on is the heat absorbed, Qn. Let me therefore define the efficiency of an e, as
ID: 1781024 • Letter: O
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on is the heat absorbed, Qn. Let me therefore define the efficiency of an e, as the benefit/cost ratio: benefit W cost Qh stion I would like to ask is this: For given values of Th, and Tes what is the m possible efficiency? To answer this question, all we need are the first and laws of thermodynamics, plus the assumption that the engine operates in eturning to its original state at the end of each cycle of operation. first law of thermodynamics tells us that energy is conserved. Since the the engine must be unchanged at the end of a cycle, the energy it absorbs precisely equal to the energy it expels. In our notation, (4.2) e this equation to eliminate W in equation 4.1, we have for the efficiency Qc (4.3) e efficiency canno roceed further we must also invoke the second law which tells us that the tropy of the engine plus its surroundings can increase but not decrease. e state of the engine must be unchanged at the end of a cycle, the entropy must be at least as much as the entropy it absorbs. (In this context, as on 3.2, I like to imagine entropy as a fluid that can be created but never d.) Now the entropy extracted from the hot reservoir is just Qh/Th, while opy expelled to the cold reservoir is Qe/Te. So the second law tells us er tha can equ or (4.4) Te Th Qh Th g this result into equation 4.3, we conclude the ma (4.5) Th desired result. So, for instance, if T-500 K and T, = 300 K, the max- ourExplanation / Answer
From the equation (4.5) we can conclude that the maximum efficiency of engine that can be obtained is less than or equal to 1-Tc/Th . Hence, a 100% efficient engine is not possible theoritically as well which is very clear with the expression as Th>Tc .
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