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72

There's two main misconceptions in your question that cause your confusion. First, pressure doesn't cause higher temperature. This misconception is probably a result of a massive oversimplification with relation to the ideal gas equation. The actual relation is "increasing the pressure of an ideal gas while volume remains constant increases the temperature ...


61

It's not so much the pressure, but rather compression that creates heat. Heat is a measure of increased kinetic energy as molecules are forced into a smaller space. Water is not very compressible, and water at the bottom of the ocean is not confined to a significantly smaller space under pressure. The kinetic energy of water molecules at the bottom of ...


26

We can understand all of this business if we visit the statistical mechanics notion of temperature, and then connect it to experimental realities. Temperature is a Lagrange multiplier (and should have dimensions of energy) First we consider the statistical mechanics way of defining temperature. Given a physical system with some degree of freedom $X$, ...


22

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 ...


20

Colder water is denser until it reaches a temperature a couple degrees above freezing, then it gets lighter again. So the water at the bottom is at the specific temperature where it is densest: any heating makes it rise. Any further cooling makes it rise. See Why does the ocean get colder at depth? This further points out that without ocean circulation ...


13

I personally find the terms consistent. Think of the entropy as Boltzman proposes: $S=k \, \ln W$ Meaning high entropy states can be realized via many different configurations. Truly ordered state (assume you arrange a sculpture from atoms) can be realized via much smaller number of microscopic states. So again, equilibrium is not order - it is a mess.


13

TL; DR The material with the greater effusivity will be more likely to burn you upon contact. Analysis of a simplified case First consider the case where your palm comes into contact constant temperature wall. Often, we can consider your palm as a semi-infinite solid. The requirement for the semi-infinite approximation is that $T(x \rightarrow \infty, t) = ...


8

What you are missing is the microscopic definition of entropy, once you know that, you will understand why people say that entropy is disorder. Equilibrium as order First, let's address your valid intuition that equilibrium as a form of order. Indeed, if everything is in thermal equilibrium, you just need to measure the temperature somewhere, and then you ...


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 ...


6

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 ...


6

It has been covered above that there is little scope for water to be heated by compression to begin with. Another aspect is that the water at the bottom of the ocean has been there for a considerable length of time. Hence, if it had heated up to any large extent when the oceans were formed, there has been ample time for the heat of compression to have ...


6

First of all as stated by Madan Ivan: equilibrium is not order. But you can get certain systems that are in a meta-stable "local" equilibrium (here meaning that you need some energy to move it from there), for example a crystal. These can be highly ordered. Intuitively: if you smack the crystal with a hammer it breaks to pieces. This brings your closer to ...


5

I think this question can be interpretted a few ways. I will frame the argument using the example of phase transitions. 1. Does temperature need to exist (or do we really need another constant)?: No. Let us consider what it means to have a phase transition. In the broadest terms, we are introducing energy into a system and approaching a critical point ...


5

Consider that the specific heat of water is 4.147 kJ/kg. The specific heat of air is less than 1/4 of that, so for the air to heat you as much as the water, you would have to heat the air to a much higher temperature. That's one reason, along with conduction, that Arctic explorers say that water cools you 30 times faster than air. So it doesn't balance out. ...


5

For any function of multiple variables, say, $f(x, y, z, \ldots)$ you can always write the following: $df = \big( \frac{\partial f}{\partial x} \big)_{y, z, \ldots} dx + \big( \frac{\partial f}{\partial y} \big)_{x, z, \ldots} dy + \big( \frac{\partial f}{\partial z} \big)_{x, y, \ldots} dz + \ldots$ In fact, the constancy of the other variables is ...


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

There is another factor that I feel the other answers have overlooked, because there is a similar analogy with air, and air is compressible. Specifically, why is air in valleys often colder than at the top of the hill when pressure heats things? In reality, there are two different dynamics at work. One is adiabatic compression, which as has been ...


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

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

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 ...


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

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 ...


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

Yes, agitation will generally promote heat transfer and reduce heating times (although quantifying the effect is not easy). But the effect is not related to the bulk speed of the kettle. When the water is heated a diffuse (poorly defined) boundary layer is formed on the bottom of the vessel. This layer is at a temperature that is slightly higher than the ...


3

So far there is no definitive answer as to why time flows forward. While many scientists speculate what may the cause of this asymmetry in time none of the options has been accepted as the one answer. Often these ideas lack a complete understanding of the underlying more complex principles that we cannot yet understand but here is one of the most common ...


3

Water has a very narrow range of temperatures over which it expands when cooling rather than contracting (IIRC +2 to 0 Celsius). This occurs due to the way the highly polar molecules "line up" with each other near the freezing point. For a more interesting example, consider a number of rubber compounds which shrink as they warm up. In this case it's due ...


3

The Planck temperature isn't the hottest possible temperature in the same sense that zero Kelvin is a theoretical minimum. It is simply the temperature at which it's black body radiation is of the order of the Planck length. The Planck length is the length scale at which it is theorised that quantum-gravitational effects become significant. Quantum-gravity ...


3

Due to the cylindrycal shape of the lamp, pressure forces act compresing the glass. Since the compressive strength of glass is quite high, $\approx 10\,000$ bar, atmospheric pressure is not enough to break it.


3

For an irreversible expansion or compression, let $P_B(t)$ represent the gas pressure on the portion of its boundary where the work is being done, and let $T_B(t)$ represent represent the gas temperature on the portion of the boundary where heat transfer is taking place, with $t$ representing the time during the irreversible change. If the change is ...



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