If i had a black hole of 20 solar masses, how much energy would it radiate?

AI Thread Summary
A 20 solar mass black hole would radiate essentially no energy through Hawking radiation, as it is too massive and cold, being below the temperature of the cosmic microwave background. The temperature of such a black hole can be calculated using specific physical constants, revealing that larger black holes are colder than 2.75 Kelvin. Consequently, they act as net absorbers of energy rather than emitters. Discussions suggest that smaller black holes would need to be considered to achieve a noticeable radiation rate. Overall, the energy output of a 20 solar mass black hole is negligible under isolated conditions.
rubecuber
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If i had a black hole of 20 solar masses, how much energy would it radiate?
 
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If you are referring to Hawking radiation, essentially none. Black holes are "observed" by their effect on accreting matter, which is very energetic.
 
rubecuber said:
If i had a black hole of 20 solar masses, how much energy would it radiate?

Let's assume your 20 solar-mass black hole is totally isolated, i.e., nothing is falling in.

The temperature (Kelvin) of the black hole is given by

T = \frac{\hbar c^3}{8 \pi k G M}.

Using the stuff below, can you work out the temperature of the black hole?

c=3.00\times10^8m/s[/itex]<br /> G=6.67\times 10^{-11} m^3 kg^{-1} s^{-2}<br /> \hbar=1.054\times 10^{-34} J.s<br /> k=1.381\times 10^{-23} J/K<br /> M_{Sun}=1.989\times10^{30} kg
 
rubecuber said:
If i had a black hole of 20 solar masses, how much energy would it radiate?

mathman and George Jones already answered. the smaller and less massive holes are hotter and radiate more watts

big massive ones (like solar mass and up) are colder than the 2.75 kelvin microwave background----they'd be net absorbers under most circumstances

you could turn the question around and ask how small would a BH have to be for it to have an interesting temperature, and radiate heat at some noticeable rate. you know, could you heat a cup of coffee with one? I don't know the answer but George Jones has provided the necessary equations I think.
 
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