CarnotTheorem

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

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Unformatted text preview: 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 Kelvins 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 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....
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CarnotTheorem - Lecture 10: Carnot theorem Feb 7, 2005 1...

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