# Tag Info

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 ...

6

You're right that the classical idea of radiation emission from an accelerated charge cannot be applied to electrons in orbit around nuclei, and thus they do not emit radiation (unless they're in an exited state and decay to a lower state). The same thing does not apply to the nuclei. As you suspect, they will, over time, lose energy and vibrate less and ...

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

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

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 ...

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

There seems to be some confusion here - the perception of "heat" from an IR lamp relates to the amount of energy absorbed by the body. This depends on the reflectivity of the skin at that particular wavelength; also, you would have to normalize this in some way. Now we know from Planck's Law that there is a distribution of wavelengths from a black body: the ...

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 ...

2

Since the question is rather vague, I will just give you some key points: Debye's model treats oscillation modes of a solid as sound waves (phonons) with frequency $\omega(\mathbf{k})=v|\mathbf{k}|$ ($v$ the sound velocity). As a result, with this model, Debye shows how the heat capacity is directly related to the rate of change of the energy expectation ...

2

As march pointed out, your reasoning is incorrect for finite $Q$ and $\Delta S$. However, the equation is true for differentials, so we have to address that. Possible 'conceptual' answers: Stat mech: the increase in $S$ is related to how much extra disorder is produced. When there are less particles, they each get a bigger share of the $dQ$, so they each ...

2

To add a little detail about radiative thermal equilibrium: As atoms at nonzero temperature collide with each other, they do emit electromagnetic radiation, and if they were in an empty universe, they would approach zero temperature. However, since the universe isn't empty, they also absorb electromagnetic radiation coming from all the other atoms around. ...

1

Cubero et al. 2007: Thermal equilibrium and statistical thermometers in special relativity (http://arxiv.org/abs/0705.3328) came to the conclusion that 'temperature' can be statistically defined and measured in an observer frame independent way. With fully relativistic 1D molecular dynamics simulations they verified that the temperature definition ...

1

By the Ideal gas law, $PV=nRT$, or "pressure times volume equals the number of molecules times a constant times temperature". So, all else being the same, as the temperature goes up, the pressure goes up in an exact ratio. However, all else does not have to be the same. So, for instance, if you reduce the number of molecules in a container ($n$), the ...

1

Of course, they are relate to each other but that doesn't mean they are the same things. Temperature is the average kinetic energy of the molecules while pressure is the force they exert perpendicularly on any surface. Of course, more the temperature, more would be the pressure. While the former is related to the energy, the later is related to the ...

1

Because conventionally we assume constant temperature, and length and density are also assumed to be constants for a given resistor. Of course, this is not true. In some circuit designs I have to pay very careful attention to resistance changes with temperature, and indeed this is sometimes used to provide temperature measurements in the form of RTD ...

1

The electric fan increases the velocity and hence the kinetic energy of the molecules in the air. this would mean that the temperature has increased. I think that there is a bit of a problem here. Kinetic energy is not quite the same as thermal energy and temperature. Thermal energy and temperature is a measure of random, thermal motion of atoms or ...

1

Temperature is connected with the kinetic degrees of freedom of the atoms or molecules . In a gas , temperature is given by directly proportional to the root mean square of velocity, an average kinetic energy. Raising the temperature raises the probability of atoms/molecules scattering against each other. Attraction with neutral atoms and molecules ...

1

Background Let us assume we have a function, $f_{s}(\mathbf{x},\mathbf{v},t)$, which defines the number of particles of species $s$ in the following way: $$dN = f_{s}\left( \mathbf{x}, \mathbf{v}, t \right) \ d^{3}x \ d^{3}v$$ which tells us that $f_{s}(\mathbf{x},\mathbf{v},t)$ is the particle distribution function of species $s$ that defines a ...

1

On the phase diagram of methane, you can see that at RT (20°C), methane can only be gas (or super-critical if pressure is enough). The pressure can be calculated with the ideal gas equation $\frac{pV}{T}=nR$ You need to calculate quantity of n (in moles, given the volume, density of liquid methane, and the weight of the molecule). R is a constant, V is ...

1

If we were to send a unmannedspaceship through the Sun. What material can survive? A wide variety of materials might survive passing through the outer layer of the sun, but only if the spaceship is big enough, fast enough and has a thick enough sacrificial/ablative shell. If the ship is slow, it doesn't really matter if the hull survives, most stuff ...

1

Writing $(Q_m)$ (and so on) for the numerical value of the constants in the units stated, \begin{align} \frac{Q_m}{\omega_b \rho_b c_b} &= \frac{(Q_m)\mathrm W/\mathrm m^3}{(\omega_b)\frac{\mathrm{ml}}{\mathrm{ml\: s}} \times (\rho_b) \frac{\mathrm {kg}}{\mathrm m^3} \times (c_b)\frac{\mathrm J}{\mathrm{kg}\:^{\circ}\mathrm C}} \\ & = ...

1

Can a single particle be “heated” by radiation? Not really, because heat is an emergent macroscopic property of an ensemble of particles. But since you put the word "heated" in quotes, we can allow a yes of sorts. Take a look at the Wikipedia temperature page, where we can see this picture: CCASA image by Greg L, see Wikipedia It's to do with the ...

1

The conventional method of nuclear waste disposal is just fine. Just bury it deep into some geologically inactive rock. Consider this: If you dig a deep hole into ground in some geologically inactive area, you will find rocks and minerals that have been there for millions of years. If you put your nuclear waste there, there is no reason why it wouldn't ...

1

The energy per unit area radiated by an object at a temperature $T$ is given by the Stefan-Boltzmann law: $$J = \varepsilon\sigma T^4 \tag{1}$$ where $\sigma$ is the Stefan-Boltzmann constant and $\varepsilon$ is the emissivity. A spaceship in a vacuum can only lose heat by radiation, so it will heat up until the energy loss given by equation (1) is ...

1

Planck temperature is the maximum temperature on the Planck scale. Some sources call it absolute hot. The Planck temperature is defined as $$T_P = \frac{m_P c^2}{k} = \sqrt{\frac{\hbar c^5}{G k^2}} = 1.416833(85)\times10^{32} \text{K}$$ where $m_P$ is the Planck mass, $c$ is the speed of light, $\hbar = h/2\pi$ is the reduced Planck constant, $k$ is ...

1

Ultimately, Newton's law of cooling is a simplification that can be obtained from the full heat equation, i.e. $$\rho c\frac{\partial T}{\partial t} = - \kappa \nabla \cdot T.$$ The heat equation itself can be derived from first principles, assuming Fourier's law for heat flow, namely that it is proportional microscopically to the difference in temperature ...

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