Part of my question on Can a low-energy source/object heat a higher-energy object via radiative heat transfer? asked if a 100W lightbulb could heat a blackbody in a vacuum indefinitely, presuming the blackbody is perfectly insulated against the radiation it emits (which insulation lets the flashlight's energy through).

The up-voted answer was "Yes, the blackbody's temperature should rise indefinitely (or at least until the system melts down into a liquid)."

I'm trying to reconcile this with the fact that, contrary to this, a flame can't heat something hotter than itself (see: can a flame raise the temperature of an object higher than itself?).

What's the distinction? And, if both are true, couldn't we "hack" it with a system like this:

  • Have a flame continuously heat a piece of metal.
  • The flame will reach equilibrium with the metal at some point, e.g. when flame temperature equals metal temperature.
  • At this point (as well as throughout the process) the metal will be emitting thermal electromagnetic radiation, as all hot things do. It will be emitting energy some amount of watts of energy.
  • Direct this radiation towards the same apparatus as in the first question. Now we have an equivalent situation (the 'flashlight' being replaced by 'the EM emissions of a heated piece of metal'). Therefore the target blackbody will heat up indefinitely, i.e. hotter than the flame.

This feels like cheating, and from what I understand, you can't cheat the thermodynamic Gods. My question then is where does this break down?

EDIT: to clarify my question is, if both are true it seems we can rig the flame to a system that emits radiative energy and then heats a black body to hotter than the flame itself. Can it really be so? If yes then why doesn’t this violate any laws, and if no then why not?

  • $\begingroup$ Energy is not temperature, a low energy source can have high temperature and vice versa. A large lake whose temperature is just infinitesimally above $0C$ has an enormous internal energy that is released in the air upon freezing when it falls infinitesimally below $0C$. Whether a body can transfer thermal energy to another via conduction or radiation depends on their relative temperatures not on their energies they contain. Given the temperatures the amount of energies they communicate depends on how much internal energies they have, the ability to communicate depends on their temperatures. $\endgroup$
    – hyportnex
    Commented Mar 6, 2023 at 1:51

4 Answers 4


The First Law tells us that energy can't be lost, and the Second Law tells us that entropy can't be destroyed.

Heating transfers entropy, broadly inversely proportional to the temperature, and since a cooler object directly heating a hotter object, with no other factors, would imply entropy destruction, we know that process can't occur.

This precludes a passive arrangement of lenses/filters/mirrors focusing energy to heat a target to a higher temperature than its source. A coupled law of nature is conservation of etendue; the most we can accomplish with such an apparatus is to make one glowing object appear to be surrounded by another glowing object.

Direct this radiation towards the same apparatus as in the first question. Now we have an equivalent situation (the 'flashlight' being replaced by 'the EM emissions of a heated piece of metal').

But this is not an equivalent situation.

A flashlight connected to a power source doesn't have a well-defined thermodynamic temperature because it's far from equilibrium. The chemical energy in a battery, for example, could be used to heat an arbitrarily small electrical resistor to an arbitrarily high temperature.

In contrast, a hot piece of metal at uniform temperature is easily modeled as an equilibrium system, with a well-defined thermodynamic temperature. This is true even if the sample is periodically returned to a flame, which has a well-defined adiabatic temperature because of the constraint—not applicable to an electrical current—that a combustion reaction must heat its own combustion products.

The First and Second Laws therefore preclude a flame on its own being used to heat a separate object to hotter than the flame temperature.

But an interesting thought experiment emerges in the comments:

What about taking the energy from a large flame, using it to power a laser, and using that laser to heat a small object to hotter than the flame temperature?

Perhaps surprisingly, this is not generally prohibited by the Second Law, with one caveat: There must always be a region colder than the flame to serve as the cold reservoir for the heat engine. Energy rejected to this cold reservoir increases its entropy substantially, more than enough to compensate for the entropy decrease when energy from the flame, transferred as heat, ultimately heats a hotter object (the target of the laser).

Sometimes the Second Law gives us positive news, rather than a constraint, and this is one of these cases.

  • $\begingroup$ See also "Can a flame raise the temperature of an object higher than itself?". $\endgroup$ Commented Mar 5, 2023 at 21:07
  • $\begingroup$ I linked to that same question in my question so I don’t see how it answers it 😅. To follow up with your answer here. This heated piece of metal will be emitting thermal radiation. Say it emits 100W of thermal radiation. If we direct it at the insulated black body - 100W is 100W isn’t it? It doesn’t matter where it comes from. So the black body will heat forever, ie even hotter than the flame temperature. But then aren’t we using the flame to heat something hotter than itself (albeit indirectly)? $\endgroup$
    – Cloudyman
    Commented Mar 5, 2023 at 21:22
  • 1
    $\begingroup$ The contradiction there lies in your definition of "insulation." You already said in the other question that the insulation reflects the blackbody spectrum of a hot material. It can't simultaneously pass the blackbody spectrum of the hot metal you describe. It could, however, pass a certain laser wavelength, for instance—but as noted above, this doesn't violate the Second Law because the thermodynamic temperature of the laser light is far, far higher than the physical temperature of the bulk laser apparatus. $\endgroup$ Commented Mar 5, 2023 at 21:34
  • $\begingroup$ Hmmm I see. I guess it's like saying, "we could use a flame to charge a battery that we then use to power a laser that heats some metal up really hot". We could, and the flame is "heating" something to a higher temperature to itself... but it's because we're storing the energy over time and converting etc. So I think my answer is, the flame can't heat something it's directly heating to a temperature hotter than itself --- this 'law' doesn't apply when using the flame to store energy and use it in other ways, it's a different situation. $\endgroup$
    – Cloudyman
    Commented Mar 5, 2023 at 21:43
  • 1
    $\begingroup$ To use the flame to power the laser, you must have a cold reservoir (some region that remains cooler than the flame) to operate a heat engine. This violates no physical law because the heat engine isn't a passive lens/filter/mirror. In theory, one can use a persistent 1°C difference between two regions to heat another region to 1000°C or higher, even in real time. $\endgroup$ Commented Mar 5, 2023 at 22:15

The point is that a flame has its own temperature limit, because you are continuously feeding it with "cold" carburant and comburant, which react and produce energy. So you cannot heat it more that the amount of produced energy divided by the provided materials thermal capacity.

On the other side, with EM waves (and with and kind of electrical heating), you do not continuously provide new "cold" material. And for this reason, the only source of cooling is the EM radiating of the material you are heating. So, in reality, you won't be heating it indefinitely as you say, but the limit will be much higher.


Regardless of the mechanism of heat transfer (conduction, convection, and/or radiation), heat transfer only naturally occurs from a higher temperature object to a lower temperature object. Once the temperatures of the objects are equal, net heat transfer ceases. I say "net" because each body still radiates heat to the other. But once the temperatures are the same, the rate of heat transfer from the black body to the bulb will equal the rate of heat transfer from the bulb to the black body for a net heat transfer rate of zero.

Hope this helps.


According to Kirchhoff's law of radiation: At a given temperature, the ratio of the emissive power of a body to its absorptive power is constant and is equal to the emissive power of a black body at the same temperature.

The apparatus that you suggest only absorbs, but does not emit. So, its existence contradicts Kirchhoff's law

  • $\begingroup$ You’re referring to the black body in the insulated sphere? Let’s say the insulation reflects 99.9999% of the radiation the black body emits instead of 100% . Now it emits something. The effect will be it will emit 100 W once it reaches equilibrium , but instead it’ll be the equivalent of 100,000,000 W (ie a multiplier of 1/(1-0.999999)) $\endgroup$
    – Cloudyman
    Commented Mar 5, 2023 at 21:26
  • 2
    $\begingroup$ It does not matter what. Both the black body alone or the system back body + insulation. In your case the system black body + insulation only absorbs, but does not emit. So it is not compatible with Kirchhoff's law. Though, I do not see immediately why by using a a glass, which is only transparent one way and reflective the other way, such a system cannot be constructed $\endgroup$
    – Dr.Yoma
    Commented Mar 5, 2023 at 21:34
  • $\begingroup$ Can you elaborate what you mean by "does not emit" ? I didn't follow. Emit on the same wavelength? Glass for example can let visible light through but reflect IR. So we can use visible light to heat something and then that heated object gives off IR that is then reflected back (or not emitted as much).. $\endgroup$
    – Cloudyman
    Commented Mar 5, 2023 at 21:46
  • $\begingroup$ For any wavelength. Your system "black body surrounded by insulation" absorbs light, as light from outside goes in and heats the black body. Right? At the same time, it does not emit, as according to your supposition, you designed insulation in a way that it prevents any light from going out from the system "black body surrounded by insulation". This contradicts Kirchhoff's law $\endgroup$
    – Dr.Yoma
    Commented Mar 6, 2023 at 7:10
  • 1
    $\begingroup$ This should not be allowed, as the body should absorb proportionally to what it emits with the fixed proportionality coefficient -- the same as for the black body. What I still wonder is why your design is impossible: it looks technically feasible. You can take insulation in the form of a half-transparent glass, that is the one that is transparent from one side and reflective from another. We know that you can do that. But then, it seems, you should be able to design an object that contradicts Kirchhoff's law. Strange. I am a bit rusty with this. I think the solution should be simple $\endgroup$
    – Dr.Yoma
    Commented Mar 6, 2023 at 9:44

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