# Tag Info

17

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 ...

6

if the pressure is the same, there in no net force on the piston, so it will remain at rest, so neither gas will expand, and this will stay this way because there is no heat exchange that could change the pressure on either side. Ask your teacher to return his diploma.

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: ...

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 ...

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 ...

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.

2

Your teacher is wrong... in a way. What he probably meant was same mass, different temperture.

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 ...

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 ...

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 ...

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 ...

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 ...

2

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 ...

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 ...

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 ...

1

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.

1

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 ...

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.

1

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 ...

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 ...

1

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

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 ...

1

It doesn't. They'd have to exchange heat through the piston.

1

The best vacuum we can make is about 10^-12 Pa. Atmospheric pressure is about 10^5 Pa, so it's 17 orders of magnitude lower pressure. The best vacuum recorded is the intergalactic void, at about 10^-17 Pa. Even if you managed to remove all matter, there would still be energy from any light or electric fields.

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 ...

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. ...

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