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While going through the idea for the hawking radiation,the argument was

"As T is inversely proportional to the product of GM(planck), so as black hole radiates energy ,then temperature will increase and mass will decrease,thus we say that black hole evaporates".

But my question is if some body is heated and source of heat is taken away,then as it is radiating energy and so temperature must decrease,but here opposite is happening. Why.

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  • $\begingroup$ Time to google negative specific heat. $\endgroup$ – thermomagnetic condensed boson Mar 20 at 11:17
  • $\begingroup$ It is certainly interesting question.In classical gas radiation means the particles become more and more bound and from the potential energy they loose they radiate. At some point (unelastic scattering) they form a (rigid) body and cool down. It is in some sense "binding energy" as known from nuclear physics. In the latter case nickel is formed, somehing that stays and resists. In case of black hole the black hole disappears. What a paradox! In what form is the potential energy (from which radiation is crated) stored in black hole? Presumabely the virial theorem is not to be applied here. $\endgroup$ – F. Jatpil Mar 20 at 11:54
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The radiation mechanism is quite different.

The thermal radiation temperature of a hot "black body" is governed by the surface temperature of the body.

The radiation temperature of Hawking radiation from a black hole is governed by the curvature of spacetime at its event horizon. The smaller the black hole, the tighter the curvature and the more energetic the radiation quanta, i.e. the higher the radiation temperature.

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if some body is heated and source of heat is taken away,then as it is radiating energy and so temperature must decrease

That is only true for small amounts of heat energy that curve space-time just a small amount.

If we consider a suitably large amount of heat energy inside a suitably small volume, we can observe that as energy is radiated away the radiation temperature of the radiation increases.

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