Blackbody radiation of beam of light

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SUMMARY

The discussion focuses on the physics of blackbody radiation in relation to a 100W beam of light illuminating a blackbody with a mass of 2e-3 kg for 10e4 seconds. The total energy absorbed by the blackbody is calculated to be 1e6 J, with a momentum of 0.0033 kg·m/s. The final velocity of the blackbody is determined to be 1.65 m/s, resulting in a final kinetic energy of 0.0027 J. The discrepancy between the kinetic energy and the total energy of the absorbed photons is attributed to energy loss through temperature increase and radiation.

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Homework Statement


a 100W beam of light is shone onto a blackbody of mass 2e-3 kg for 10e4 seconds. The blackbody is intially at rest in a frictionless space

a) Compute the total energy and momentum absorbed by the blackbody from the light beam b) calculate the blackbody's velocity at the end of the period of illumination, and c) compute the final kinetic energy of the blackbody. Why is the latter less than the total energy of the absorbed photons?


Homework Equations





The Attempt at a Solution



a) find energy - easy 100 J/s * 10e5 s =1e6 J absorbed

momentum - here's the part I'm slightly unsure of. At first I thought that I needed the wavelength, but that's only if I want to know the momentum of an individual photon.

So here's what I said sum of photon momentum = 1/c * sum of absorbed energy
so sum of absorbed p = 0.0033 kgm/s

b) 0.0033 = (2e-3)*v , v =1.65 m/s

c) (1/2)*(2e-3)*1.65^2 = 0.0027 J and I think this is less than the total energy of the absorbed photons because much of the energy goes into raising the temperature of the blackbody and much of the energy is radiated out.

Is this correct? I'm not sure if I have a really good picture of what is going on in my head.
 
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