Energy yield of a meteor blast - how to explain the discrepancy? The energy yield of the Chelyabinsk meteor blast is estimated to be about 500 kT TNT, or $2\cdot 10^{15}$ joules. When compared that to its estimated speed ($\sim 2\cdot 10^4$ m/s) and mass ($1.3\cdot 10^7$ kg), we see that this is about all of the kinetic energy is had upon entering the atmosphere.
As far as I understand, the mechanism of the blast is that kinetic energy quickly turns into heat. This means that every kg of the meteor would receive $2\cdot10^8$ joules. Dividing by a specific heat capacity, say, of $2\cdot 10^3$ J/(kg*K), this means that the meteor would be heated to $10^5$ K. This looks unreasonable, as is seems that something dramatic must happen at much smaller temperatures (e. g., the entire meteor would vaporize, as melting and vaporization only requires $\sim 10^6$ J/kg).
So, what is going on here? Could it be that 500 kT figure refers to the entire energy, and the energy yield of the blast is much smaller? Then, how come the blast produced the shock wave capable of breaking windows a dozens of kilometers away? Does part of the shock wave come directly from the meteor going through atmospehre? 
 A: I am hoping that someone who understands the behaviour of meteors quantitively can give a better answer than this rather hand-waving one.
First of all, energy is conserved so the kinetic energy of the meteor as it enters the atmosphere is going to go somewhere.  The only question is how fast the energy gets turned into heat, really.  And I think there turn out to be only two options once the meteor has made it intact into the lower atmosphere:


*

*it could survive mostly intact until it hits the surface, dumping its energy at the surface in an extremely short period of time, which will result in a significantly explosive event;

*it could dump its energy, rather rapidly, in the atmosphere, which is also an explosive event as it will heat the atmosphere dramatically and also turn most or all of itself to very hot gas.  This is what happened to the Chelyabinsk meteor, of course.


The reason I think these are the only two options is by considering a third one and showing that it turns into the second one.  The third proposed option is that it breaks up in the atmosphere and dumps its energy relatively gradually, as the fragments slow down, get hot and 'burn up'.  Consider what happens in this case: the meteorite fragments (relatively gently) into a number of smaller objects, But these smaller objects have a significantly greater surface/volume ratio than the original meteorite since they are smaller, and they are travelling at hugely supersonic speeds in the lower atmosphere as it was, since they carry essentially all its kinetic energy.  So they are going to lose kinetic energy (and heat themselves and the air they are travelling through) much more rapidly than the original object.  If they themselves fragment, then their yet smaller fragments will heat still more rapidly as their surface/volume is still larger.  So if the object fragments in the lower atmosphere the fragments will, extremely quickly, get turned into meteor-vapour and will dramatically heat the air they pass through in the process.  This is an explosive event, almost by definition.
So this third option in fact isn't an option: it's the same as the second one, and my guess is that it's how the second one happens.  See this SE answer which describes this phenomenon rather better and has references (I wasn't aware of it when writing this answer).
So I think the only way the thing can not lose all its KE in an essentially explosive way is if it fragments very high in the atmosphere, which would allow the fragments to lose KE much more slowly: once it's made it to the lower atmosphere it's either going to dump all its KE into the atmosphere explosively or it's going to hit the surface, which is a different kind of explosion.
A: Here’s one effect that contributes.  When air is heated, either by the shock wave, or viscosity acting on the turbulent wake of a meteor, or by soft x-rays from a nuclear blast, its pressure is initially raised.  It will then expand back to normal ambient pressure, more or less adiabatically, according to the isentropic rule $P\sim {{\rho }^{7/5}}\sim {{T}^{7/2}}$ .   But the air will still be warmer than ambient due to increased entropy, and the residual heat will slowly be lost via conduction and/or convection, without conversion into mechanical blast.  
