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In order for the black hole to evaporate it must have a temperature greater

than that of the present-day black-body radiation of the Universe.

Cosmic microwave background radiation temperature:

[tex]T_u = 2.725 \; \text{K}[/tex]

Hawking radiation temperature:

[tex]T_H = \frac{\hbar c^3}{8 \pi G M k_B}[/tex]

Hawking radiation temperature is greater than or equal to cosmic microwave background radiation temperature:

[tex]T_H \geq T_u[/tex]

[tex]\frac{\hbar c^3}{8 \pi G M k_B} \geq T_u[/tex]

Hawking total black hole mass:

[tex]M_H \leq \frac{\hbar c^3}{8 \pi G k_B T_u} \leq 1.226 \cdot 10^{23} \; \text{kg}[/tex]

[tex]\boxed{M_H \leq 1.226 \cdot 10^{23} \; \text{kg}}[/tex]

Earth total mass:

[tex]M_{\oplus} = 5.9722 \cdot 10^{24} \; \text{kg}[/tex]

[tex]\frac{M_H}{M_{\oplus}} = 0.007539 = 0.754 \; \text{%}[/tex]

According to reference 4 - p. 2, eq. 1:

[tex]\frac{M_H}{M_{\oplus}} = 0.8 \; \text{%}[/tex]

Are these equations correct?

Reference:

Cosmic microwave background radiation - temperature - Wikipedia

Hawking radiation - black hole evaporation - Wikipedia

Earth mass - Wikipedia

The Last Eight Minutes Of A Primordial Black Hole - Joseph Kapusta

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# Hawking radiation and cosmic microwave background radiation

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