Entropy Increases in a Coin Toss

Noting that the number of microstates is labeled $W$ in Table 15.4 for the 100-coin toss, we can use $\mathrm{\Delta }S=S_{\text{f}}-S_{\text{i}}=k\text{ln}{W}_{\text{f}}-k\text{ln}{W}_{\text{i}}$ to calculate the change in entropy.

The change in entropy is

where the subscript i stands for the initial 60 heads and 40 tails state, and the subscript f for the final 50 heads and 50 tails state. Substituting the values for $W$ from Table 15.4 gives

This increase in entropy means we have moved to a less orderly situation. It is not impossible for further tosses to produce the initial state of 60 heads and 40 tails, but it is less likely. There is about a 1 in 90 chance for that decrease in entropy ($–2\text{.}7\times {\text{10}}^{–\text{23}}\text{J/K}$) to occur. If we calculate the decrease in entropy to move to the most orderly state, we get $\mathrm{\Delta }S=–\text{92}\times {\text{10}}^{–\text{23}}\text{J/K}$. There is about a $1\text{ in }{\text{10}}^{\text{30}}$ chance of this change occurring. So while very small decreases in entropy are unlikely, slightly greater decreases are impossibly unlikely. These probabilities imply, again, that for a macroscopic system, a decrease in entropy is impossible. For example, for heat transfer to occur spontaneously from 1.00 kg of $\mathrm{0º}\text{C}$ ice to its $\mathrm{0º}\text{C}$ environment, there would be a decrease in entropy of $1\text{.}\text{22}\times {\text{10}}^3\text{J/K}$. Given that a $\mathrm{\Delta }S{\text{ of 10}}^{–\text{21}}\text{J/K}$ corresponds to about a $1\text{ in }{\text{10}}^{\text{30}}$ chance, a decrease of this size (${\text{10}}^3\text{J/K}$) is an utter impossibility. Even for a milligram of melted ice to spontaneously refreeze is impossible.