Assuming a Dyson sphere is absorbing all of the energy of the sun it surrounds, then eventually that energy will be dissipated into space, presumably by radiating it away with some equivalent black body temperature under equilibrium. What temperature would that be (on average) presuming some given parameters, e.g., sun output and sphere radius?
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$\begingroup$ Millions kelvin if you absorbs gamma too ;p $\endgroup$– user6760Commented Feb 29, 2020 at 8:08
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$\begingroup$ @user6760 wrong. You haven't really thought about the surface area of a dyson sphere, or the actual amount of energy coming out of the star, have you? $\endgroup$– Starfish PrimeCommented Feb 29, 2020 at 13:00
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A simplified calculation is rather easy:
- Each unit surface of your sphere will receive an amount of energy equal to $I_0$. let's assume it will entirely absorb it, behaving like a black body
- The device, assumed to have a yield of $\eta$, will produce $\eta I_0$ and discard $(1-\eta)I_0$
- The equilibrium temperature of the outer side will be such that $(1-\eta)I_0=\sigma T^4$ where $\sigma$ is the Boltzmann constant and $T$ the absolute temperature
