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Let’s say I have 1kg of matter at room temperature (300K), held in a spherical configuration. I symmetrically squeeze this until it forms a black hole. What formula (formulae) would I use to (or how can I) calculate the new temperature of the 1kg of matter once it reaches its event horizon?

user131744
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    A 1 kg BH has a Schwarzschild radius of 1.48517E-27 m, a trillion times smaller than a proton, and a lifetime of 4.65118E-17 s. See https://www.vttoth.com/CMS/physics-notes/311-hawking-radiation-calculator – PM 2Ring Dec 21 '21 at 20:27

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When the air in a room is at $300 \mathrm{K}$, objects placed in that room only acquire a temperature of $300 \mathrm{K}$ as well if they can come to equilibrium with it. This will not be the case for a black hole. The purely classical prediction is that all of the air molecules will eventually fall into the black hole making it an infinitely long lived object with temperature zero. An improved analysis (due to Hawking) shows that if a black hole sits around for a long enough time without colliding with anything, it will evaporate. This means it has a nonzero temperature but one determined entirely by its mass.

The order of magnitude for this temperature can be read off from a previous question. Since you said $1 \mathrm{kg}$ instead of $0.2 \mathrm{kg}$, the temperature will be $1.2 \cdot 10^{23} \mathrm{K}$.

Connor Behan
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