The thernodynamic definition for.entropy change is only valid for constant temperature. But when we heat up something and its temperature does not rise what happens then to the added heat. If we assume that the system has a infinitely high heat capacity then the heat added must be do something. But what? Or is it “hidden“ in the gazillion microstates and too small to create a measurable change in the average kinetic energy?
Entropy – Is Entropy Change at Constant Temperature the Amount of Lost Heat?
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There are two definitions of entropy, which physicists believe to be the same (modulo the dimensional Boltzman scaling constant) and a postulate of their sameness has so far yielded agreement between what is theoretically foretold and what is experimentally observed. There are theoretical grounds, namely most of the subject of statistical mechanics, for our believing them to be the same, but ultimately their sameness is an experimental observation.
(Boltzmann / Shannon): Given a thermodynamic system with a known macrostate, the entropy is the size of the document, in bits, you would need to write down to specify the system's full quantum state. Otherwise put, it is proportional to the logarithm of the number of full quantum states that could prevail and be consistent with the observed macrostate. Yet another version: it is the (negative) conditional Shannon entropy (information content) of the maximum likelihood probability distribution of the system's microstate conditioned on the knowledge of the prevailing macrostate;
(Clausius / Carnot): Let a quantity $\delta Q$ of heat be input to a system at temperature $T$. Then the system's entropy change is $\frac{\delta Q}{T}$. This definition requires background, not the least what we mean by temperature; the well-definedness of entropy (i.e. that it is a function of state alone so that changes are independent of path between endpoint states) follows from the definition of temperature, which is made meaningful by the following steps in reasoning: (see my answer here for details). (1) Carnot's theorem shows that all reversible heat engines working between the same two hot and cold reservoirs must work at the same efficiency, for an assertion otherwise leads to a contradiction of the postulate that heat cannot flow spontaneously from the cold to the hot reservoir. (2) Given this universality of reversible engines, we have a way to compare reservoirs: we take a "standard reservoir" and call its temperature unity, by definition. If we have a hotter reservoir, such that a reversible heat engine operating between the two yields $T$ units if work for every 1 unit of heat it dumps to the standard reservoir, then we call its temperature $T$. If we have a colder reservoir and do the same (using the standard as the hot reservoir) and find that the engine yields $T$ units of work for every 1 dumped, we call its temperature $T^{-1}$. It follows from these definitions alone that the quantity $\frac{\delta Q}{T}$ is an exact differential because $\int_a^b \frac{d\,Q}{T}$ between positions $a$ and $b$ in phase space must be independent of path (otherwise one can violate the second law). So we have this new function of state "entropy" definied to increase by the exact differential $\mathrm{d} S = \delta Q / T$ when the a system reversibly absorbs heat $\delta Q$.
As stated at the outset, it is an experimental observation that these two definitions are the same; we do need a dimensional scaling constant to apply to the quantity in definition 2 to make the two match, because the quantity in definition 2 depends on what reservoir we take to be the "standard". This scaling constant is the Boltzmann constant $k$.
When people postulate that heat flows and allowable system evolutions are governed by probabilistic mechanisms and that a system's evolution is its maximum likelihood one, i.e. when one studies statistical mechanics, the equations of classical thermodynamics are reproduced with the right interpretation of statistical parameters in terms of thermodynamic state variables. For instance, by a simple maximum likelihood argument, justified by the issues discussed in my post here one can demonstrate that an ensemble of particles with allowed energy states $E_i$ of degeneracy $g_i$ at equilibrium (maximum likelihood distribution) has the probability distribution $p_i = \mathcal{Z}^{-1}\, g_i\,\exp(-\beta\,E_i)$ where $\mathcal{Z} = \sum\limits_j g_j\,\exp(-\beta\,E_j)$, where $\beta$ is a Lagrange multiplier. The Shannon entropy of this distribution is then:
$$S = \frac{1}{\mathcal{Z}(\beta)}\,\sum\limits_i \left((\log\mathcal{Z}(\beta) + \beta\,E_i-\log g_i )\,g_i\,\exp(-\beta\,E_i)\right)\tag{1}$$
with heat energy per particle:
$$Q = \frac{1}{\mathcal{Z}(\beta)}\,\sum\limits_i \left(E_i\,g_i\,\exp(-\beta\,E_i)\right)\tag{2}$$
and:
$$\mathcal{Z}(\beta) = \sum\limits_j g_j\,\exp(-\beta\,E_j)\tag{3}$$
Now add a quantity of heat to the system so that the heat per particle rises by $\mathrm{d}Q$ and let the system settle to equilibrium again; from (2) and (3) solve for the change $\mathrm{d}\beta$ in $\beta$ needed to do this and substitute into (1) to find the entropy change arising from this heat addition. It is found that:
$$\mathrm{d} S = \beta\,\mathrm{d} Q\tag{4}$$
and so we match the two definitions of entropy if we postulate that the temperature is given by $T = \beta^{-1}$ (modulo the Boltzmann constant).
Lastly, it is good to note that there is still considerable room for ambiguity in definition 1 above aside from simple cases, e.g. an ensemble of quantum harmonic oscillators, where the quantum states are manifestly discrete and easy to calculate. Often we are forced to continuum approximations, and one then has freedom to define the coarse gaining size, i.e. the size of the discretizing volume in continuous phase space that distinguishes truly different microstates, or one must be content to deal with only relative entropies in truly continuous probability distribution models Therefore, in statistical mechanical analyses one looks for results that are weakly dependent on the exact coarse graining volume used.
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.
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That is not correct. A differential change in entropy is defined for a reversible transfer of heat as follows
$$ds=\frac{\delta q_{rev}}{T}$$
So the change in entropy between two equilibrium states 1-2 is
$$\Delta S_{12}=\int_1^2 \frac{\delta q_{rev}}{T}$$
Only if $T$ is a constant does it come out of the integral. But it does not have to be constant.
In the absence of work, if the temperature does not change as a result of heat transfer it is because the heat capacity of the substance is very large compared to the amount of heat being transferred. The heat increases the internal energy of that something but not enough to be detected as a measurable temperature increase. We call this something a "thermal reservoir", and we call the heat transfer process an isothermal (constant temperature) process. For isothermal processes $T$ comes out of the above integral and the change in entropy is the heat transferred divided by the constant temperature.
The heat capacity does not have to be "infinitely high". It just has to be high relative to the amount of heat being transferred. The relative heat capacity of a cup of water at room temperature to the heat capacity of a hot grain of sand is great enough that while dropping the grain of sand into the water increases the internal energy of the water, the change would not result in a measurable change its temperature. Yet the heat capacity of the cup of water is not "infinitely high".
Entropy, and the second law, was defined like it is in order to account for the fact that heat can only flow naturally from a warmer to a cooler body, and never in the reverse direction. The new property and law was needed because the first law (conservation of energy) and the property of internal energy would not preclude heat from naturally flowing from the cooler to warmer body.
For a description of how the property came to be defined, see the following: https://en.wikipedia.org/wiki/History_of_entropy
First of all, entropy is not “energy per kelvin lost to do useful work”. It is entropy generated due to an irreversible process that results in less energy available for doing useful work. In going from equilibrium state 1 to 2 the total change in entropy of the system is
$$\Delta S_{tot}=S_{2}-S_{1}= S_{transferred}+S_{generated}$$
Since entropy is a system property, the total change in entropy between two states is the same for all paths (processes) between the states. For a reversible process, the entropy generated within the system is zero and the total entropy change is just the entropy transfer at the boundary with the surroundings. For an irreversible process additional entropy is generated within the system. It is the entropy generated within the system in an irreversible process that results in less energy to do useful work. The generated entropy internal to the system equals the total change in entropy minus the entropy transferred at the boundary.
An example for a reversible (Case 1) and irreversible (Case 2) isothermal process with the diagram below illustrates the how an irreversible process results in less work than a reversible process although the entropy change between the same two states is the same.
CASE 1: Reversible Isothermal Expansion
The process 1-2 in the diagram is a reversible isothermal expansion of an ideal gas (pV=constant).
For an ideal gas the change in internal energy depends only on temperature change. Since the temperature is constant, the change in internal energy between state 1 and 2 is zero. From the first law $Q=W$ and for one mole of gas we have
$Q=W=(RT)$ln$\frac{V_2}{V_{1}}$
The total increase in entropy between states 1 and 2 is
$\Delta S=S_{2}-S_{1}= \frac{Q}{T}$
Where $Q$ is the heat added during the expansion and $T$ is the equilibrium temperature of the gas and the surroundings. Combining this with the previous equation for one mole of gas
$\Delta S=S_{2}-S_{1}=$ R ln$\frac{V_2}{V_{1}}$
So we see that for the reversible process the work done is the sum of the light and dark grey areas and equals the heat added.
CASE 2: Irreversible Isothermal Expansion
Process 1-1a-2 is represents an irreversible isothermal process connecting the same initial and final states. The process is carried out as follows:
At state 1 the gas is in equilibrium with the surroundings as was the case for the reversible process. However, instead of gradually lower the external pressure as was done in the reversible expansion, it is abruptly lowered from 1 to 1a to equal the final pressure. From 1a to state 2 the gas is allowed to expand irreversibly against constant external pressure $P_{2}$ and constant temperature $T$ (at the boundary between the system and surroundings) and allowed to re-equilibrate with the surroundings at point 2.
In going from 1 to 1a, because the pressure drop is sudden, there is no time for heat transfer to occur. Consequently for that part of the path $Q=0$.
The lack of heat transfer between 1 and 1a results in the lost opportunity to do work equal to the light grey area under the reversible expansion 1-2.
For the irreversible process with the ideal gas, the first law still applies between states 1 and 2, so the heat transfer at the boundary equals the work done, or
$$Q=W=P_{2}(V_{2}-V_{1})$$
Which is the dark grey area.
The entropy transfer at the boundary is
$$S_{transferred}=\frac{Q}{T}=\frac{P(V_{2}-V_{1})}{T}$$
So the entropy generated internal to the system is
$$S_{generated}=R ln \frac{V_2}{V_1}-\frac{P(V_{2}-V_{1})}{T}$$
Conclusion: Although the change in entropy from state 1 to state 2 is the same, less work is done by the irreversible expansion than the reversible expansion due to the entropy generated within the system during the irreversible process.
Hope this helps.