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27

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 ...


7

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

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 ...


5

The simple answer to this question is that the specific heat capacity of a hot solid body you touch really doesn't matter too much for whether you burn yourself; what is much more important is its heat conductivity. The reason for this is that the surface of a hot body with low heat conductivity will rapidly cool down when you touch it (the blood in ...


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

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 ...


4

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 ...


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 ...


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

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 ...


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

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 ...


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

It's not the question asked, but looking at the power requirements might give some insight. Raising the water temperature requires a specific amount of energy, and the time constraint gives a required power. $$P = \frac{m C T} { t}$$ $$P = \frac{300\text{gallon }(1000\text{kg/m^3})(4.186\text{J/g K}) 14\text{degF}}{1 \text{hour}}$$ $$P = ...


2

Assuming no heat is lost to the environment the heat balance on adding some boiling water ($212\:\mathrm{F}$) is given by: $$m_{bath}cT_{bath}+m_{added}cT_b=(m_{bath}+m_{added})cT_f$$ where: $m_{bath}=300\:\mathrm{Gall}$ is the initial amount of water, $T_{bath}=32.2\:\mathrm{Celsius}$, $m_{added}$ the amount of boiling water added, ...


2

Your error in method 1 was assuming the U = U(T) and not U = U(T,A). But, you would have been much better off starting out directly with entropy S = S(T,A), so that $$dS=\frac{C_A}{T}dT+\left(\frac{\partial S}{\partial A}\right)_TdA$$Then from the Maxwell relationship, $$\left(\frac{\partial S}{\partial A}\right)_T=-2\left(\frac{\partial \sigma}{\partial ...


2

In the method 1, you expressed increase of energy during adiabatic process as $$ dU=C_A dT $$ where $C_A$ is presumably capacity when the area $A$ is constant: $$ C_A = \left(\frac{dU}{dT}\right)_A. $$ But $C_A$ is not the right factor to use for adiabatic process, because in this process $A$ is not constant, but $S$ is. The energy increase formula is ...


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,$$\frac{\partial^2 z}{\partial x \partial y}=\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$$


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 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 ...


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

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.


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

DanielSank's answer is 100% rigth about the temperature issue. The question deserves much more argumentation because the book you use, and many others, are plain wrong. I will use numbers and the laws of physics to show my point: it is not true that If the mass of an electron, the Planck constant, the speed of light, or the mass of a proton were even ...


1

A typical hot tub will be coming to an equilibrium based on some forcing term $F$ adding heat to the system plus some proportional response $\lambda$ which loses heat to the environment:$$\rho \frac{dT}{dt} = F - \lambda (T - T_0)$$This is a linear ODE whose equilibrium temperature is $T = T_0 + F/\lambda.$ To increase $T$ as fast as possible you should: ...


1

Phase changes depend not only on temperature but also pressure. Like any substance, if you get the right combo of pressure and temperature you can keep it at a solid, liquid or gas or if you're really good keep it at the triple point and get all three phases in equilibrium with each other. For C02, the liquid phase is very likely outside of most temperature ...



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