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0

The Equipartition theorem states that: $$ v_{rms}= \sqrt{\left \langle v^2\right \rangle} = \sqrt{\frac{3 k_B T}{m}}$$ where the averaging process is taken over the distribution of velocities of the gas: $$\left \langle v^2\right \rangle = \int d v \ v^2f(v).$$ For and ideal gas at the equilibrium the H-theorem states that the distribution is a Maxwellian: ...


0

In first place, the temperature is a quantity that measure thermal equilibrium by the zeroth law of thermodynamics. We have the contact with this quantity by with a thermal equilibrium can do. For example, the Celsius units is constructed by define $0°C$ as the volume of mercury in contact with freezing water and $100 °C$ as the volume of mercury in contact ...


3

The filament will be a reasonable approximation to a black body emitter, so it's spectrum will be given by Planck's law: $$ B = \frac{2hc^2}{\lambda^5} \frac{1}{e^{\frac{hc}{k\lambda T}} - 1} $$ So just measure the radiance of the light from the filament for a range of wavelengths and do a fit to Planck's law by varying $T$. This will give you an excellent ...


2

You really are asking two questions. First - how do we calculate the temperature: At the typical temperatures of a halogen bulb, the large majority of heat loss is due to thermal radiation (although there is some conductive loss in a halogen bulb as the bulb is not evacuated). Because of this, the most important factor is the "apparent size" of the ...


0

The Kelvin temperature scale is an absolute temperature scale. That is 0 K is absolute zero. It also has the property that temperature intervals on the Kelvin scale are the same as on the Celsius scale. That is a decrease or increase of one degree Kelvin is the same as a decrease or increase of one degree Celsius. To meet these two requirements it is ...


5

If we assume you are a sphere in space, at the same distance from the sun as Earth, then we can calculate the heat absorbed - and we can calculate how hot you need to be so heat in = heat out (assuming uniform surface temperature, and radiative heat transfer only). For this, we need the Stefan-Boltzmann expression for total emission at a given temperature: ...


0

Since the heat capacity of water is 10x that of steel, the same energy input to equal masses of steel and water will raise the temperature of the steel 10x more. So to raise a certain mass of water from 0 °C to 100 °C (boil it) and raise the same mass of steel from 0 °C to 1000 °C (melt it) will take approximately the same amount of time, assuming the same ...


0

Short answers: no, they are not the same; they are somewhat related. A more detailed discussion follows. Indeed, in most QFT books zero temperature is usually assumed. However, if one is interested in energy scales that are way beyond the temperature of the system, the zero-temperature approximation is a valid one. For example, the thermal energy at room ...


2

As a physics problem in a textbook, you could get somewhat close. Both the water and the steel have a heat capacity that relates the amount of temperature rise that would accompany an input of energy. If you make a few assumptions, you can relate the two. The problem with a real-world application is that those assumptions may be far from valid. The two ...


4

The Weights and Measures Act (the origin of the Imperial Units) does not speak of temperature. It was intended to create a uniform system for trade. You don't sell temperature, in the way you sell a pint of milk or a yard of cloth. And frankly, when it was first conceived (before Magna Carta, which already stated: "There shall be but one Measure ...


2

According to the wiki page on Imperial and US customary units Fahrenheit is part of both the Imperial and US customary system. I can't think of any reason it wouldn't be included in the Imperial system. Note that in the wiki page on Imperial units it is mentioned that the weight's and measures act (which defined the Imperial system) explicitly used the ...


2

How's this for simple and intuitive? A gas is separate particles moving around at great distances from each other. As you compress the gas, the particles get closer together. The particles of a liquid are in contact with each other, but as you heat it, there is more and more space between them as they zip and jiggle more and more. The combination of heat ...


1

The air in the freezer is a poor conductor of heat. The greater surface area of the porous paper along with its heat conducting water allows it to act as a heat sinc. It transfers heat from the liquid to the glass or aluminium and then to the water in the paper which sheds the energy by conducting it to the surrounding materials such as the air and the ...


0

"Which reading is correct?" That is called measurement uncertainty and that's why we have people called metrologists. Is there a naturally occurring phenomenon with a known sound pressure level to which we can compare a sound pressure level meter's reading? Probably not, at least not any practically feasible one at the current level of technology. ...


1

Define the efficiency of a heat engine as \begin{equation} \eta = \frac{\text{Net work out}}{\text{Heat in}} \end{equation} The Second Law of Thermodynamics tells us that the efficiency can't be greater than the Carnot efficiency: \begin{equation} \eta \leq 1 - \frac{T_L}{T_H} \end{equation} where the $T$'s must be in absolute units (e.g. Kelvin) for the ...


1

To understand negative temperatures, it is best to consider the inverse temperature $\beta \equiv (k_B T)^{-1}$, which appears in the expression of the partition function $Z\equiv \sum_n e^{-\beta E_n}$. A system 1 is hotter than system 2 (i.e., if brought into thermal contacts, heat will flow from 1 to 2) if $\beta_1 < \beta_2$. For systems with an ...


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It is because by bathing in cold water the heat energy of the body is given to the cool water.This rises the temperature of the air inside the bathroom which we can not feel because our body has already attained the thermal equilibrium with the air inside.


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First of all you have to have the Mass, Heat capacity and the Temperature of each fluid. then you can calculate the final temperature. $Q = M.C.Delta(T)$ where $M$ is the mass, $C$ is heat capacity and $Delta(T)$ is the amount of changes in the temperature. first calculate how much energy do they take to get to the 0 temperature. let me show you in an ...


0

$1+{R\over C_v}$ is always bigger than 1 as $R$ and $C_v$ are positive. Although I'm not sure you can necessarily put $C_p=C_v+R$, if you can then you can probably say that $R<C_v$ so that $$ 1<1+{R\over C_v}<1+{C_v\over C_v}=2 $$


1

The joule Thompson coefficient is negative as the gas is coming out through the small hole, for negative coefficient the tempr. decreases and for positive coefficient the tempr. increases.


2

Actually, that's exactly what James P. Joule demonstrated in 1845: the Mechanical Equivalent of Heat. He used brass paddles to mix a barrel of water. He found a direct relationship between his calculated amount of work and measured temperature. Boiling point is highly improbable. For a simply detectable change, Oocities.org reckons: 'The amount of heat ...


0

When I wrote a units package, I put degFinterval and degCinterval as units, specifically units that are intended to be squared and multiplied and divided. I put "interval" in the name to make it absolutely clear to users that they should only be multiplying temperature differences (which don't have a fixed origin) rather than raw temperatures (which do). (In ...


2

Yes, you will definitely increase the temperature of the water. Lets say that we increase the speed of the container, with an amount of energy of 1kJ. If we stop the container (does't matter if we do it suddenly or progressively), the 1kJ speed energy will be converted to heat, both to the water inside and to the device used to stop it (e.g. a wall or our ...


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In practice, no. In theory, also no. The Universe is filled with photons with an energy distribution corresponding to 2.73 K. Every cm$^3$ of space holds around 400-500 of them. That means that if you place your "stable body" in an ever-so-isolated box, the box itself will never come below 2.73 K, and neither will the body inside. It will asymptotically go ...


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It´s impossible to drop a body temperature to absolute 0K. You must notice that, at the same time the body is radiating energy from its own temperature, it is also receiving temperature from other sources (regardless the distance of the source) like distant stars.


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It greatly depends on what you need to do with your temperature. In almost all physics applications, the thermodynamic temperature is the only one that is meaningful. A notable exception is in linear heat flow simulations, where temperature differences only are important (e.g. between a point in the simulated region and the "ambient"). The squared ...


0

Thermodynamics is concerned with the study of equilibrium states of macroscopic matter. Those equilibrium states are defined so that they can be characterized by just some few extensive parameters like internal energy, volume and number of moles. Now, the point is that we have then some measured degrees of freedom like energy and volume, but we still have ...


2

If you are using an absolute temperature, you should use Kelvin. For instance, when using the Stefan-Boltzmann Law, $$P=A\epsilon\sigma T^4$$ it wouldn't make sense to have units of $^\circ C^4$; only units of $K^4$ physically make sense here. However, if you are using a temperature difference, then both Celsius and Kelvin are equally valid because a ...


0

we feel warm after the bath in the bathroom because as the temperature of the water is cool when compare to temperature of the body.So,that the body cools & the surrounding temperature become hot or warm(humidity).


1

What people don't understand is that the laws of thermodynamics are not exact in the same way as, for example, energy conservation is. They are only quite probable, meaning that for a finite system there always exists a non-zero propability to break them. So, even though it's quite improbable to reach $T= 0 \ \textrm{K}$, in principle it is possible. With ...


2

The unit $^\circ\mathrm{C}^2$ does make sense. It represents a difference in temperature squared. You are worried that it is invalid because the origin is shifted. This is not a problem though, because we use $\mathrm{m}^2$ all the time, and there is no natural origin at all for positions.


1

The relativistic doppler shift of frequency is given by $$f_o = \frac{f_ s}{\gamma (1 + (v/c) \cos \theta)},$$ where $f_o$ is the observed frequency, $f_s$ is the emitted (source) frequency, $\gamma = (1 - v^2/c^2)^{-1/2}$ and $\theta$ is as you define. This is a standard result - e.g. ...


0

The standing waves you introduced are models for how air moves as it resonates. In fact, each is called a mode (of oscillation). In the continuum approximation of air being an infinitely divisible, continuous fluid, you need infinitely many of them to simultaneously model arbitrary resonating movement. In practice, you can hope to neglect all but low ...


0

The wavelength is fixed by the dimensional length of the tube, but since wavelength is equal to the speed of sound divided by the wave frequency, temperature will affect the frequency because temperature determines the speed of sound.


3

The amount of energy liberated per gram of material per second in the fusion reactions depends on the density, the mass fraction (hydrogen, $X$, helium, $Y$, and all others $Z$) and temperature: $$ \epsilon = \epsilon(\rho,X, Y, Z, T) $$ Typically we express the energy generation rate as a power law, $$ \epsilon\propto\rho^\alpha T^\delta. $$ though the ...



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