Different statements of the second law of thermodynamics This may sound trivial but I am having a hard time linking different statements of second law. What I get from second law is that heat cannot be completely converted into work and hence efficiency of heat engine is always less than one. Please explain what it has to do with entropy. Also from second law, entropy of any isolated system never decreases. How are the two statements similar? Please use simple layman terms, I am simply trying to grasp the idea.
 A: You are referring to the Kelvin statement of the second law, which can be stated more precisely as

`no cyclic process is possible in which heat can be taken from a hot source and converted completely into work'.

To see how this is related to the more common definition of the second law

`The entropy of the universe increases in the course of any spontaneous change'.

we need to consider the system and its surroundings. If we remove the cold sink in the heat engine and imagine that all the heat extracted from the hot source is converted into work then the entire change in entropy is $dS=-dq_{\text{hot}}/T_{\text{hot}}<0$, i.e. a violation of the second law, where $dq_{\text{hot}}$ is the heat extracted and $T_{\text{hot}}$ is the temperature of the hot source. Including the cold sink, the change in entropy becomes $dS=-dq_{\text{hot}}/T_{\text{hot}}+dq_{\text{cold}}/T_{\text{cold}}>0$
where the work done is $dW=dq_{\text{hot}}-dq_{\text{cold}}$. For a change in entropy to be positive, the minimum amount of heat that must be discarded into the cold sink must be such that $dq_{\text{hot}}/T_{\text{hot}}=dq_{\text{cold}}/T_{\text{cold}}$, which gives for the work $dW=dq_{\text{hot}}\left(1-T_{\text{cold}}/T_{\text{hot}}\right)$, which gives for the efficiency of the conversion $\epsilon=\left(1-T_{\text{cold}}/T_{\text{hot}}\right)$, which is Carnot's formula.
In short, for there to be an increase in entropy, our heat engine must have a cold sink where some heat can be discarded to provide a positive contribution to the change in entropy, which implies Kelvin's statement.
A: 
What I get from second law is that heat cannot be completely converted
into work and hence efficiency of heat engine is always less than one.

That's not quite correct. Heat can be completely converted to work in a reversible isothermal process. But that would violate the Kelvin statement of the second law:
It is impossible to convert the heat from a single source into work without any other effect.
The key restriction is "without any other effect". After the isothermal process is competed the engine is not in its original state, which means an "other effect" has occurred. In order for the engine to return to its original state it must complete a cycle.
The Kelvin-Planck statement of the second law is

No heat engine can operate in a cycle while transferring heat with a
single reservoir

You will find that in order to perform net work in a complete cycle some of the heat absorbed by the system must be rejected. So while you can completely convert heat into work in a process, you cannot in a cycle.

Please explain what it has to do with entropy.

Whenever heat is transferred, entropy is transferred. The differential change in entropy $dS$ is defined by
$$dS=\frac{\delta q_{rev}}{T}$$
Where $\delta q_{rev}$ is a reversible transfer of heat and $T$ is the temperature at which the transfer occurs.
The maximum efficiency of a cycle is when the cycle is reversible. When a cycle is reversible, the entropy transferred into the system from the surroundings equals the entropy transferred out of the system to the surroundings for a total entropy change of zero. If it is irreversible, additional entropy is generated in the system which must then be transferred to the surroundings in the form of heat. That leaves less heat available to do work.

Also from second law, entropy of any isolated system never decreases.

The only way to decrease the entropy of a system is to transfer mass or heat to the surroundings. An isolated system can't interact with the surroundings in any manner. So the entropy of an isolated system cannot decrease. Therefore the entropy of an isolated system can either stay the same or increase. It will stay the same if the system is in internal equilibrium. It will increase if it is not in internal equilibrium which results in spontaneous processes. Spontaneous processes generate entropy.

How are the two statements similar?

To obtain an understanding on how these and other statements associated with the second law, I suggest you check out this site: https://en.wikipedia.org/wiki/Second_law_of_thermodynamics
Hope this helps.
