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The outside (not inside) of black holes, the event horizon, are apparently extremely hot, and get even hotter the smaller in size and mass they are. Being completely black, are they generally more hotter than normal stars? Are black holes hotter than our sun?
An in-detail comparison of their heat and the heat of the sun would be appreciated, if not only temperature but electromagnetic radiation measurements.
Heat in black holes is due to Hawking radiation, which means in particular that the temperature you measure will depend on the observer you are considering. For an observer at infinity, the temperature is given by the Hawking temperature, $$T_H = \frac{\hbar c^3}{8\pi k_B G M}.$$ Or, in units with $\hbar = c = G = k_B = 1$, $$T_H = \frac{1}{8\pi M}.$$ However, this is the temperature measured by static observers at infinity. In other words, it is the temperature that observers at a fixed, infinite "radial distance" to the center of the black hole measure. Other observers measure other temperatures. If you are free falling, you won't measure any temperature at all: it is $T = 0 \text{K}$. If you are at a fixed, finite radial distance $r$, then the temperature is $$T_H = \frac{1}{8\pi M \sqrt{1 - \frac{2 M}{r}}}.$$ Notice that $r \to +\infty$ recovers the previous formula. Furthermore, for $r \to 2M$ (i.e., close to the horizon), the temperature gets arbitrarily large. Hence, yes, the black hole will be hotter than the Sun if you are hovering sufficiently close to it. Furthermore, there is no limit on the black hole's mass: this will be true for a black hole of any size, as long as you are close enough.
Notice an important point: the temperature depends on the observer measuring it. A free falling observer will be freezing with the cold, but a hovering observer will be burning with the heat.
As for a more detailed comparison on heat spectrum, black holes are black bodies (okay, there are some caveats), while the Sun is not, and its emission is altered by its composition.
