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39

With the same argument, I could deduce (and I know that it's wrong) that the cold air above is denser, so it will go down, pressing the hot air away sideways. Replace your hot air with a helium balloon. You can see there's no force on the balloon to push it sideways. The buoyancy forces it to accelerate upward (and some cool air around it to ...


9

Welcome to the magic of convection cells =) The first thing to remember is that you're working with a large number of gas molecules. The effect theoretically occurs no matter how many particles you have, but the effects are much easier to describe using bulk terms that handle many molecules at once, rather than trying to track each molecule. As BowlOfRed ...


7

If you've ever flown a glider, either a radio controlled model glider or a full-size man-carrying glider, then you'd be aware that cold air do in fact fall in a column. Glider pilots call hot air rising "thermals" and cold air falling "sink". Both move in columns, bubbles, sheets etc. For every "shape" hot air moves in upwards, cold air can also move in the ...


7

You ask why a column of hot air (as in a chimney) rises, given that denser cold air is above it, pushing down. It rises, because denser cold air around the BOTTOM of the chimney is under higher pressure than the cold air at the top of the chimney. The extra pressure due to a chimney-height of cold air is pushing it down. The lesser density of hot air means ...


7

Actually, the process you described is not irreversible. What has happened is that potential energy has been converted to kinetic energy. It is possible to reverse the process by using a roller coaster kind of arrangement where the ball is redirected to go up a hill. In the end, its kinetic energy is converted back into potential energy. There is no ...


4

You have to distinguish between "different states" and "number of states" - or, in the words of @Numrok, between "macrostates" and "microstates". The fundamental theorem refers to "accessible micro states". If I have three white balls and two buckets to put them in, I could put two balls in one and one in the other (that is a macro state); there are in fact ...


4

Background Generally, all phase transitions require some input energy in order for the transition to occur. For instance, the transition from solid-to-liquid or vice versa requires what is called the enthalpy of fusion or latent heat of fusion. This is the amount of energy needed to change the total interal energy (i.e., enthalpy) of a substance in order ...


4

I'll start by saying that correctly stating Kirchhoff's law is quite tricky. "Emissivity equals absorptivity" in a certain sense, but they may depend on wavelength, and angle of incidence (or emission), and polarization. In magneto-optic materials, you can have high absorptivity from one direction balancing high emissivity into a different direction!! (This ...


3

The ratio of emissive power to the absorbitity is constant when the substance is at thermal equilibrium with surrounding. Or The emissive power of a substance is equal to its absorbtivity under the same conditions.


3

Your question is how the differential quotient $\frac {dU}{dS}$ can mean anything in equilibrium when the quantities $U$ and $S$ are supposed to be constant, it is equilibrium after all... Indeed, $dU$ or $dS$ do not mean changes over time in a physical sense, ie., over time during some process, Instead they mean the differentials of the respective ...


3

I know it's frustrating because it's nonintuitive, but it really is true. You can read about etendue at wikipedia or your friendly neighborhood optics textbook. I recommend Smith's "Modern Optical Engineering" . Or you can settle for the 2nd Law of Thermodynamics (aka "TINSTAAFL"), as GLR wrote in that what-if. It's a matter of surface brightness, and ...


3

There are two things at work here - gravity, and gas pressure. In the beginning, due to gravity, denser air is at bottom and lighter air is at the top. You may ask why it is this way to begin with and the answer is in the word "denser". At every level, there is an equilibrium density in the beginning and it is higher at the bottom and lower at the top. When ...


3

A more thorough explanation involves the energy of the atoms/molecules in air rather than the density. What happens is a combination of the two following phenomena: 1) When you heat the air above the plate, the atoms in the air get energised depending on the temperature. This additional energy is manifest as the velocity of the air atoms. In doing so, they ...


2

Whether one sees a solid or a liquid or a gas is determined by the correlation properties of the molecules in the substance. Solids (in crystals) have strong and long-range correlations, liquids have no long-range but strong short-range correlations, while gases have almost only weak intermolecular correlations. In the part of the phase diagram below the ...


2

The issue with your model is you're only considering the density of the particles rather than their motion. If you heat up a plate, the molecules above it gain thermal energy and begin to move about faster and faster. They bump into each other and the colder particles at the boundary of the column of air above plate. Since the pressure decreases slightly ...


2

The statement "hot air rises" is not in general true, although often used. Instead, Less dense air rises Now usually, locally heated air will expand (because pressure will be similar to the pressure of the surrounding air) according to the universal gas law $PV=nRT$, and less dense air will experience buoyancy from the surrounding more-dense (cooler) ...


2

If you carry out an irreversible adiabatic process starting out at state A and ending at state B, you will not be able to identify a reversible path between the same two states that does not involve an exchange of heat with the surroundings. In other words, there is no adiabatic reversible path between states A and B. Here is the simplest example I can ...


2

I'll show you how the book derives that solution. First, we start with the definition for energy fluctuation: $$ \langle(\Delta E)^2\rangle =\frac{\partial^2}{\partial \beta^2}\ln(Z) $$ where $Z$ is the Canonical Partition Function. The heat capacity, $C$, can be defined as: $$ C=\frac{1}{k_BT^2}\langle (\Delta E)^2\rangle $$ Next, let's calculate $Z$ ...


2

If z is a function of x and y, then:$$dz=\frac{\partial z}{\partial x}dx+\frac{\partial z}{\partial y}dy$$So, $M=\frac{\partial z}{\partial x}$ and $N=\frac{\partial z}{\partial y}$. Since the order of taking partial derivatives is immaterial,$$\frac{\partial^2 z}{\partial x \partial y}=\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$$


1

This equation describes 3D transient heat conduction in a material, where $\alpha$ is the thermal diffusivity. The thermal diffusivity is related to the thermal conductivity, the heat capacity, and the density by $\alpha=\frac{k}{\rho C}$. The equation is derived by performing a differential heat balance on an infinitesimal volume of the material. It can ...


1

It depends on gas whether its monoatomic diatomic nonrigid ,rigid as $\alpha$ is degree off freedom which depends on gas for mono it s $3/2$


1

The answer to your query is Equipartition of Energy. Equipartition Theorem: At temperature $T\,_,$ the average energy of any quadratic degree of freedom is $\frac{1}{2} kT\;.$ For each degree of freedom, the ideal gas molecule can store $\frac12 kT$ of energy on average. For monatomic ideal gas molecule, there are only three degrees of freedom: ...


1

You say: We shouldn't care about how we reached that equilibrium In fact the entire point of the example is that we do. I shall try to explain why. Jaynes and Gull both work in the framework of Bayesian inference (I can recommend the introductory text: http://www.amazon.co.uk/Data-Analysis-A-Bayesian-Tutorial/dp/0198568320. The title may seem ...


1

Here you are considering 2 different processes: in case 1 the process occurs at constant pressure (that of the atmosphere); in case 2 it occurs at constant volume (that of the pot). The coefficients that relate the heat exchanged in the process with the change in temperature are the (molar) heat capacity at constant pressure $c_p$ and constant volume $c_v$ ...


1

The formula is valid, but you have to consider $Q$ to be the net amount of heat transferred into the water. In both cases, as the water gets hot, it will radiate some heat to the environment (because the region above the water is cooler). In addition, the lidless pot will lose heat (and some mass) through evaporation. The problem is that the amount of ...


1

No it doesnt necessarily mean that. In general, the internal energy is of the form,$$\d U = T\d S - P\d V + \sum \mu_{i} \d n_{i} + J.dx$$ where $J$ refers to a generalised force and $dx$ to a generalised displacement, and even $-P.dV$ can be written as product of a generalised force and displacement with pressure serving as the former and volume serving as ...


1

You need to collect some data to answer this question. Heat pumps can provide more heat to a building than the electricity they use. Whether yours does is something that needs to be measured. Real world efficiencies are often different from lab efficiencies. The radiator coils can get dusty and be installed in places with poor air flow, for example. You ...


1

The key to understand the complex first formula is that air is compressible, that is, its density changes with pressure. Your second formula, the simpler one, assumes a constant density in the fluid. It is useful when the medium is not compressible (such as the sea), or when the difference in height, thus in pressure, thus in density is small (such as from ...


1

For a height difference of 0.18 m the density of air changes very little so $h\rho g$ can be used. As the Wikipedia article says "At low altitudes above the sea level, the pressure decreases by about 1.2 kPa for every 100 meters and this is the constant density of air (1.23 kg/m$^3$) approximation. The full blown formula is for much higher altitudes. If ...


1

For a compressible gas, your elementary physics formula changes to: $$\frac{dp}{dz}=-\rho g$$whre z is the elevation (above ground level) and $\rho$ is the density of the gas. From the ideal gas law, $$\rho=\frac{pM}{RT}$$ where M is the molecular weight. If we combine these two equations, we get the "barotropic" ...



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