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My understanding of the 2nd law of thermodynamics is, partly that, basically a 'cold' object cannot heat a 'hot' object.

Case in point: if I take a 1500W hot plate and put it on full, the plate will get to a certain temperature, say 130°C.

If I now place a glass dome on top of the plate, the plate will now heat the air inside the dome, and the glass dome as well. But the temperature of the air inside will never exceed 130°C, and the hot plate itself won't get hotter than 130°C. That's because the heat is flowing from the 'hot' surface (130°C) to the 'colder' objects (air/glass dome) and not the other way around.

But if I model this with blackbodies... an emission 1500W/$m^2$ for a 1 $m^2$ surface area is 130°C. With no other objects around it, this object is in equilibrum at 130°C. But if I now place a second blackbody near it, the first blackbody will start to heat that second blackbody.

Further once that second blackbody heats up up, it, too, will have a temperature and thus be emitting a certain W/$m^2$ back towards the first blackbody. A new equilibrium will be reached where the first blackbody's temperature increases, as a result of contact with the second blackbody's temperature. The second blackbody won't reach higher than 130°C of course... but it will increase the first blackbody's temperature.

This contradicts common sense/experience, as the hot plate won't get hotter than 130°C... plus it seems to violate the 2nd law of thermodynamics. If reality worked this way it seems you could have a sufficiently reflective/resistent dome surface and get the hot plate to much hotter than 130°C (e.g. reflect most of its energy back into it, now it has almost double the input, 1500W from its source plus almost 1500W reflected back, and it could get to 200°C... plus now it's hotter so it's emitting more energy and thus even more is reflected back, etc.)

Where does it break down, where's the disconnect between the theoretical blackbody and lived experience? And doesn't the fact that the hotter blackbody get heated up to higher than its equilibrium temperature mean it violates the 2nd law?

Cloudyman
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2 Answers2

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"My understanding of the 2nd law of thermodynamics is, partly that, basically a 'cold' object cannot heat a 'hot' object."

That's only true in some situations. This is a problem of over-simplified explanations in physics education. A situation is described - spontaneous heat flow between otherwise-isolated bodies at different temperatures - and the second law says that in this situation the net heat flow is from the hot to the cold, so the hot body cools down and the cold body warms up. It's never in the opposite direction, net heat flow from cold to hot, cooling the cold body and warming the hot one. But it only applies to spontaneous heat flow, isolated bodies, and is only talking about the net flow.

The obvious counterexample is a refrigerator, that pumps heat from the cold bodies inside to the warm outside. That particular phrasing also proves hugely confusing in the study of black body radiation, because heat is transferred in both directions, both hot to cold, and cold to hot. The net flow is still hot to cold, but the loose wording misleads a lot of students into thinking the flow of black body radiation from cold to hot cannot happen.

"Case in point: if I take a 1500W hot plate and put it on full, the plate will get to a certain temperature, say 130°C."

There are two ways this can happen. You can hold the external power input fixed (1500 W) and work out the equilibrium temperature as it escapes. Or you can use a thermostat to turn the power on and off to hold the body at 130°C. What happens depends on which you are doing.

If you are using a thermostat, then the hot body stays at 130°C regardless. Put an insulating dome over the top, and instead of running full time at 1500 W the heater starts turning on and off, putting less power in. The air and the dome heat up. The outside of the dome radiates to the void, but can never reach 130°C because it's surface area is bigger than 1 m$^2$ and it's got a maximum of 1500 W input even with the heater running full blast. Since the heater isn't running all the time, the outer temperature is even lower.

This is the situation of a human covering themselves with a blanket. The body has a thermostat, and can supply about 100 W when at rest to keep itself warm. If the surroundings are cold enough that 100 W is insufficient, skin temperature starts to drop. A blanket resists and slows the heat flow. The body warms the inner side of the blanket, which radiates back and reduces the outflow. The body needs less heat to maintain its temperature. The outer side of the blanket is colder than the naked body, and radiates far less heat to the cold surroundings. Blankets mean people can maintain their body temperature with less power expenditure.

The other situation is where the heater generates 1500 W, no matter the temperature. In that case, the dome radiates more heat back, and the temperature of the heater rises far above 130°C. The temperature keeps rising until the outside of the dome is radiating 1500 W to the void. The outside of the dome will be at a bit less than 130°C, because it has a larger area. But the heater in the middle will be a lot hotter than 130°C, so that the net heat flow from heater to dome will remain at 1500 W.

A nice analogy for this situation is to have a hose pipe pouring water at a fixed rate into a leaky bucket. The water level rises until the same amount is pouring out of the holes as the hose pipe is supplying. If you put one leaky bucket inside another, you can get the water level higher, as the flow is determined by the difference in water height between the inside and outside of each successive bucket. If the bucket never leaks or spills, there is no limit to how high you can get the water. But you can never get water to spontaneously flow up hill, from a low water level to a higher one. You have to pump it.

Similarly, you can get even higher temperatures if you feed the power into a volume with a much smaller surface area. Consider an electric arc welder, where we feed 1500 W into a volume a few millimetres across. The temperature needed to blackbody radiate 1500 W through such a small area is considerably higher than 130°C! The power input on its own does not set a limit on temperature - it depends on how easy it is for the heat to escape.

Another good example is the sun. The heat generated by fusion in the sun's core is roughly the same per unit volume as a human body. But the sun is big! The total heat generated increases as the cube of radius, but the surface area only increases as the square. So big things tend to have more difficulty losing heat. The centre of the sun is so much hotter than its surface (10 million degrees compared to 6000 degrees) because the centre is thermally insulated by the mass surrounding it. If you cut out a human-sized chunk of the sun's core, somehow kept it compressed at the same pressure, but able to radiate freely to space, it's temperature would soon be the same as that of a human in the same circumstances. It's the incredibly thick 'dome' of outer material radiating heat back that keeps it so hot.

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In order for your 130C body "all by itself" to be "in equilibrium" at 130C, it needs to be in a 130C environment, like a box whose walls are 130C. If it's fully alone in a 0K universe (i.e. there is no constant-temperature photon gas like the 2.7K photon gas in our universe), its blackbody radiation will cause it to lose heat over time until it too is 0K. It is emitting radiation - photons carry energy away. This is a loss of heat.

Your reflective surface is "insulation" it does the same thing as a sweater - or really exactly the same thing as the reflective foil called MLI used in physics experiments to decrease the heat exchange due to blackbody radiation between a cold experiment (say liqiud helium cooled at 4K) and the 300K (25C) lab. It doesn't heat the body, it just reduces how much of the radiation flies away - it decreases the rate at which the temperature goes down.

And finally, when you have a second blackbody next to the first, the 130C blackbody is losing heat, and some of it is absorbed by the other blackbody. The other blackbody is also emitting radiation which is absorbed by the 130C blackbody. But the 130C blackbody loses more heat to the colder blackbody than the colder blackbody sends back (because the colder blackbody sends back generally lower frequency photons and less photons). Again, the second blackbody's heat just decreases the rate at which the 130C blackbody cools down - but considerably less than the reflective surface.

AXensen
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  • Ah I am presuming the 130C blackbody is not just 'by itself' but has some heat or energy source that keeps it at 130C (analogous to the electric current feeding the hot plate and keeping the hot plate hot) – Cloudyman Mar 23 '23 at 12:39
  • ahh sorry. Then yes. It will get hotter from the reflective surface. If the hot plate alone heats it to 130C, thats an equilibrium between heat in from the hot plate and the heat out from radiation. If you decrease the heat out from radiation, the new equilibrium will be hotter. Like how your oven is hotter when the door is closed. – AXensen Mar 23 '23 at 12:49
  • Ah hmm... but don't we then have a colder object heating a hotter object? i.e. the hot plate itself is 130C, and now we add some other object near it (the glass/reflective dome), and now the hot plate gets hotter than 130C... logically it seems like it would work but I feel like something is amiss, that we could then use mirrors/reflections to get a heat source object to any temperature we want, which for example violates conservation of étendue (?). I find this surprisingly unintuitive (I flip between both answers 'seeming right' or 'seeming wrong'). – Cloudyman Mar 23 '23 at 12:58
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    No it's true that if you perfectly insulate something but provide heat to it its temperature will be infinitely high. It won't violate any thermodynamic laws because you're assuming you have a source of infinite power (like the wall outlet). You arent just moving heat from a cold system to a hot system (which would reduce the entropy of the universe). Somebody in a powerplant is making tons of entropy so you can have the energy to do the work needed. Not sure what you meant by conservation of entendue - isn't that some kind of optics thing? – AXensen Mar 23 '23 at 13:03
  • eg https://what-if.xkcd.com/145/ , you can’t take moonlight and arrange a system of lenses to reach a temperature hot enough to burn paper. But it sounds like by focusing moonlight on a spot and insulating it you would be able to … the moonlight has some temperature and keeps adding energy to a system so therefore it would get infinitely hot? – Cloudyman Mar 23 '23 at 13:12
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    Gotcha. Yeah im not sure what the end result is but there are difficulties doing this with light. How are you gonna keep the blackbody radiation from escaping your body but have an opening for light from the moon? It seems that if you have a surface where light can come in, blackbody light will also escape that hole. – AXensen Mar 23 '23 at 13:19
  • Hmm I’m not sure the reflective dome would heat the hot plate hotter. Consider: A stovetop that is off will be at room temperature (~20C) which is about 400 W/m^2. So by this logic if I just leave it off and put the dome on it then it’ll heat it up , without having to even turn the power on … – Cloudyman Mar 24 '23 at 00:23
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    You are still neglecting the role of the environment in preventing objects from reaching 0K by emitting blackbody radiation. The insulation slows down the rate at which a body hotter than the environment comes to equilibrium with the environment. If the object is in equilibrium with the environment already, it receives just as much heat from other objects in the room as it emits. Then the insulation just makes the cold plate see more of its own radiation, and less of the radiation from other surrounding objects - which doesn't change the fact that its radiating just as much as it receives. – AXensen Mar 24 '23 at 11:09