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1

Short answer: use your technique, but use scale factors of 0.690 for L, and 0.348 for M (instead of 0.542 and 0.575), and you will reproduce the luminosity function. Long answer: You're on the right track. I tried to find 'official' scale values for the LMS curves online to combine into the luminosity function, but couldn't quickly get them. You are ...


1

Solar radiation is pretty close to $1kW/m^2$. That's already borderline problematic for a human who is completely exposed. I would guestimate with a giant foam hand that $2kW/m^2$ for more than a few minutes is more than enough to give humans a bad time. You could, of course, test this experimentally with a 100W light bulb. What's the closest distance that ...


3

It has to be an error. Rayleigh-Jeans theory fails at high frequency, but it's a good approximation for low frequency. Infrared catastrophe has nothing to do with the black body radiation problem. Only at high frequencies it was evident that Rayleigh-Jeans model failed. In the following plot you can see how at low frequencies (where infrared belongs), the ...


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Without seeing your code, it is hard to know where you went wrong. I did write a few lines of code myself to see what might be going on - and found that the values converge quite nicely as you approach higher temperatures (for the lowest temperatures of 500 K and 1000 K, significant amounts of energy do indeed extend beyond 10,000 nm). updated There is a ...


3

Imagine the light coming from a hot body entering a prism This graph shows you (at least in the visible spectrum) the relative brightness of each color as leaving the prism. Obviously, the relative proportion of light at invisible frequencies has to be measured by some other device such as a spectrometer. The best way to read the graph is probably to ...


1

The dependence of spectra radiance (y-axis) on wavelength (x-axis) is as follows (Planck's law): $$I(\lambda,T)=\frac{2hc^2}{\lambda^5}\frac{1}{e^{\frac{hc}{\lambda k_BT}}-1}$$ As you can see, $I$ also depends on temperature T. For each T, we have corresponding curve. You should pay attention to the peak first. The peak will tell you at what wavelength ...


1

The x-axis on the graph shows the wavelength, while the y-axis shows the corresponding intensity for that wavelength (all for a given temperature of the black body, in this case 5000K). Also note that this graph shows a distribution of the various wavelengths and their intensities at that temperature. It is a distribution graph like the "bell curve" as ...


0

There are 3 methods of thermal energy transfer!!! [radiation, convection and conduction] Given that energy always obeys the law of entropy and is always moving towards lower levels. Left alone hot things cool. As space is an ideal insulator, the only means of cooling is radiation. On earth the atmosphere serves to wick away thermal energy from the can and ...


0

The heat from Sun, comes in the form of radiation. To be specific, the infrared part of the EM light is responsible for heating. Infrared frequency is closest to the resonant frequency of most molecules. Thus, when these molecules absorb the infrared radiation i.e. the molecules' electron clouds oscillate at the infrared frequency, there's resonance, which ...


14

You probably are under the misconception that heat travels only via molecular interactions. (i.e, heat transfer by conduction, which needs a medium of sorts). Heat also transfers by radiation, which the sun is an enormous source of. Electromagnetic radiation does not need a 'medium' to travel through. All types of electromagnetic radiation carry energy, ...


4

The heat 'comes' as electromagnetic radiation, that is light. Light from the sun is electromagnetic radiation, that is a wave having energy and momentum or a very big amount of quantum particles, photons, that have energy and momentum. The interaction of this electromagnetic interaction is what heats the earth. Hope this helps.


1

The problem is quite complex to solve quantitively and requires a differential calculus of multi-variable functions, but I'll try to simplify it. Imagine that the object consist of many thin slices across the temperature gradient. Every second slice is a heat container with heat capacity $c \left[ \frac{J}{K}\right]$ and the remaining are heat conductors of ...



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