whatever, would just see a lot of empty space. The first thing to know is that the heat death is not a single event. The universe, after heat death, is dead in the sense that nothing is ... 0 Most physics education teach learners about slow vs rapid expansion of gases in thermodynamics. This is often misleading, but only because many miss the point that slow expansion accounts for heat flow to and from the gaseous system and is assumed that the gas reaches thermodynamic equilibrium with the external system; and that rapid expansion assumes heat ... 4 I would say it looks like nothing. The heat death requires that the whole universe is thermodynamically homogeneous, and that the universe has reached its maximum entropy. This means that every thing becomes a disordered lump of very sparse matter, without anything to see whatsoever. It's as if the universe is in a state akin to the "chaotic nothingness" ... 13 From a thermodynamical point of view, living beings are able to reduce their entropy by exporting entropy to the external world. This does not contradict the 2nd principle, since living beings are open systems. For this reason, in a thermodynamically homogeneous universe (heat death), no change in the entropy can occur, and consequently no living beings (nor ... 1 Here's a hands-wavy explanation, which might help you're intuition. The alternative to step 4 and 5 would be to vent the steam to the atmosphere. But the steam is still much hotter than the atmosphere, so a lot of energy will be wasted. So you want to recapture that energy, and the logical place to put it is back into the boiler. However, taking it out ... 1 Your model is correct. The speed of the transfer depends on the difference in temperature and the thermal resistance of the glass and the water, which is constant. Interestingly the speed is not constant. The closer you get to the final temperature the more it slows down. Technically you never really get there exactly. There is no finite time where you can ... 1 If you place water (or other material) in a pressure-tight container, the water will change as heat and pressure cause its molecules to become more or less energetic and the bonds among its molecules to become more or less stable, or begin breaking apart. These changes are summarized in a chart called a phase diagram. Here is a simple phase diagram for ... 0 First of all, this is called flow work and it is calculated in this way when we have when the system is irreversible in nature when work crosses the boundary when the thermodynamic system is an open system e.g. turbines, compressors, pumps etc etc then we use integral -vdp to calculate the flow Work 1 Without knowing a lot more about the details of the entire system, we can't say exactly what the difference will be between the AC1+AC2 case and the AC2-only case. However, there are a few things we can work out from simple energy balance. In the following, I'm considering the air as an ideal gas with a fixed specific heat capacity at constant pressure ... -1 The mass of the vegetables would be the same, but there might be some water (extra mass) on the surface of the vegetables. They can't loose mass if they're at some recipient that keeps everything together. The properties of the vegetables susceptible to change when you put them inside a fridge are: temperature, volume, density/elasticity and stiffness. 0 If you're objective is to cool down the office, and you have a machine blowing air into the office which is warmer than the current (indoor) ambient temp. then I would by all means turn the damned thing off! air conditioners are heat pumps, just like a fridge. They take in the ambient air, pass it over a cool surface (to absorb the energy from the air) and ... 0 By saying that it is insulated they are saying that heat can't enter or leave the system. So you have no change in entropy due to$\Delta S = \Delta Q/T$. So if the entropy is constant throughout the process you can use the reversible adiabatic formula. And yes the temperature and pressure do change as you compress the gas (they increase). Strictly speaking ... 0 Thermal energy is exactly the average (with respect to the time interval of your measure) of the overall translational kinetic energy of all the particles of your system. This, in turn, can be related to the temperature of your system in case the Hamiltonian is separable into the coordinates of each one of your particles (the equipartition theorem). In ... 4 Checking for electron degeneracy is a matter of comparing the Fermi kinetic energy with$kT$. If$E_F/kT \gg 1$, then you may assume the electrons are degenerate. The central density of the Sun is around$\rho=1.6\times 10^5$kg/m$^3$and the number of atomic mass units per electron is around$\mu_e =1.5$. The number density of electrons is therefore ... 2 That is the heat equation in polar coordinates with axial symmetry. The (isotropic) heat equation without sources or sinks is $$\frac{\partial U}{\partial t} - K\nabla^2U =0.$$ If you look up the Laplacian operator in cylindrical coordinates, you will find that your expression matches this exactly. 0 Wondering if this falls in the category of convection as a mode of heat transfer? Yes. You are discussing heat transfer due to air traveling through a crack in a wall. Any heat transfer due to a moving fluid is convective heat transfer. If there is wind, then it is further categorized as forced convection. If there is no wind, just bouyancy ... 0 My guess, if the source of heat is in the centre of the room, is a mixture of radiation directly to the walls, plus convection of the air to the walls. When the heat gets to the walls, it's conduction through the walls, unless there are holes, cracks or vents in the walls, when it's convection of air through the walls. Warm air moves outwards, from the ... 7 As mentioned in the comments, this is an instance of supercooling. When you cool a liquid below its freezing point, the molecules are still moving around quite a lot and any two that stick together are likely to be broken up by a subsequent impact. Liquids freeze better when the molecules have something to latch onto -- either a block of the same ice they ... 0 The answer to the question you are trying to ask is yes, but I will need to do a bit of explanation. First, be aware the entropy is like mass or volume, in the sense that if you have two copies of something, the two copies will have twice as much entropy as the single object, exactly the same way they have twice the mass or volume. Because of this, you ... 3 The first part is handled well by Chris Drost - the kinetic particle energies are a lot larger than their interaction energies, so the gas can be considered (approximately) ideal. The last part - yes, as long as the Coulomb energy is a lot lower than the thermal energy then the protons or He ions can be considered an ideal gas with the appropriate average ... 3 since P is a constant and can be taken outside of the integral There is no reason whatsoever why$p$should be a constant, unless specified so; in particular, in your exercise the task is to find a solution for isothermal transformations. For gases and fluids$p$is a function of the volume and other variables as well, therefore the equation becomes $$... 7 According to a NASA page, the density in the middle of the Sun is about 150 g/cm3. That's about 9 × 1025 protons in a 1cm3 box, or 450 million to a side, and using that spacing for a voltage calculation reveals a typical interaction energy of 65 eV or so. (If you've never seen this unit before, that is the energy used by a 1V battery to move an electron's ... 1 Summary: Based on provided data you will be able to comfortably bring the water to boiling point in 1 hour. The energy input required to heat the water alone is significantly less than the available energy and the difference is greater than typical thermal losses. ie Water_mass x delta_temperature x Water_specific_heat < energy input Details below. ... 1 You may have seen the reasoning to follow in most textbooks already but apparently it is not emphasized enough so I will say it again here. The crucial starting point is the second law of thermodynamics that claims that the entropy change of the universe \Delta S_{univ} is either zero or strictly positive for any physical change that occurs in it. I ... 2 In today's understanding of Nature, there is nothing completely isolated. So technically there will always be interaction with the surrounding, at least from a quantum physical perspective. Here vacuum is not empty i.e. it does allow for electromagnetic interaction and there will be heat loss due to these vacuum effects. Furthermore also the other concepts ... 1 This phenomenon is due to the presence of air bubbles in the water. First note that the solubility of air in water decreases as temperature increases. Therefore when water is heated, less air can be dissolved in the water and when the water leaves the highly pressurized pipes, the air within the water is able to form bubbles and escape into the surrounding ... 0 \beta is not a property of the rod alone. It depends on the rod's interaction with the rod's surroundings. Let us consider some of the possibilities: The rod is surrounded by a fast moving fluid. This situation is called forced convention. \beta will depend on the speed of the fluid as well as several fluid properties such as its viscosity and ... -1 Ah, but who says that negative absolute temperatures exist at all? This is not without its controversies. There's a nature paper here which challenges the very existence of negative absolute temperatures, arguing that negative temperatures come about due to a poor method of defining the entropy, which in turn is used to calculate the temperature. Other ... 4 In a given orbital, electron motion has nothing to do with temperature. Atoms do have a variety of electronic states and, at higher temperatures, the higher energy states are more likely to be populated. Temperature, however, is most commonly determined by the translational motion of the nucleus of the atoms. Let v be the speed of a nucleus of an atom ... 1 Statistical Equilibrium: That state of a closed statistical system in which the average values of all the physical quantities characterizing the state are independent of time. In other words when systems do not evolve in time i.e. when they are in steady state. In thermodynamic equilibrium there are no net macroscopic flows of matter or of energy, either ... 0 TVOC 1.pdf Snell's Law.pdf Optical WedgeA.pdf http://www.raydextech.com/led-collimation.html 3 Since state \sigma is not in thermal equilibrium I don't think one can use your definition of "thermodynamical" entropy. In fact, one should instead use Von Neumann entropy, which is a correct measure of statistical (so not quantum!) uncertainty. There is no other "classical" or "thermodynamical" entropy in quantum systems. As you mentioned, for a thermal ... 0 When a billiard ball is pushed by a cue, the K.E. of the ball increases as it gains a velocity from rest ( supposing it is at rest). Well, then the internal energy of the ball increases. Thus the work done is done by you on the ball and definitely non-zero. 0 The First Law formula you cite is for work done BY a system. If the system is a billiard ball and you hit it with a cue, you do work ON the system. The formula in that case would be delta U = Q + W. The work increases the internal energy of the billiard ball. 0 To speed up the ball, you must move it over a short distance x while applying a force F. Thus you add work W=Fx. Kinetic energy is internal energy. The work you do raises \Delta U in the shape of kinetic energy K. In general the 1st law \Delta U=Q-W consists of: W work: When a force causes motion (like expansion/compression, ... 1 In mechanics, a mass m experiences a force \textbf{F} along some path C. The work done on the mass is given by$$ W = \int_C \textbf{F} \cdot d\textbf{r},$$such that the energy of the mass increases by$W$. Positive work corresponds to energy being added to the system in question (which is inevitably taken from the surroundings). Edit: To answer ... 1 According to the phase diagram of diamond (see for example http://files.umwblogs.org/blogs.dir/6093/files/2011/10/carbon_phase_diagram2.jpg) there is a region where diamond is stable and graphite is metastable, at pressures and temperatures in the range you are asking about: Given the slope on the diagram, you would actually expect that this reaction is ... 1 Chemical vapor deposition is used to produce diamonds at low pressure https://en.wikipedia.org/wiki/Synthetic_diamond#Chemical_vapor_deposition 0 If you take liquid water for example and start heating it at atmospheric pressure, the temperature increases. It corresponds to following a horizontal line along$p=p_{atmospheric}\$ on the above graph. Under liquid state, the volume is small hence it corresponds to the left region on the graph. If you keep heating, the liquid will start turning into ...

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