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1

Dr. Robitaille says that blackbody radiation is not universal, even inside cavities where the surfaces are all at thermal equilibrium. That is highly controversial since the electromagnetic fields in a cavity are usually considered as an additional substance, called a "photon gas" which is also at thermal equilibrium and hence has a temperature. This ...


0

Both expressions are equivalent; they are simply calculating slightly different quantities. The first one gives the energy density per unit frequency, and the second one gives the energy density per unit wavelength. These two differ by a square of the frequency / wavelength because one unit of frequency represents a larger range of wavelengths at low ...


0

Rest has already been covered however for basic students....the crux to understand is....Infra Red or commonly referred as IR detects the heat generated by a body( which of course is there due to atomic vibrations), and generates a picture through a IR conversion tube. whereas Thermal Imaging device or commonly referred as TI detects the picture based upon ...


2

The connection of heat to entropy in thermodynamics is through: where S is entropy Q is heat T is temperature, and it is through differential changes. This in no way means that heat is entropy . The easiest way to acquire an intuition of entropy is to read up on the statistical definition which can be proven to be the same as the thermodynamic ...


0

There are 3 forms of heat transfer: radiative, convective, and conductive heat transfer. All 3 forms of heat transfer are normally operating at the same time.


1

An open cup of hot water loses a lot of its heat through evaporation. You can easily test this: prepare two cups of hot water, and then float a thin layer of cooking oil on top of one. Come back in a half hour: you'll find that the cup without oil is a lot cooler, and the water level has also significantly gone down. Edit: I did the experiment: I poured ...


0

Thermal expansion means that the distance between atoms increases with temperature. Picture the ring of atoms surrounding the hole: the distances between them increases, hence the circumference of the hole increases. I hope this makes it clear that the radius of the hole increases with temperature.


2

When the metal is heated, all inter-atomic distances increase by the same factor. This drawing may help understand why the hole also increases in size. Here, I increased all distances by a factor a two. Replace the atoms with galaxies, and you have a model of the expanding Universe, which may help understand why an observer in any galaxy will see herself ...


0

If the object is in free space, the hole shall expand proportionally as you expect. However, a different situation is when the object is somehow restrained to expand; then the external force simultaneously causes elastic compression which limits the expansion. If the restriction of expansion is stiff enough (such as placing it in an infinitely rigid box), ...


2

You've got your work cut out for you, so this answer can only point you in a direction. This calculation can be difficult because there is a profusion of pieces of terminology with subtly different meanings. Find a book/other source (but for the straight dope I suggest a book) which discusses radiometry and the differences between radiant flux, radiance, ...


2

I will give a closed form for the integral in Chris Cundy's answer. Doing the substitution $u=nx$, we get $$ \sum_{n=1}^{\infty} \int_{x_{min} \cdot n}^{\infty} \frac{1}{n}\left(\frac{u}{n}\right)^3e^{-u}\mathrm{d}u$$ $$ \sum_{n=1}^{\infty} \frac{1}{n^4} \Gamma(4,x_{min}\cdot n)$$ where $\Gamma$ is the upper complete gamma function. We write $a=x_{min}$ ...


40

The formula you want is called Planck's Law. Copying Wikipedia: The spectral radiance of a body, $B_{\nu}$, describes the amount of energy it gives off as radiation of different frequencies. It is measured in terms of the power emitted per unit area of the body, per unit solid angle that the radiation is measured over, per unit frequency. $$ ...


11

The wavelengths of light emitted can be calculated using planks law and the temperature of the object. For your average 100W incandescent light bulb, the filament is 2823 kelvin according to google. The spectral radiance, $B$, is equal to $$\frac{1.2\cdot10^{52}}{\mathrm{wavelength}^{5}\cdot ...


3

To a good approximation the radiation emitted from a hot piece of steel will be the same as emitted by a black body. The relationship between the wavelength $\lambda$ of light emitted by a black body and the temperature $T$ is given by Planck's law: $$ B = \frac{2hc^2}{\lambda^5} \frac{1}{e^{hc/\lambda k_b T} - 1} $$ where $B$ is the spectral radiance, $h$ ...


5

An object at 1000K glows red because that is basically dominating the radiation that it is emitting that we can see. Most of the radiation it emits emerges in the infrared part of the spectrum to which our eyes are not sensitive. Your edited question asks about the appearance of hot metals. The chart below is an example of a "colour-temperature" chart for ...


-1

Your argument is wrong in that the second body receives Qin=σT41A1. You can direct radiant energy at it - does not mean it will receive it. Heat does not flow from cold to hot. Temperature wins every time. Double the heat at the same temperature is still the same temperature.


3

Mirrors and lenses cannot do what you ask. Light has a specific intensity, meaning a power per unit area per unit solid angle per unit wavelength. Mirrors and lenses can never increase the specific intensity. As an example, consider using a lens to focus sunlight onto a target. The specific intensity of the sunlight is the same with or without the lens. ...


1

I am unsure of the solution below because of my free manipulation of P and Q, so if anyone has a comment or better way, let me know. Some assumptions: I assume that in the question you meant that we have a body with a surface area $A$, which is at an initial temperature $T_{0}$ and is losing energy by means of radiation, for which you used the ...



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