{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

CarnotTheorem - Lecture 10 Carnot theorem Feb 7 2005 1...

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Lecture 10: Carnot theorem Feb 7, 2005 1 Equivalence of Kelvin and Clausius formulations Last time we learned that the Second Law can be formulated in two ways. The Kelvin formulation: No process is possible whose sole result is the complete conversion of heat into work. It can be also expressed in a slightly different form: It is impossible to transform heat systematically into mechanical work in a cycle involving only one bath The Clausius formulation: No process is possible whose sole effect is transfer of heat from a colder to a hotter body. It was mentioned that these two formulations are equivalent, which can be formally proved as follows. Suppose the Second Law in the Kelvin’s formulation does not hold, i.e. one can produce some work W using only one bath of some temperature T 1 . This work can be transformed into heat, for example by friction. So using the work which the engine produces one can increase the temperature of any other system of arbitrary temperature T 2 . Since T 2 may be higher then T 1 it turns that one transfers heat from colder system to the hot one, which is violation of the Second Law in the Clausius formulation. For instance, one can use the work W to run another engine which operates the inverse Carnot cycle (see the figure below). This combined machine produces no work and its net effect is transferring heat from cold bath to the hot one. The combined machine absorbs positive heat Q c from the cold bath, and delivers the heat Q d = Q h - Q x to the hot bath. Clearly, Q d > 0 because Q x = W = Q h - Q c . (Note: in this lecture we use symbols Q , W , etc. for absolute values of corresponding quantities.) 1
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Q x W Q h Q c T c T h Q c d Q =Q - Q h x T c T h Let us made the opposite assumption, i.e. that, in violation of the Clausius formulation, one manages to transfer a certain amount of heat Q from cold bath to the hot one. Suppose it is exactly the heat another heat engine absorbs from the hot bath during isothermal expansion. Then the hot bath receives and gives up the same amount of heat, and remains virtually unchanged. As a result, the net effect is producing work at the expanse of heat extracted from a single (cold) bath, which is a violation of the Second Law in the Kelvin formulation. The above argument prove that the Clausius and Kelvin statements are equivalent.
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern