Ok. Entropy $S$ is somehow related to energy dispersal and thus "hides" some part of the energy from being converted to work $W$, that is, the $Q$ you've received from hot sink $T_h$ will not be fully converted to $W$, because there should be "heat dump" onto cold sink at the temperature $T_c$.
Question 1: This (above) is a principle of "limit to a heat engine's efficiency" in the action. Right
That is correct as it pertains to a heat engine operating in a cycle, because any such cycle requires rejection of heat to a lower temperature body (see answer to question 2, below). This precludes the possibility of a 100% efficient engine operating in a cycle.
Question 2: What if there were no such principle? I know the universe
will disintegrate or something... but I care only about one thing.
Would this mean that there is no entropy in some sense? Since there is
no limit to heat engine efficiency, then could there be an engine,
which takes $Q$ from $T_h$ and just completely converts $Q$ into $W$
without any cold sinks $T_c$
The problem is, there IS such a principle. If there were no such (Kelvin-Plank) principle, then the heat engine you describe would constitute a perpetual motion machine of the second kind. It would be capable of continuously spontaneously transferring heat from a single thermal reservoir and convert it into work. No one has been able to create such a machine.
In order for a heat engine to operate in any cycle and perform net work, it will need to reject some heat in order to complete the cycle.
For example, take again the case of the Carnot cycle. Heat is absorbed from the high temperature source during the reversible isothermal expansion and the gas does work. But for the device to perform this work in a cycle you need a path to get back to the initial state. You could reverse the process to isothermally compress the gas and get back using the same single temperature source, but the work done on the gas during the compression would equal the work done by the gas during the expansion and no net work would be done! Therefore, in the case of the Carnot cycle, you need to isentropically expand the gas to get to a lower temperature and then isothermally compress the gas rejecting heat to a second, lower temperature sink. The cycle then wraps up with an isentropic compression to the original state and net work is done.
Can you envision any thermodynamic cycle, reversible or not, in which net work is done without involving at least one process where heat is rejected? I can't.
Question 3: then entropy should somehow be a consequence from Kelvin-Plank statement (see quote below) and it maybe possible to derive entropy existence from this statement, right?
Yes, or the other way around, to the extent that entropy is associated with efficiency limits and the Kelvin-Plank statement prohibits a 100% efficient heat engine operating in a cycle. The example of a Carnot cycle can help to show this.
A Carnot cycle consists of two isothermal processes (expansion and compression) involving two thermal reservoirs (heat source and sink sufficiently massive that their temperatures remain constant for a given quantity of heat transfer). Let:
$T_H$ be the temperature of the high temperature thermal reservoir.
$T_L$ be the temperature of the low temperature thermal reservoir (heat sink).
$Q_H$ be the heat transferred from the high temperature reservoir to the system during the isothermal expansion (Work done by the system).
$Q_L$ be the heat transferred from the system to the low temperature reservoir (work done on the system).
Since the heat transfers occur isothermally (at constant temperature), the entropy changes of the two thermal reservoirs are as follows.
$\frac {-Q_H}{T_H}$ is the change in entropy of the high temperature reservoir (a reduction in entropy)
$\frac {+Q_L}{T_L}$ is the change in entropy of the low temperature reservoir (an increase in entropy). So the total entropy change is given by:
$$\Delta S_{total}=\frac {+Q_L}{T_L}+\frac {-Q_H}{T_H}$$
The second law tells us that the total change in entropy must be equal to or greater than zero. The equality applying to both transfers being reversible. Keep in mind that the change in entropy of the system has to be zero since it has undergone a complete cycle which restores all its properties, including entropy, to their original states.
Now, in order to maximize the net work, $W$, done in this cycle you would want to make the heat rejected, $Q_L$, be as low as possible relative to $Q_H$. In your case you are asking that $Q_L=0$ (no heat sink) so that you have 100% efficiency. But if $Q_L=0$, then from the previous equation $\Delta S<0$ in violation of the second law.
The best we can do is the case where $\Delta S=0$ (a reversible cycle), in which case we have:
$$\frac{Q_L}{T_L}=\frac{Q_H}{T_H}$$
or
$$\frac{Q_L}{Q_H}=\frac{T_L}{T_H}$$
This last equation basically limits the maximum possible efficiency of all heat engines to convert heat into work in a thermodynamic cycle. Recalling that the definition of efficiency of a heat engine is given by
$$ζ=\frac{W}{Q_H}$$
And
$$W=Q_H-Q_L$$
We get
$$ζ=1-\frac{Q_L}{Q_H}=1-\frac{T_L}{T_H}$$
Which is the Carnot cycle efficiency.
Question 4: Can someone provide a link where entropy is derived as a consequence of Kelvin-Plank statement or as a consequence to "the limit to heat engine efficiency".
I have not been able to find a link that directly connects the two. But hopefully the example I gave above can help you connect entropy and the Kelvin-Plank statement.
Hope this helps.
