Why is such huge mount of heat released by brute-force GPG cipher? Please see GnuPG FAQ. I'm really interested in how to derive to "boil the ocean" by using the Landauer bound and the Margolus–Levitin limit.
Any suggestion is welcome.
 A: I'll take a crack at this:
To brute force this, you'd have to, in the worst case, try every one of the possible combinations of keys. Let's take their example of a 128 bit key.
The Landauer limit states that at least $kT \ln2$ energy must be generated to erase a bit. You have to store your key somewhere so everytime you try one, you'll have to erase the old key with your new one, consisting of 128 bits.
Margolus–Levitin tells us that there is a maximum processing rate for computation. The time required to go from one state to an orthogonal state is at least time $\frac{h}{4E}$. The states in our system are the bits, which are potentially all orthogonal (e.g. $0000\dots \rightarrow 1111\dots$).
Putting these together, using the minimum amount of heat generated for each bit, and the maximum computation rate, let's be generous and say we can take 10 years to brute force a key.
Then to test a key that only differs by one bit from the previous takes:
$$t \approx \frac{h}{4E} \approx \frac{h}{4 kT \ln2}.$$
Solving for the temperature we find:
$$T \approx \frac{h}{k}\frac{1}{t}\frac{1}{4\ln2}.$$
Remember that we need to do this for each $2^{128}$ combination of bits, and the temperature change will linearly increase with every attempt so we have (using 10 years):
$$T \approx \frac{h}{k}\frac{1}{t}\frac{1}{4\ln2} 2^{128} \approx 1.3 \times 10^{19}K$$
I haven't checked if this would actually char the planet, but this is obviously a huge number and likely where GnuPG drew their conclusions from.
Note: this is totally ignoring dissipation of the heat generated, so take this with a grain of salt.
A: That GnuPG FAQ entry doesn't make much sense.  The minimum heat released by a computer brute-forcing a 128-bit cipher can be calculated just from the Landauer limit, the Margolus-Levitin limit doesn't factor in at all.  From Landauer's Principle we know that each bit erasure requires $kT \ln 2$ of energy to be released, so $2^{128}$ bit erasures  at a temperature of $295 K$ will cause $9.6 * 10^{17} J$ to be released.  This is a lot of heat, perhaps too much for a brute force attack on a 128-bit cipher to be reasonable, but certainly not enough to boil the Earth's oceans!
The Margolus-Levitin limit can be used to calculate the minimum time required, but that time is extremely small - a $1 \ \text{kg}$ computer can process a maximum of about $5.4 * 10^{50}$ operations in a second, so $2^{128}$ operations would take roughly $6.3 * 10^{-13}$ seconds.
My guess is that the GnuPG FAQ cribbed the "boil the oceans" idea from Jeff Bontick, who made the claim that fully populating a 128-bit storage system (which would actually have $2^{140}$ bits) would require boiling the oceans, and makes the argument for that claim here.  For his argument he actually uses a bound on storage capacity found by Warren Smith and Seth Lloyd, not the Landauer or Margolus-Levetin limits.  His argument is also flawed, in that he finds the minimum mass required by the limit (136 billion kg) and then says to operate at the limit, "the entire mass of the computer must be in the form of pure energy".  Even if this is true, there is no requirement to operate at the limit; a storage device of 272 billion kg of regular matter operating at half of the Smith-Lloyd limit would be extreme, but more palatable than half that mass of pure energy, and wouldn't require boiling the oceans.
