![]() Both are delivered at relatively high temperature with respect to the cold reservoir outside the engine, and the two air-masses (the relatively hot and the relatively cold) will mix. The reconciliation is to look at the heat dissipation and exhaust gas release. ![]() This is OK, but what makes the people perplex is the meaning of entropy increase that is in general viewed as a lost chance to use some heat energy in doing work. Carnot combustion cycle has zero entropy resultant as entropy increase in the isothermal expansion (leg 1) is equal to the entropy decrease in the isotherm compression (leg 3) and there is no change of entropy in the adiabatic expansion (leg 2) nor in the adiabatic compression (leg 4) since there is no Q term in any adiabatic process at all. The unique increase of entropy takes place out of the internal ignition engine, not inside the engine. The point is to turn as much of this heat into macroscopic work. ![]() So yes, high T hot would produce huge entropy but also would give much energy at the same time. ![]() You must produce heat and the entropy but you must try to turn as much of it into work done by a piston. It is almost as if you would lower the entropy by means of seting the heat to zero. So i believe that if you try to lower the entropy produced because of heating the gas, you wont get a thing. Saying that temperature of the gas is the same as the temperature of the hot reservoir is the statement of perfect flow of heat to the gas from the hot reservoir. Idea in the text quoted is to try to turn as much heat into work and in that way minimize entropy.Another thing is, if you always keep gas temperature that is in contact with the hot reservoir very high and the gas which is in contact with the cold reservoir very low, you get the most out of your machine. It seems strange to try to lower the Q hot in order to do this because this simply decreases the heat flow. The idea is to maximize the heat transfered into mentioned macroscopic kinetic energy in order to increase machines effectivness. It is a simple consequnce of a molecular chaos which heat actually is. The problem of any heat engine is that not all of the heat that is absorbed can be transformed in a macroscopic kinetic energy of a piston. Hot reservoir is giving away its energy to the gas in the form of heat. In some sense, this part really doesn't have much to do with entropy at all, because from the thermodynamic perspective, entropy production (which is the increase in entropy of an isolated system) is a measure of how much work we could have done if we had done the process reversibly, but we have already designed the perfect engine operating between those two particular temperatures above, so entropy doesn't have anything else to say. So the entropy change for the hot reservoir is $\frac$, being the difference between the heat flows, must go up if, say, we lower $T_C$ (because then $Q_C$ goes down) or if we raise $T_H$ (because then $Q_H$ goes up). This is the heat that is given to the gas in my engine. The heat that leaves my hot reservoir is $Q_h$. Okay, so now let's try to minimize the entropy that is created. This seems to suggest that to maximize the efficiency of the engine, one should minimize the entropy produced during the process. The work produced by the engine is the difference between the heat absorbed and the waste heat expelled. To get rid of the entropy, every heat engine must dump some waste heat into its environment. The reason is that the heat, as it flows in, brings along entropy, which must somehow be disposed of before the cycle can start over. Unfortunately, only part of the energy absorbed as heat can be converted to work by a heat engine. I'm reading Schroeder's An Introduction to Thermal Physics.
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