swansont

It’s not a vacuum, right? So work is being done against atmospheric pressure, i.e. there is a pressure on the plunger. There is a difference between expansion in those two conditions

They do annihilate. Particle/antiparticle pairs can form bound states before doing so. Electrons and positrons form positronium, only annihilating after they’s dropped into the ground state. The article says this particle could be a loosely bound pair of mesons. The J/psi (charm/anticharm), which is what it decays into, has a lifetime less than 10^-20 seconds

I did not claim otherwise. But where would entropy come into play in a multi-particle Feynman diagram?

Your threads get shut down because you don’t present evidence to back up your claims. In January you wrote ”I have a framework for a theory of everything that is worthy of discussion except the moderators keep shutting it down. They say there is no evidence for it but there is, its just that the discussion is usually stopped before it gets there. I do not think our current scientists are stupid but their philopsophy is no where near the level of their mathematics, their philosophy is poor.” Not getting to the point of posting the evidence was a poor strategy (several if your threads went 4 or 5 pages) And by doing it repeatedly you wore out any goodwill and patience that might be afforded you

CPT is often applied to single particles, which are in 1 state. There’s no way for entropy to change, so there’s no application of the 2nd law.

Put another way, the diagrams with Q’s and W are schematics, not maps. They show a concept, not a physical path.

One common mis-step in problems like this is to try and substitute numbers in too early. You haven’t shown your work, as studiot has suggested, so nobody can be sure if that’s the case. The angle is not given because you can solve this without it (i.e. anywhere the angle might show up, it will cancel once you start simplifying the math, and with other methods it is absent altogether) The answer depends only on given quantities

How does the image “show” this? A thermal imaging camera captures thermal radiation, which indicates temperature, not heat. Do you understand what a reservoir is in thermodynamics? This picture is perfectly in accord with the cold plate operating at Tc, exactly as the theory states. IOW, the thing supposed to be at Th is at Th, and the thing supposed to be at Tc is at Tc. Hardly an indictment of the theory. It does not support your misunderstanding of thermodynamics. Heat is not a substance. It does not have a temperature. Whatever you think heat is, it is not what thermodynamics defines as heat. And that stands in the way of any fruitful discussion.

You can ask questions about thermodynamics, or you can explain some idea you have about the subject. But you can’t do both.

That’s weird. I could have sworn you mentioned ice a dozen or so times, and linked to a bunch of videos that mentioned it. That would be helpful, but the ship had sailed, so to speak. how can it? Heat flows from hot to cold. In a Stirling engine the working gas is sealed in the engine. Heat cannot just be exhausted as in an internal combustion engine. If the temperatures are equal there is no heat flow from hot to cold as there is no hot or cold. But there is hot and cold. The one side in contact with water that was brought to boil. The other side is at ambient. You even calculated the efficiency using the temperatures. The temperature gradient means there is heat transfer across the engine. As you said, let’s focus on this example. Nobody is asserting this. You own mental model is flawed. Heat isn’t a substance; it doesn’t have a temperature. When the gas strikes the “cold” plate, it transfers energy to it. That’s the heat transfer taking place at Tc. The plate at ambient temperature seems to remain there because the reservoir is large. In reality the reservoir temperature would be increasing, but this is infinitesimal.Over a long time, with a finite reservoir, the temperature will increase (and Th will decrease) but this is generally ignored in a simple analysis This system is ~20% efficient because ~80% of the energy goes into heating the cold plate and the reservoir it’s coupled to. The rest goes into doing work.

Why? No... If no heat were rejected, your efficiency would be 100% You need to explain why you think this is so. IOW, explain why your ice melts, if no heat is flowing Again, you need to explain why you think this is so Actual thermodynamics doesn’t work this way.

Work is not in your equation, so “not really” isn’t something you show here. And since W = Qin -Qout, my statement is correct. It’s not like this is an independent step. The rejected heat is into a reservoir at Tc. So it’s true the temperature won’t go lower, since you can’t have spontaneous heat flow from cold to hot. But caloric theory is wrong, so understanding thermodynamics this way will lead to problems. Which means it must reject heat.

Some things have to be done at the WH level, because presidential authority is required. Compelling companies to do certain manufacturing, for instance.

It means that heat will be rejected. Qin > W Sure. But I don’t see how this applies to your discussion thus far. You have a reservoir at a higher temperature than ambient. It’s also clear that you haven’t studied physics in any depth; your use of non-SI units and misuse of terminology. You can’t expect understand what’s going on without a better grasp of the basics. studiot has started to explain this; you should take them up on the offer. To clarify, this translational KE is of the individual atoms, not of the sample as a whole, what is being described is conduction, which is not the only heat transfer process (there is also convection and radiation)

There are dangers, but that does not mean the risks are the same.

“Spreading radioactive material” and “nuclear reaction explosion ” are not the same thing. The devices are not the same, so their response to a chemical explosion would not necessarily be the same. Devices can also be designed to withstand such explosions and similar accidents.

It’s the second law of thermodynamics dynamics; no conversion of het to work will be 100% efficient. Not really vague, and not dependent on caloric theory Of course heat is getting through. Insulation reduces heat flow, it does’t stop it. When you reduce losses, more of the energy is available to do work. When you reduce heat flow, it takes longer to deplete the energy. Also, you need to recognize that increasing RPM is not a continuous increase in energy loss. Once the system has sped up, its rotational energy is constant. Heat is energy transfer owing to a temperature difference

Why is the risk the same? Which risk(s)?

The angular size is about the same, which is why we get solar eclipses. The sun is ~400x bigger but also ~400x further away.

No, such an act does not make you a bad person. You are free to not interact with whomever you choose (except you can't ignore staff, IIRC). You do not owe anyone your attention. We're all here voluntarily. However, if this is because they are flouting the rules and not a matter of personal preference/style, we ask that you bring this to the attention of the staff via the report post function.

Absolutely

They were banned, but I don't think it's a criminal offense. It was thought they could record conversations, which would be illegal in secure spaces, but they can't.

https://www.wave3.com/2020/07/28/kentucky-town-hires-social-workers-instead-more-officers-results-are-surprising/ "Instead of hiring an additional officer and taking on the added expenses of equipping that officer, the police chief at the time hired a social worker to respond in tandem with officers. ... Instead of working at another agency and waiting for a referral from a police department after a crisis, Pompilio works side-by-side with officers to respond as calls come in. ... After four years on the job, Pompilio said there has been a significant drop in repeat 911 calls with approximately 15 percent fewer people going to jail." They saved $45,000 - $50,000 year, from reducing the policing burden and because they didn't have to spend money on all the peripherals a police officer needs

-40ºC = -40ºF

You keep adding restrictions to your problem, which is a tad frustrating. I don't think this has ever been a problem with my use of an ultrasound, so that's not something I considered. (Curious as to how much energy you think they will absorb and why this is a problem) Good luck with your search.