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Credit for use of the phrase "anti-perpetual motion guesswork" in reference to the Carnot efficiency equation, used in several of his papers:

https://www.researchgate.net/profile/Bo-Miao-4

Don't know what's "cold" about 100°C Anyway, your making up numbers or "rocks" out of thin air imagination proves not anything at all. The fact is the "Carnot engine" and/or cycle, as published in Carnot's book in 1824 "transported" ALL the "Caloric"...

Perhaps you make a valid argument. The same could be said, however, of Kahn's Carnot engine. https://youtu.be/aAfBSJObd6Y If not, why not? He removes rocks to expand the gas then replaced all the rocks to compress the gas. If my "ideal" engine is bogus then the Carnot engine also does zer...

Tom, by "removing rocks" you are reducing the inside pressure. .... ...That is why it will generate zero CYCLIC work. ... Even your description conveys zero work. Rocks are removed, forward stroke. All the same rocks are put back on, return stroke. Equal and opposite "rock" work...

.... You still are, mechanically getting zero cyclic work out. That is also part of the Carnot theorem. Work out equals Th-Tc . You've described nothing more than an adiabatic steam spring. ... Actually, work out is Qh-Qc. That represents joules of heat transfered in less joules of heat transfered ...

...Boiling removes energy from the liquid and adds it to the gas, zero heat. All done with work input. ... The transfer of energy: "Boiling removes energy from the liquid and adds it to the gas " by definition, IS heat . https://m.youtube.com/watch?v=glLPMXq6yc0 You still are, mechanicall...

I've even demonstrated this recently. https://youtu.be/KBtv3skwfDw With water held at the boiling/condensation point just a very slight pressure change causes a transition between condensing and vigorous boiling. Do you really imagine such a rapid pressure change has any significant effect on the te...

Maybe you are forgetting temperature is a measure of the average kinetic energy of a substance?

The following quote from the following website : http://www.iapws.org/faq1/boil.html In the following table, we list the vapor pressure for water as a function of temperature as taken from the latest IAPWS formulation for general and scientific use. Temperature (degrees Celsius) Vapor Pressure (MPa...

... Only during an adiabatic process does Caloric/heat not "fall"/flow,. You have described an adiabatic process. That is why it will generate zero CYCLIC work. ... No I haven't 1st convenient search reference: Isothermic reactions are characterized as reactions that occur in constant tem...

Tom, by "removing rocks" you are reducing the inside pressure. Reduced pressure on water reduces the temperature at which it boils. That is a delta T. Sorry. ... No it isn't. Reducing the boiling point only means the water boils/changes phase to a gas much more readily at 100°C it does no...

The essence of "Caloric theory" is that "the fall of Caloric" (temperature change) is the only metric and that without that there is zero work, zero efficiency. Such a view is based on obsolete, incomplete knowledge. I could, for example, envisage a "perfect" or "i...

The rocks are the load... .... The magic answer is the flywheel ... Sure. The flywheel however is only a very small part of "the load". A "reversible" part. Around 7:00 I think, Kahn mentions that the gas does work putting "potential energy" into the rocks. The potenti...

OK, well, suppose we all watch this video: https://youtu.be/aAfBSJObd6Y What is the taking away and replacing of the "pebbles" supposed to represent? This seems to be essential to the operation of the engine but in the real world people don't remove weights from the piston in their engine ...

Right, so here is your typical Carnot cycle PV diagram: Carnot_cycle_p-V_diagram.jpg Perhaps one of you guys can determine the external load or shaft work output for this PV representation and explain how you arrived at that determination or value for mechanical torque, watts generated or whatever.