Dependence of change in entropy on temperature

The thermodynamic definition of entropy is expressed as 
$$dS = \frac{dq\_{rev}}{T}$$
I understand that it depends on amount of heat transferred because heat transfer can be understood as the degree of dispersion of energy but what does the temperature dependence actually mean?




An analogy that I like is one paraphrased from Peter Atkins' great book "The Laws of Thermodynamics - a VSI": 



> 
> Imagine two rather different places in terms of noise - a silent
>  library and a busy train station. Imagine that the temperature is the amount of noise at the given place - a measure of the "state of noise". Let's say that you laugh out loud,
>  corresponding to supplying energy as heat. 
> 
> 
> In the library, where there is little background noise (low $T$) this would cause a large amount of disturbance and
>  disorder - corresponding to a large increase in entropy. On the busy
>  train station, on the other hand, you laughing would go largely
>  unnoticed - there is lots of background noise already (high $T$). The (relative) amount of additional disturbance and
>  disorder would be much smaller, corresponding to a smaller increase of
>  entropy.
> 
> 
> 


In a very cold system, even a tiny amount of supplied energy in terms of heat would cause a large change in entropy. On the other hand, supplying the same amount of energy to a really hot system - with lots of molecular motion going on - would cause a minuscule change in entropy. 


**Disclaimer:** Note that the analogy and explanation above are two ways of convincing yourself that the entropy change due to added heat is inversely proportional to the system temperature. They do not claim to be correct or physical in any other sense, so read them with a (large) pinch of salt. 


 4

What this equation is saying essentially is that the level of disorder can be modeled as the number of configurations if you have a number of particles and a number of quanta of energy. It's like having three cups and three balls; all the balls can occupy one cup or one ball can occupy one cup and two can occupy another etc. Hence an increase in energy would result in an increase in entropy (more balls and more configurations). Temperature is defined as the average energy of the system. Hence if energy is concentrated (all the balls are in one cup) then the average temperature per particle will be lower and the entropy will be lower which makes sense since the system is more ordered with all the balls in one cup.


I hope this helps answer the question somewhat


 1

If in defining entropy we did not divide the heat by T, heat could flow from a cold body to a hot body with no change in entropy i. e. without breaking the second law. 


This does not happen. 


Likewise, in order for heat to flow from a hot body to a cold body, the loss of entropy of the hot body has to be smaller than the gain of entropy of the cold body. 


 1

I am also often confused and have no clear answer. Consider melting of ice. If we do not allow any physical change (no movement, no pressure change, no volume change), it will not happen. Without volume change when heat is added to gas, entropy cannot increase. We may say the heat was transferred to internal energy...
I think the temperature dependance of entropy is not the nature of entropy Itself but it is caused by the changes of parameters affecting entropy that are affected by temperature such as volume, pressure.. Most thermodynamic equations are for gases. For liquid and solids the final equations may not be the same as those for gases although all definitions should be the same. Based on statistical definition of entropy, $S=k\_\mathrm{B} \ln{W}$, it is obvious that there is no temperature term but it is assumed that particles can freely interchange positions. If particles cannot move ($0\, \pu{K}$), $W=1$ and entropy will be 0 regardless of how many different particles are three. When we calculate entropy changes for melting we just consider $\Delta H$ and T but we do not think about volume changes or anything else. All such changes are already included in the delta H value we use and assumptions. If molecules are not allowed to move, melting will no happen because crystalline state is that molecules cannot move. All reversible processes in thermodynamics allow changes associated With them as long as they are reversible. So, I would prefer not thinking too much about details. Leave them to physicists...


 0

This is a bit of a "chicken-and-egg" problem, as those ultimately dealing with definitions tend to be. The concepts of entropy and temperature were originally developed empirically and were later modified and merged into a coherent theory of thermodynamics. It turns out that temperature can be defined as follows (this follows from the combined 1st and 2nd laws of thermodynamics):


$$T^{-1}=\left( \frac{\partial S}{\partial E} \right)\_V$$


$T^{-1}$ describes the sensitivity of the system's entropy to the addition of energy at constant volume.


What the equation means can also be interpreted if we define entropy as follows (modifying a statement in the OP): 



> 
> *entropy* can be understood as a measure of the degree of dispersion of energy (the number of possible configurations of the system given the current energy)
> 
> 
> 


When a fixed amount of energy is added to a system more configurations become accessible (invoking concepts from statistical mechanics). At a lower temperature the number of additional configurations - in proportion to the initial number - increases by a greater amount than at a higher temperature. This can be restated as "a higher entropy change results when a very small fixed amount of energy is added to a system at low temperature". This explanation of the difference between the entropy change at low and high temperatures is captured by the silent library/busy train station analogy presented in another answer.


 0