# Tag Info

28

That is simply convection. The wick does suck molten wax, and it goes up by capillary to the middle of the flame, but that movement is way to slow to explain the fast particles in your video. Moreover they are moving in the opposite direction! Convection happens because the wick is hot, and it makes the wax around also hot, so the wax expands a little and ...

23

Well first you have the energy in the form of kinetic energy of the spinning water. Once you let that water settle, it DOES get hotter. The only problem is that water has a high specific heat (it takes a LOT of energy to heat up water), so you don't notice the water getting hotter since the amount it's heating up is not very noticeable. Coincidentally, it ...

22

I think you are right. A perhaps more precise relation between temperature and velocity is the Maxwell–Boltzmann distribution: \begin{equation*} P(\textbf{v}) = \left( \frac{m}{2\pi k_B T} \right)^{3/2} \text{exp} \left[-\frac{m ( \textbf{v} - \textbf{v}_0)^2}{2 k_B T} \right]. \end{equation*} where you see that the mean velocity $\textbf{v}_0$ and the ...

12

I think your view is correct, and you can think about the following real word example. In labs here on earth, we can use laser cooling techniques to cool atoms to $\mu$K scales in the lab frame. But the lab is on earth, and the earth is moving very fast around the sun, and the sun is moving very fast around the galactic center and so on. We don't take ...

10

When two identical gases mix, the state is generally indistinguisable from the previous state. If one molecule from the left of the partition changes places with a molecule from the right of the partition, does the mixture actually look any different? If the left and right molecules are identical, you would never know which ones started where. So entropy ...

9

Suppose the separated volumes of identical gas are a low-entropy state and the mixed volume is high-entropy. Imagine the reverse process from mixing. You have a single tank full of helium. You insert a partition, so now you have two half-tanks of helium. This can be done reversibly, but it takes you from the high-entropy to the low-entropy state. The entropy ...

8

Squeezing the wavefunction means confining it to a smaller space. It takes more energy to confine something within a small space than within a big one. 2, 3: These are consequences of the quantum adiabatic theorem: if you take a system in state $n$ of some system, and act on the system sufficiently slowly, it ends up still in state $n$ of the new system. ...

6

I think that the key requirement is that the material be in local thermodynamic equilibrium. Even if it is a dynamic situation with mass flow or shock waves running through the material under consideration, if the material is in local thermodynamic equilibrium at every instant in time, then equilibrium thermodynamic concepts such as temperature, pressure, ...

6

Indistinguishability There is something that is lurking in the background here, which is an important statement of physics: all helium atoms are indistinguishable. You cannot tell two helium atoms apart from each other, they are so identical, they are basically the same helium atom, twice. Now let's use that to understand what is going on. Two experiments ...

5

When two quantities of water ($m_1$ and $m_2$) at different temperatures (resp. $T_1$ and $T_2$) are mixed in adiabatic conditions (no heat loss and no external heating during mixing) the temperature $T$ of the resulting mixture can be calculated from the heat balance (no heat is lost or added so the heat contained in both masses is found again in the ...

4

Its because if we divide the container in two halves then the volume of the gas will also get half. But the pressure applied on the walls of both the containers will be same.

4

One of the reasons that makes you not to believe that one cannot heat up water by stirring it, might be that we usually experience the opposite effect. Namely, one usually stirs a hot tea or soup to cool it down. Why a cup of hot tea or a bowl of warm soup cools down when one stirs it? The reason is that the liquid/air interface where the heat exchange ...

4

You seem to only have a blurry idea of the hydrodynamic approach, so I will add a tad more about the whole idea, mainly to give you a better intuition. Hopefully this will be a useful addition to Samuel Weir's wonderful answer. A hydrodynamic state is described by the variables: mass density field, energy density field and momentum density field. These are ...

4

Why does heat added to a system cause an increase in entropy that is independent of the amount of particles in the system? Short answer: it doesn't The systems won't end up with the same entropy. Your intuition is correct that the change in entropy depends on the number of particles. The reason why you can't just reason directly from $dS = \delta Q/T$ ...

4

A single particle can be a 'system' within itself having modes depending on the particle ' s structure. These modes may be in 'tune' with the incident radiation and thus capture the energy which can increase the particle ' s momentum and therefore its velocity. We never say the particle's temperature has increased but rather it's momentum. When a system of ...

3

From what I've gathered, I think my initial guess is correct. Air tries to maintain a constant pressure. According to the ideal gas law, there are two ways to maintain the same amount of pressure with an increasing volume: 1) increase the amount of gas, and 2) increase the temperature of the gas.

3

Do the maths and calculate how much energy is needed to raise the water's temperature by 1K. If you have a fast moving stirrer, you should be able to measure the increase in temperature of a liquid in an isolated pot. By the way: the microwaves in your microwave oven turn around the water molecules very fast and heat up your food this way.

3

Let $m_1$ be the mass of the peas, which we'll assume to be mainly water, $m_2$ the mass of boiling water added and $m_p$ the mass of the pan. We'll assume no heat losses. After adding everything to the pan and allowing for thermal equilibrium to be established the temperature of pan, water and peas is $T$. $T$ can be calculated from the adiabatic heat ...

3

From the ideal gas equation, $$P=\frac{nRT}{V}$$ Now assuming the gas is uniformly distributed over space (has constant density for a given temperature), halving the number of moles will divide the volume by the same amount. Essentially, if we divide the number of moles by any number, we will end up dividing the volume by the same number to maintain ...

3

The temperature is really only negative in the sense of the classical definition of temperature. What is actually happening in a population inversion is the particles aren't following Boltzmann distribution of energies anymore. Comparing the temperature of a Boltzmann distributed system to a non-Boltzmann system might not be meaningful at all. People say ...

3

At the classical framework , i.e. no General relativity and astrophysical observations of the 18th century , this is a valid question. When talking of a "Universe" one must have a model , and the model depends on the state of physics knowledge at the time of the model. The second law states that entropy always increases or stays the same. One can make a ...

2

The minimal counterexample seems to me to be the following: Take two materials, placed next to each other: ____________________ | | | | Material|Material | | 1 | 2 | ____________________ E1 _ _ _ E0 _ _ They have energy levels as indicated above- both have states at E0 and E1, but one has two excited states. ...

2

As rockets have finite fuel capacity and important parameter is how much rocket velocity can be achieved in a given situation per unit of fuel mass consumed. The higher the exhaust velocity the more effectively the fuel mass is utilised. Issues such as energy required are also important but generally exhaust velocity or "specific impulse" is amongst the most ...

2

Basically a "molecule" of water cannot heat up ice. I think what you are trying to say is, how does heat transfer take place on a molecular level? If that's the case, then its something like this. In the interface between water and ice, water molecules are moving, while ice molecules are static. on contact, some molecules of ice acquire velocity (due to no ...

2

This discussion feels familiar; I figure this question is a follow up of some comments to an answer i provided to one of your previous questions. Specifically: thanks for the great answer. Although I have to say i find steady state to be a bit confusing here since this in fluis mechanics normally means $∂_t v=0$ but I venture you mean steady mass state? ...

2

The kinetic energy of the gas molecules is of order $kT$, so when two gas molecules collide there is up to around $2kT$ of energy available. If the spacing of the vibrational energy levels is less than or equal to $2kT$ then there is a non-zero probability that the kinetic enegry could be used to excite a vibrational transition. This would cool the gas ...

2

One mole of oxygen is 32 grams. So the figure you cite tells you that 23.5 cm$^3$ of solid oxygen weighs 32 grams making the density about 1.36 g/cm$^3$ or 1360 kg/m$^3$. For comparison the density of air is about 1.225 kg/m$^3$ and only 23% (by weight) of the air is oxygen. So the density of the oxygen in air is around 0.28 kg/m$^3$, which is a factor of ...

2

It's simple. You can think of temperature as being the standard deviation of KE among all components (atoms) of a mass. This is significant because KE is a relative quantity, but temperature is absolute, and this relationship makes that possible. If all atoms are moving uniformly in the same direction, then the temperature would be 0.

2

Magnetic fields certainly can influence thermal conductivity. This shows up, not surprisingly, when there is a strong influence of the magnetic field on other properties, particularly electronic ones. One (non-metal) example is 'Thermal conductivity tensor in YBa$_{2}$Cu$_{3}$O$_{7-x}$: Effects of a planar magnetic field' by R. Ocana and P. Esquinazi, Phys ...

2

Therefore, when we say, for example, that the energy of the ideal gas at temperature T is E=32NkBT, we should really be saying "the energy of the ideal gas immersed in a heat bath at temperature T"? Is this reasoning valid? This is true. What is also true is that you can also say that the temperature of the (completely isolated) gas of energy E is T=2/3 ...

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