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Assuming no energy input, the lifetime of a black hole is related to its mass by: $$ T = \frac{5120\pi G^2}{\hbar c^4}M^3 $$ There is a nice summary of the derivation of the lifetime on this web site. I make the condition assuming no energy input because for large black holes the Hawking temperature is less than the temperature of the cosmic microwave ...


When the surface area of liquid is increased, more molecules of liquid can go or evaporate at the same time. Because it also follow the law of physical equilibrium. The vapor pressure of liquid remains constant, then if surface area increases then evaporation also increases.


The chance that water will evaporate is greater than that of steel because of boiling point. The chance that a water molecule will get "bumped" to boiling point is higher than that of steel because steel has a higher boiling point. This doesn't mean that metals don't evaporate. They just do so very slowly.


If you assume adiabatic compression, the equations are $$PV^\gamma = K \text{ (constant)}\tag1$$ Where for air (mostly a diatomic gas), $\gamma=\frac75$. This means that the final volume $V_f$ when the pressure is 50 bar (50x atmospheric) is $$V_f = V_i \left(\frac{P_i}{P_f}\right)^\frac{1}{\gamma} = 61.2 L$$ The temperature of the gas is given by the ...

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