A bit of computer memory is some physical object that can be in two different states, often interpreted as 0 and 1. A byte is eight bits, a kilobyte is $1024\left(=2^{10}\right)$bytes, a megabyte is 1024 kilobytes, and a gigabyte is 1024 megabytes.

(a) Suppose that your computer erases or overwrites one gigabyte of memory, keeping no record of the information that was stored. Explain why this process must create a certain minimum amount of entropy, and calculate how much.

(b) If this entropy is dumped into an environment at room temperature, how much heat must come along with it? Is this amount of heat significant?

Computer memory can be classified in two different states, often defined as 0 and 1 .

$$\begin{gathered} 1byte=8bits \\ 1kilobyte=1024(=2^{10})bytes \\ 1megabyte=1024(=2^{10})kilobytes \\ 1gigabyte=1024(=2^{10})megabytes \end{gathered}$$

The expression for entropy with multiplicity $\Omega$is given as:

$S=k\ln \left(\Omega \right)\text{.........(1)}$

Where,

$k$= Boltzmann constant

$N$= the number of atoms

$\Omega$= multiplicity

If a single bit of computer memory is taken to be a particle with two states, then a collection of $N$bits has a particle $2N$, and its entropy may be determined using equation (1)

$S=k\ln \left(\Omega \right)=Nk\ln \left(2\right)\dots \dots ...\left(2\right)$

If a gigabyte $\left(2^{30}\right.bytes\approx 2^{33}bits,so\left.N=2^{33}\right)$is used to store certain information and later erased without a backup, $2^{33}bits$of information is lost indirectly. If the original information is replaced with a known pattern, it appears that the entropy hasn't changed because the state has only changed. If the original information containing the random pattern is deleted, entropy is likely to increase.

The amount of entropy generated by randomizing gigabytes can be given as:

$S=Nk\ln \left(2\right)=2^{33}k\ln \left(2\right)\dots \dots \dots .\left(3\right)$

By substituting the value of $k$in the above equation, we get,

role="math" localid="1647267704138" $$\begin{gathered} S=\left(2^{33}\right)\left(1.38\times {10}^{-23}\right)\ln \left(2\right) \\ S=8.22\times {10}^{-14}{\mathrm{JK}}^{-1} \end{gathered}$$

Hence, the required entropy is$8.22\times {10}^{-14}J{K}^{-1}$.

Room temperature $=T=298K$

Entropy created$=∆S=8.22\times {10}^{-14}J{K}^{-1}$

The amount of heat is given as:

$Q=T\Delta S$

By substituting the values in the above equation, we get,

$$\begin{gathered} Q=298\times 8.22\times {10}^{-14} \\ Q=2.45\times {10}^{-11}\mathrm{J} \end{gathered}$$

Hence, the amount of heat generated is$2.45\times {10}^{-11}J$.