A "cold" subject,$T_1=300K$, is briefly put in contact with s "hut" object,$T_2=400K$, and$60\text{J}$of heat flows frum the hot object io the cold use. The objects are then spiralled. their temperatures having changed negligibly due ko their large sizes. (a) What are the changes in entropy of each object and the system as a whole?

(b) Knowing only this these objects are in contact and at the given temperatures, what is the ratio of the probabilities of their being found in the second (final) state for that of their being found in the first (initial) state? What dies chis result suggest?

The infinitesimal change in entropy dsrelates to the infinitesimal amount of heat transferaccording to the equation:

$dS=\frac{dQ}{T}$

Where is the temperature of the object or system considered. If the process is reversible, then the change in entropy solely depends on the initial and final states. Thus:

$\begin{array}{rcl}{\int }_i^{f}dS& =& {\int }_i^f\frac{dQ}{T}\\ \Delta S& =& \frac{\Delta Q}{T}\\ & & \end{array}$

Let us now calculate the change in entropy of the cold object $\Delta S_1$since the cold object receives energy via heat transfer, then $\Delta Q_1$must be positive. Thus, plugging in $\Delta Q_1=60\text{J}$and $T=300\text{K}$into the equation for $\Delta S$we get:

$\begin{array}{rcl}\Delta S_1& =& \frac{d{Q}_1}{T}\\ & =& \frac{60}{300}\\ \Delta S_1& =& 0.2\text{J}/\text{K}\\ & & \end{array}$

Contrastingly, since the hot object gives off energy via heat transfer, then $\Delta Q_2$must be negative. Plugging in $\Delta Q_2=-60\text{J}$and $T=400\text{K}$we thus get:

$\begin{array}{rcl}\Delta S_2& =& \frac{d{Q}_2}{T}\\ & =& -\frac{60}{400}\\ \Delta S_2& =& -0.15\text{J}/\text{K}\\ & & \end{array}$

The change in entropy of the system is just the sum:

$\Delta S=\Delta S_1+\Delta S_2$

Plugging in our previous results, the change in entropy of the system must be:

$$\begin{gathered} \Delta S=0.2+(-0.15) \\ \Delta S=0.05\text{J}/\text{K} \end{gathered}$$

Hence the chance in entropy of each system and whole system are, $\Delta S_1=0.2\text{J}/\text{K}$, $\Delta S_2=-0.15\text{J}/\text{K}$and$\Delta S=0.05\text{J}/\text{K}$ .

The ratio of their probabilities is given by the ratio of their macro-states $W\_\left\{2\right\}/W\_\left\{1\right\}$Luckily, the relationship between the probability ratio and the entropy change is given by Boltzmann's formula:

$\Delta S=k_B\ln \frac{W_2}{W_1}$

Re-arranging the above equation we obtain:

$\begin{array}{rcl}\ln \frac{W_2}{W_1}& =& \frac{\Delta S}{k_B}\\ \exp \left(\ln \frac{W_2}{W_1}\right)& =& \exp \left(\frac{\Delta S}{k_B}\right)\\ \frac{W_2}{W_1}& =& e^{\Delta S/k_B}\\ & & \end{array}$

Substitute$\Delta S=0.05\text{J}/\text{K}$ , and $k_B=1.38\times {10}^{-23}\text{J}/\text{K}$we thus obtain for the probability ratio:

$$\begin{gathered} \frac{W_2}{W_1}=e^{0.05/1.38\times {10}^{-23}} \\ \frac{W_2}{W_1}=e^{3.62\times {10}^{21}} \end{gathered}$$

As we can see, the ratio of the system being found in the final state to that of the initial state is an enormous number. This can be interpreted as follows: The probability of the system found in the second state is approximately equal to one whereas the probability of the system found in the first state is nearly zero. This means that the likelihood for the two objects to exchange heat is almost guaranteed to happen