4 Correct grammo.
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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 $𝑄_𝐢^E= 0$ so $𝑄_𝐻^𝐸 = π‘Š$, where $𝑄_𝐢^E$ 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 $π‘Š^𝐸 = π‘Š^R$, 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 $$\oint\frac{\delta Q}{T} = \begin{cases} 0, & \text{for a reversible cycle} \\ \ge 0, & \text{for an irreversible cycle} \end{cases}$$

One way in which to show the two statements are equivalent is to 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 $𝑄_𝐢^E= 0$ so $𝑄_𝐻^𝐸 = π‘Š$, where $𝑄_𝐢^E$ 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 $π‘Š^𝐸 = π‘Š^R$, 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 $$\oint\frac{\delta Q}{T} = \begin{cases} 0, & \text{for a reversible cycle} \\ \ge 0, & \text{for an irreversible cycle} \end{cases}$$

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 $𝑄_𝐢^E= 0$ so $𝑄_𝐻^𝐸 = π‘Š$, where $𝑄_𝐢^E$ 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 $π‘Š^𝐸 = π‘Š^R$, 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 $$\oint\frac{\delta Q}{T} = \begin{cases} 0, & \text{for a reversible cycle} \\ \ge 0, & \text{for an irreversible cycle} \end{cases}$$

3 Added more details and improved format
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One way in which to show the two statements are equivalent is to 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 $𝑄_𝐢^E= 0$ so $𝑄_𝐻^𝐸 = π‘Š$, where $𝑄_𝐢^E$ 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 $π‘Š^𝐸 = π‘Š^R$, 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 $$\oint\frac{\delta Q}{T} = \begin{cases} 0, & \text{for a reversible cycle} \\ \ge 0, & \text{for an irreversible cycle} \end{cases}$$

One way in which to show the two statements are equivalent is to 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 $𝑄_𝐢^E= 0$ so $𝑄_𝐻^𝐸 = π‘Š$, where $𝑄_𝐢^E$ 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 $π‘Š^𝐸 = π‘Š^R$, 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.

One way in which to show the two statements are equivalent is to 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 $𝑄_𝐢^E= 0$ so $𝑄_𝐻^𝐸 = π‘Š$, where $𝑄_𝐢^E$ 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 $π‘Š^𝐸 = π‘Š^R$, 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 $$\oint\frac{\delta Q}{T} = \begin{cases} 0, & \text{for a reversible cycle} \\ \ge 0, & \text{for an irreversible cycle} \end{cases}$$

2 Minor format changes
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One way in which to show the two statements are equivalent is to 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 perfectperfect heat engine: forFor each joule of heat extracted from a hot reservoir, exactly one joule of work would be done. For such a heat engine $𝑄_𝐢^E= 0$ so $𝑄_𝐻^𝐸 = π‘Š$, where $𝑄_𝐢^E$ 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 $π‘Š^𝐸 = π‘Š^R$, 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.

One way in which to show the two statements are equivalent is to 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 $𝑄_𝐢^E= 0$ so $𝑄_𝐻^𝐸 = π‘Š$, where $𝑄_𝐢^E$ 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 $π‘Š^𝐸 = π‘Š^R$, 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.

One way in which to show the two statements are equivalent is to 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 $𝑄_𝐢^E= 0$ so $𝑄_𝐻^𝐸 = π‘Š$, where $𝑄_𝐢^E$ 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 $π‘Š^𝐸 = π‘Š^R$, 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.

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