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The statements of the second law of thermodynamics are:

  • Kelvin statement : No process is possible in which the sole effect is the absorption of heat from a reservoir and its complete conversion into work.
  • Clausius statement : No process is possible in which the sole effect is the transfer of heat from a colder reservoir to a hotter reservoir.

Are both these statements equivalent? If so, how? And also how are they related to entropy?

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closed as off-topic by Jon Custer, peterh, user36790, AccidentalFourierTransform, John Rennie Dec 8 '16 at 9:26

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    $\begingroup$ Did you consider looking this standard question up anywhere? $\endgroup$ – Sanya Dec 7 '16 at 13:54
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The "sole effect" in the two statements is a wrong requirement that converts them into true but trivial propositions, having nothing to do with the second law of thermodynamics. Consider the following scenario. An operator carries out heat and work exchanges between the system and the environment (which is the only heat reservoir). If, at the end of the cycle, the system is in its initial state, a weight is lifted in the environment, and an equivalent amount of heat is missing in the environment, the second law is obviously violated. Yet, according to the Kelvin statement you refer to, the second law is not violated - the "sole effect" requirement is not obeyed. Apart from the lifted weight and the missing heat, which are allowed by the "sole effect" requirement, there are also unallowed but unavoidable changes in the operator.

The "sole effect" requirement is much more pernicious in the Clausius case - the statement becomes equivalent to

Heat cannot move SPONTANEOUSLY from cold to hot

which is true but trivial and has nothing to do with the second law.

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One way in which to show the two statements are equivalent is to show that if one of them is violated then so is the other. We show here that if the Kelvin statement is violated, then so is the Clausius statement.

Imagine there were such a things as a perfect heat engine: For each joule of heat extracted from a hot reservoir, exactly one joule of work would be done. For such a heat engine $𝑄_𝐶^E= 0$ so $𝑄_𝐻^𝐸 = 𝑊$, where $𝑄_𝐶^E$ is the cold reservoir for the perfect heat engine $E$ and $𝑄_H^E$ is the hot reservoir for the perfect heat engine $E$ and $W$ is the work done in extracting heat for each joule.

Now suppose the work from this heat engine is used to run a refrigerator – not an ideal refrigerator, just a normal one, which can have any coefficient of performance at all – which discharges its heat into the same hot reservoir.

Let’s say we can adjust the engine so that it produces exactly the same amount of work required to run the refrigerator so $𝑊^𝐸 = 𝑊^R$, where the superscript $R$ is used to denote the regular (not ideal) refrigerator.

Using the conservation of energy we have $Q_C^R=Q_H^R-Q_H^E$

The net and sole effect of this combination is that heat is transferred from the cold reservoir to the hot reservoir. This violates the Clausius statement therefore showing the two statements are equivalent.


For your last question; The Clausius inequality shows the link between entropy and is given by $$\oint\frac{\delta Q}{T} = \begin{cases} 0, & \text{for a reversible cycle} \\ \ge 0, & \text{for an irreversible cycle} \end{cases}$$

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