      # Carnot's Theorem

thermodynamics

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### #1 Nishan

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Posted 27 June 2019 - 10:13 PM

As I was reading through my physics book I came across a proof of Carnot's Theorem.

Consider two reversible engines A and B  working between same temperatures limits Tand T2. Source being T1 and sink T2 are coupled.

eff1(heat engine) = (Q1-Q2) / Q1 = W / Q1

eff2(refrigerator) = (Q1'-Q2') / Q1' = W / Q1'

( I )

if eff1 > eff2,

Q1' > Q1

Also Q1 - Q2 = Q1' - Q2'

i.e. Q2' > Q2

Here both Q2' - Q2 and Q1' - Q1 are positive quantity. Which means heat flows from colder body to hotter body with any external work. hence it violates 2nd law of thermodynamics .

But in ( II )

if eff2 > eff1 ,

Q1 > Q1'

also  Q1 - Q2 = Q1' - Q2'

i.e. Q2' > Q2

Here both Q2 - Q2' and Q1 - Q1' are positive quantity . Which means heat flows from hotter body to colder body.

How does this violate any law of thermodynamics to prove Carnot's Theorem

1 ) No engine working between two given temperature can be more efficient than a reversible ( Carnot's ) engine working between the same limits of temperature (i.e  between the same source and sink).

2 ) All the reversible engine working between same limits of temperature have the same efficiency whatever be the working substance.

### #2 exchemist

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Posted 28 June 2019 - 01:54 AM

As I was reading through my physics book I came across a proof of Carnot's Theorem.

Consider two reversible engines A and B  working between same temperatures limits Tand T2. Source being T1 and sink T2 are coupled.

eff1(heat engine) = (Q1-Q2) / Q1 = W / Q1

eff2(refrigerator) = (Q1'-Q2') / Q1' = W / Q1'

( I )

if eff1 > eff2,

Q1' > Q1

Also Q1 - Q2 = Q1' - Q2'

i.e. Q2' > Q2

Here both Q2' - Q2 and Q1' - Q1 are positive quantity. Which means heat flows from colder body to hotter body with any external work. hence it violates 2nd law of thermodynamics .

But in ( II )

if eff2 > eff1 ,

Q1 > Q1'

also  Q1 - Q2 = Q1' - Q2'

i.e. Q2' > Q2

Here both Q2 - Q2' and Q1 - Q1' are positive quantity . Which means heat flows from hotter body to colder body.

How does this violate any law of thermodynamics to prove Carnot's Theorem

1 ) No engine working between two given temperature can be more efficient than a reversible ( Carnot's ) engine working between the same limits of temperature (i.e  between the same source and sink).

2 ) All the reversible engine working between same limits of temperature have the same efficiency whatever be the working substance.

Because the engines can always be reversed, and then would you have the same impossible result.

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### #3 Nishan

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Posted 28 June 2019 - 10:27 AM

Because the engines can always be reversed, and then would you have the same impossible result.

Wow I was missing such a simple logic . Thank you so much

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