Doubt over Kelvin-Planck statement of thermodynamics' second law I have a doubt over the Kelvin and Planck's statement of thermodynamics' second law, in particular applied to a cycle. Let's take a Carnot cycle as an example, and let's call the first two transformations (the isotherm and the adiabatic) done. Now, isn't it obvious that the machine has to give up heat to go back to the initial state? Isn't it something that follows from the fact that the cycle has to be continuous?
 A: Every thermodynamic system satisfies the first law during a given process;
$$
  \Delta E = Q-W
$$
Here $\Delta E$ is the change in its internal energy, $Q$ is the heat transferred to the system, and $W$ is the work done by the system.  For a system undergoing a cyclic process, namely one for which it starts and ends in the same thermodynamic state, one has $\Delta E = 0$, and the first law then tells us that
$$
  Q = W
$$
Now, suppose that the system is taking some heat $Q_H>0$ from a reservoir, and turning it into some work $W$.  Let's define
$$
  Q_\mathrm{exhuast} = Q-Q_H
$$
then we can write
$$
  Q_H + Q_\mathrm{exhaust} = W
$$
The Kelvin statement tells us that we must have $Q_\mathrm{exhaust}\neq 0$ because otherwise, we would have $Q_H = W$; the sole result of the process would have been to transform the heat it took in into work.  The symbol $Q_\mathrm{exhaust}$ is what one commonly calls the exhaust heat.
A: in simple words we can expain kp statement as it is impossible to completely convert heat into work in a cycle(*).  
while in case of a single process say a isothermal expansion as dt=0 therfore ineternal energy which is a function of temp so dE = 0 neglecting k.e and p.e so from 1st law we get dq = dw .
in isothermal expansion we are able to get work but only  till the pressure of the gas do not becomes equal to the atmospheric pressure in which we can convert heat into work but not infinitely.
hence the kp statement is that we cannot completely convert heat into work in a cycle of infinitely.....
