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It isn't harder to say. The surface area of a cylinder goes up much slower than the volume as the cylinder gets bigger. The larger volume's energy would dissipate slower.

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The other comments basically reveal that every physical problem we try to understand requires approaching the solution using a mental model with built in assumptions and limitations. So, perhaps when we have a question or face a challenge when something doesn't make sense, it is good to look deep down through our pyramid of assumptions upon which we place our understanding. Unless we solve the problem, then all assumptions are suspect. For example, while others point out the probable main issue here (the metal's emissivity), there are other assumptions. There is the assumption that the metal, the other parts and air should all be at the same temperature. In fact, the problem is a bit complicated when you get into the slight differences. The objects each lose energy at a different rate by emitting (mostly) IR photons at different rates (paramaterized as "emissivity"). Additionally, the objects conduct heat at different rates. Since they all lose heat via IR to the cooler environment outside the chamber (which also irradiates the same parts, but with fewer IR photons), the parts will never purely reach the temperature of the air in the chamber. What equilibrium temperature each one reaches, down to the hundredths of a degree perhaps, depends on how quickly the parts can conductively and radiatively absorb heat compared to how quickly they lose heat via the same processes to the outer environment (via) IR, the inside air (via conduction and IR), and the parts they touch (via conduction) and the parts within line-of-sight (via IR). So, the problem is actually quite complex and only simplified if we don't care about small temperature differences. Plus, if we did care about such small differences, the measuring device would have to be highly accurate.

Heat=energy.

How much energy is required to raise the temp of silver is less than steel because one is more conductive than the other. Properties of the material being heated play a role but heat is lost over time. Assuming you are not putting any more energy (heat) into the system.

The only exception to this of course is evaporative cooling. Moving air can cool a wet surface even if the temperature of the air is above the surface temperature.

That originally what I though this question was asking but it turns out it’s just an instrument problem.

That shouldn't be true unless there's something being evaporated from the surface, in which case flowing air might speed evaporative cooling. By itself, there's no way for warm air to cool an object to below its own temperature.

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Even easier, you can buy emissivity labels (also called IR labels) and just slap a sticker on.

The ring will have its own urge to turn into moons, and if has lots of mass, that moony limit will apply.

I've seen laser thermometers that can measure emissivity, are they able to accurately compensate or do you need to know the material?

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>I was talking to a coworker and he said that the hot air in the chamber was cooling the part because it was flowing.

This can be true only for a part that's perspiring.

Not the right kind of engineer but I'll give 'er a swing. (Fiber Optic and Network engineer)

Let's say all external input variables are the same. As in everything is the same except one is occurring at a 152,000 x scale (2 cups of water vs standard pool at 19000 gallons or 340,000 cups)

My semi-educated guess: We assume conservation of momentum and the same angular velocity.

The surface-area-to-volume ratio decreases with increasing volume. Therefore the pool-sized mug will experience less force of friction relative to the increase in mass/volume.

An increase in volume increases the radius and mass. Therefore the linear velocity at the edge is much higher as mentioned in another comment causing an increase in centrifugal force causing an increase in friction for the pool.

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1 cup = 14.4375 cubic inches2 cups = 28.875 cubic inches Pool sized coffee mug340,000 cups = 4,908,750 cubic inchesYeti Coffee mug on my desk external dimensions for ratio = 3.5"W x 6.6"H (r=1.75")Right cylinder volume = V=πr2hGot busy with work, and might come back later to finish my math.

Figured out how to figure it out as long as the correct answer doesn't involve calculus cause I ain't got time for that. Do the math for two spinning cylinders with friction at the edge.

True, but because of the larger velocity, the friction from the edges of the pool would be much higher as well, right? Hard to say which effect would dominate

I have added an edit to this answer linking to your answer in that second one. I do not really follow science news so was unaware of this fiasco! I have since seen tweets of distant colleagues trying to set the record straight as well as one world leading (astro) researcher suggesting that Superman might have been involved...

Glass reflects different wavelengths of light differently. House windows may also have an infrared-reflective composition or coating (for more efficient cooling in warm sunny weather).

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Yeah, as the question that spawned this thread was more general, it seemed worth while to pick up and answer a question more specifically focused on the fundamental misunderstandings this recently published paper seems to be propagating.

Temperature is one of the most difficult things to measure. For example it is not just to put a thermometer inside an oven an do readings if you want a correct value. But if you are satisfied with an approximate value, this is ok.

Hot air cannot cool something that's at a lower temperature. Convective heat transfer can only occur when the object and the air flowing around it are at different temperatures. And heat will always flow from the hot thing to the cold thing. It will never flow the other direction. Also you should use a thermocouple for this, it will be more accurate. Just gotta make sure you dont short out the pcb with the metal thermocouple tho.

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Yeah, several "what's gonna happen in unfathomable timescale" options to choose from.

Yes. The rate of slowing of the spinning water must be a function of the kinetic energy of the water against the drag of the side walls.

Even if the max velocity of the water is the same in the pool (ie a very slow rotation rate) then the energy of the water will scale somewhat with the volume of the water, ie cubicly, while the drag of the walls will scale somewhat with the area of the side walls, ie. squared, therefore the larger the cup, the more the kinetic energy of the water outweighs the drag of the sides.

If you do actually stir at the same rotational rate, then as the other commenter says, the max velocity of the water will be very significantly higher than in the coffee cup, and as energy scales to the square of the velocity then you'll have vastly more energy proportional to drag than in the coffee cup.

Plus anecdotally, as a kid we used to run around those circular backyard swimming pools and get the water spinning faster and faster, and they kept going for ages!