Entropy is defined in my book as $\Delta\ S = \frac{Q}{T}$. To derive the formula it says that entropy should be directly proportional to the heat energy as with more energy the particles would be flying all over the place more rapidly. This makes sense. But then it says that entropy must be inversely proportional to the temperature at which it is added. To be precise
Heat added to a system at a lower temperature causes higher entropy increase than heat added to the same system at a higher temperature.
How does this make intuitive sense?
EDIT 1:
I found an answer here. I think this makes sense. Can anyone read it and tell me if it's correct or not?
Rough intuitive answer:
Adding heat to a system increases the system's total energy. This gives more kinetic energy to distribute among the particles in the system, increasing the size of the system's phase space and hence its entropy. However, since momentum is proportional to the square root of kinetic energy, it is clear that the total momentum phase space cannot grow exponentially as you add more kinetic energy; so the rate of growth of its logarithm must taper off. Since entropy is proportional to the log of the size of the phase space, we see that each additional unit of heat must increase the entropy by less and less.
Now, for technical details. The reason why the relationship is precisely $dS = \delta q/T$ rather than some more complicated relationship that depends on the nature of the system is that temperature is defined such that this is true. We know that during first-order phase transitions, the temperature actually stays constant as you add heat! This corresponds to a region where adding heat actually does cause an exponential increase in phase space. Finally, note that some systems actually have negative temperature, and adding more heat to them actually makes the temperature even more negative; so adding heat actually decreases the phase space. So the intuitive answer in the previous paragraph shouldn't be taken too seriously.
EDIT 2:
Moreover, this answer explains why there needs to be an additional factor of temperature along with heat in the definition of entropy so as to be consistent with the second law.
This is why heat flows from hot objects to cold objects: the entropy change of the hot object is negative, while that of the cold object is positive. The magnitude of entropy change for the cold object is greater.
$\Delta S = \frac{Q}{T_c}-\frac{Q}{T_h} > 0$ as $T_h > T_c$
Keep in mind that entropy increases with temperature. This can be understood intuitively in the classical picture, as you mention. However, at higher temperatures, a certain amount of heat added to the system causes a smaller change in entropy than the same amount of heat at a lower temperature.
The formula is $\Delta S = \frac{Q}{T}$. The change in entropy is related to heat. Remember that heat is a form of energy transfer, not energy; we can talk about heat only when some change takes place.
Please do point out if there is anything wrong with the answers above.
Best Answer
You asked for intuitive sense and I'll try to provide it. The formula is:
$$\Delta S = \frac{\Delta Q}{T}$$
So, you can have $\Delta S_1=\frac{\Delta Q}{T_{lower}}$ and $\Delta S_2=\frac{\Delta Q}{T_{higher}}$
Assume the $\Delta Q$ is the same in each case. The denominator controls the "largeness" of the $\Delta S$.
Therefore, $\Delta S_1 > \Delta S_2$
In each case, let's say you had X number of hydrogen atoms in each container. The only difference was the temperature of the atoms. The lower temperature group is at a less frenzied state than the higher temperature group. If you increase the "frenziedness" of each group by the same amount, the less frenzied group will notice the difference more easily than the more frenzied group.
Turning a calm crowd into a riot will be much more noticeable than turning a riot into a more frenzied riot. Try to think of the change in entropy as the noticeably of changes in riotous behavior.