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Entropy change, $\Delta{S}$, can be found from the $\frac{1}{T} - Q$ graph. When the temperature doesn't change during the dispersal of heat energy in the system, the area under the graph is more, that is, change in entropy is more. But, when the temperature increases, the area under the graph decreases, that is, change in entropy is less.

I have taken $Q$ on the x-axis and as $T$ is a function of $Q$, $\dfrac{1}{T}$ is represented on the y-axis. Now, when $T$ doesn't change, the graph covers a rectangular area but when $T$ increases, the graph covers much smaller area. So, what's the reason? Do the number of possible microstates decrease when the temperature decreases?

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Macroscopically, entropy $S$ is a function of energy $U$ and maybe some other extensive quantities (volume, particle number, etc.) Temperature can be defined as the derivative of $S$ with respect to energy with the other extensive quantities held fixed,

$$\beta\equiv\frac{1}{T} \equiv \frac{\partial S}{\partial U}. $$

So $\beta$ determines how much change in entropy $\Delta S$ you get if you add some energy $\Delta U \equiv Q$ to the system.

$$ \Delta S = \frac{1}{T} Q $$ If a certain amount of heat $Q$ is lost from a higher temperature system it loses a smaller amount of entropy (since $\beta$ is smaller) compared to the energy gained by a lower temperature system. So the decrease in entropy with temperature for fixed $Q$ is why heat flows from hot to cold.

Edit- I noticed the statement of the question has been changed since I've posted the answer. Now it is: "Do the number of possible microstates increase as temperature decreases?" The answer is no, the entropy increases with temperature, but the rate of increase of entropy with energy decreases with temperature. Notice we are talking about $\Delta S$ not the the total $S$ (and $Q$ should be understood as a transfer of energy not the total internal energy).

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  • $\begingroup$ Not this is the question - see my comment for the other answer. Paradoxically, as the temperature decreases, the disorder increases. It seems counter-intuitive. $\endgroup$
    – Sofia
    Dec 6 '14 at 3:06
  • $\begingroup$ @Sofia Read what I wrote again. As the temperature decreases the disorder gained from a fixed $Q$ does indeed increase. That is why it is entropically favorable for heat to flow to cold systems. More entropy is produced than is lost by a hot system. What is confusing you is if you decrease the temperature of a single system you are taking heat out of it - entropy does decrease. But if you put heat back into the now cold system the rate of entropy increase is higher at the low temperature then it was when it was at a high temperature. $\endgroup$
    – octonion
    Dec 6 '14 at 4:46
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Entropy $S$ is a measure of microscopic freedom, temperature $T$ the average energy per degree of freedom and heat $Q$ the amount of energy transferred microscopically.

They are related through the equation* $$ Q=T\cdot\Delta S $$ or verbally $$ \text{energy transferred microscopically} = \text{energy per degree of freedom} \times \text{change in microscopic freedom} $$ Thus, for fixed $Q$, $T$ and $\Delta S$ are inversely proportional:

If you have a given amount of energy available, dispersing it more widely requires a lower average energy.


[*] This is a simplification, as we actually have a differential relation $$ \delta Q = T\cdot \mathrm dS $$ for reversible and $$ \delta Q < T\cdot \mathrm dS $$ for irreversible processes.

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