# Tag Info

3

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 ...

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 ...

0

How do we know that the rate at which a body loses heat is proportional to the difference between its temperature and that of its environment? In classical physics this is a law. "Fourier's law The law of heat conduction, also known as Fourier's law, states that the time rate of heat transfer through a material is proportional to the negative ...

0

It appears Newton himself did some experimentation, but failed to divulge exactly what he did - though he said "he used a linseed oil thermometer" and the resulting data. Source: History of Newton's Law of Cooling

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 ...

0

One way to define temperature is via statistical physics. It is a quantity used to describe an equilibrium distribution of possible states via Boltzmann factor. Given a total energy of a classical system $$E(x_1,\ldots,x_N, p_1, \ldots p_N)$$, where $x_1 \ldots x_N$ are the position coordinates, and $p_1 \ldots p_N$ are the momentum coordinates. The ...

0

You've touched on a very important point - "negative temperature" is a misnomer. Start with a simple thought experiment. Physics tells us that temperature changes happen continuously. In order to change from 10K to 20K, we must first pass through 11K, 12K, and all other intermediate temperatures. In order to achieve a "negative temperature," you would need ...

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.

0

You are correct. The temperature is usually defined in terms of random motion of the atoms. This announcement from NIST talks of an atomic beam cooled to $30 \mu K$, so there is very little random energy. It doesn't give the speed of the beam, but that could be very high if the atoms are all moving in the same direction.

19

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 ...

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}} \\ & = ...

0

Since mL cancels with mL, J/s and N-m are compatible and therefore you should need no scaling.

0

That's an intuitive guess but you should get some good proof to support your argument. Better is to apply the principle of calorimetry to get a firm grip on your answer and reasoning: if two substances of different temperature are in thermal contact and if no heat is allowed to go out or enter into and if no chemical reactions takes place in between 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 ...

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 ...

0

No, they will cool at different rates Newton's law of cooling states that $$\frac{dQ}{dt} \propto T-T_{env}$$ where $T$ is the temperature of the water and $T_{env}$ is the temperature of the environment. As you can see, the larger the temperature difference, the more quickly the water will lose heat. We can assume that the specific heat of water will ...

-1

Even if you could cool down to absolute zero which is impossible due to third law of thermodynamics. there would still be quantum fluctuation. electron would not stop moving. it would still orbit around atom. Heisenberg uncertainty principle states that we can never precisely know the position of an electron and momentum simultaneously. so therefore ...

0

Use a table of the resistivity of wolfram, ie: http://hypertextbook.com/facts/2004/DeannaStewart.shtml If you MUST use the quadratic form, then calculate A and B from the values in the table.

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 ...

0

The sky at night (no clouds) typically measures -48c to -21c at my location (Seattle), currently -44c. The clouds measure -5c to -1c right now. The trees across the road are at 2c. The reason I can measure the clouds is that the atmosphere is largely transparent to IR, so my thermometer "sees" the radiation from the clouds, but not the atmosphere - the ...

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 ...

0

Credits given to all answers posted. They helped me figured this out. Thanks a lot. Temperature is heavily linked with Kinetic Energy. Pressure is heavily linked with number of Collisions per Time AND Kinetic Energy. Example: A gas is hot when the molecules posses high Kinetic Energy and collides with the measuring device with great force. A gas is ...

1

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 ...

0

A gas is hot when the molecules collided with your measuring device. Not quite. Gas heats your measuring device when the collisions are mostly such that the colliding gas molecule has more kinetic energy than the colliding measuring device molecule. It's instructive to think colliding molecules as sumo wrestlers: The molecule which has more ...

0

To measure somethings means to compare it with an etalon or a measurement instrument, made by the help of an etalon (or the combination of etalons). To measure the pressure of a gas inside a volume one take for example a barometer and measures the pressure difference to the outer room. The measured pressure inside the volume is the result of the hitting of ...

0

It is true that the most contribution to heat comes from compressing the air. The temperature of a falling meteor was in fact in my aerodynamics II exam where I had to predict its temperature using shockwaves. According to my estimation it was about 10,000 K. You need a proper understanding of compressible air flows in order to answer this question. And I ...

0

Pressure is a measure of force per unit area exerted on the 'measuring device', while the temperature is a measure of kinetic energy of the individual molecules of the gas. Thus, high pressure can arise when there are either many slow moving molecules with low kinetic energy colliding with the container, or a few fast moving molecules colliding with the ...

0

An example of a difference where the pressure of a reasonably dilute gas depends on something else other than the kinetic energy of the particles is actually just the air on Earth. A classic exercise in statistical mechanics is to consider an ideal gas subject to gravity and find how the pressure varies with altitude. Of course, in reality the temperature ...

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

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

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 ...

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. ...

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 ...

0

Alas, no. Unless energy is trying to pass through the gas. Then, yes. If there is no energy input/output from the gas: "energy" may be transferred from top-down, but since a density gradient also forms, that energy is shared among more molecules. Net result is that the energy per molecule becomes roughly uniform and we have a uniform "temperature". If ...

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 ...

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 ...

0

Phase and random kinetic energy are macro. If you don't call the kinetic energy of a single molecule temperature then fine but it is still kinetic energy. In a single molecule you have rotation and vibration that can be effected by radiation. You can have an exited state where an electron is temporarily bumped to a higher orbit. Some reactions only take ...

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 ...

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 ...

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 $dS = \delta Q/T$ ...

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

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

With your description, if no radiation comes back from the surface of the black hole, the temperature should be the vacuum classical temperature, 0 Kelvin. BUT Hawking predicted a radiation coming out from quantum mechanical interactions with the vacuum at the limits of the event horizon. Hawking showed that quantum effects allow black holes to emit ...

2

Your first expression is a temperature distribution: it simply tells you the temperature $T$ at each point in the $x,t$ space. Now if you, using that formula, calculated two temperatures $T_1$ (at $x_1,t_1)$ and $T_2$ (at $x_2,t_2)$, then the difference: $$\Delta T=T_2-T_1,$$ is an actual temperature difference. An infinitesimal temperature difference ...

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