# Entropy from erasing computer memory

Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problem 3.16.

If we regard a single bit of computer memory as a ‘particle’ that can be in 2 different states, then a collection of ${N}$ bits has a possible ${2^{N}}$ states, so its entropy is

$\displaystyle S=Nk\ln2 \ \ \ \ \ (1)$

If we start with a gigabyte (${2^{30}}$ bytes = ${2^{33}}$ bits, so ${N=2^{33}}$) that stores some definite information and then erase it without backing it up, we have effectively lost ${2^{33}}$ bits of information. Does this entail an increase in entropy?

This is rather unusual problem, since I would think the answer would depend on what we mean by ‘erase’. If we replace the original information by a known pattern (say, all zeroes), then it would seem to me that the entropy (at least the entropy contained in the information in the memory; clearly the electrical processes required to perform the erasing would generate entropy) hasn’t changed, since we’ve merely gone from one definite state to another, and we know the bit patterns in both cases.

However, if by ‘erase’ we mean ‘replace the original information with a random pattern’, then presumably we do increase the entropy, since the final state can now be in any of ${2^{N}}$ states. Going on that assumption, the amount of entropy generated by randomizing a gigabyte is

$\displaystyle S=2^{33}k\ln2=8.2\times10^{-14}\mbox{ J K}^{-1} \ \ \ \ \ (2)$

At room temperature, this is equivalent to an amount ${Q}$ of heat

$\displaystyle Q=T\Delta S=298\times\left(8.2\times10^{-14}\right)=2.4\times10^{-11}\mbox{ J} \ \ \ \ \ (3)$

Clearly the heat generated by a computer doesn’t arise mostly from erasing the information in memory!

## 3 thoughts on “Entropy from erasing computer memory”

1. Chase Juneau

I can’t help but notice that Eqn. (1) states that S = Nkln(2), given there are 2^N states. However, I noticed that in Eqn. (2), Eqn. (1) was incorrectly used. Based on the premise that Eqn. (1) is correct, Eqn. (2) should read:

S = 33kln(2) = 3.159x 10^-22 J/K, which yields Q as:

Q = 9.413 x 10^-20 J ~ 0.58 eV (This is roughly the rest mass of an electron!)

Nonetheless, the total heat generated by this “memory dump” is practically nothing, and doesn’t heat the computer much at all.

1. gwrowe Post author

The solution is actually correct as it stands. The number of bits is ${N=2^{33}}$, so the number of possible states for the computer’s memory is ${2^{2^{33}}}$.

2. Chase M Juneau

That is absolutely correct – thank you for your response. There are 2^33 bits in a GB, which is defining N. Each bits takes the states of either 0 or 1 from the binary system.

Thank you again – very helpful.