Heating an object with black body radiation to above the temperature of the source

Question

A recent question and its answers say that you could not use sunlight to heat something to above the temperature of the Sun. There was some discussion of the need for the bodies to be in thermodynamic equilibrium. That is not what I want to address in this question.

Thought experiment 1

A is a large black body at temperature $T_A$ in empty space. B is a small black body some distance away at the lower temperature $T_B$.

I expect that B will gradually be warmed but not to above the temperature of A.

Thought experiment 2

I place a filter between A and B. This attenuates the longer wavelengths so that the transmitted radiation has the same spectrum as a hotter object. Correct me if I am wrong but I expect that it would be possible to do this such that an observer could not tell whether or not it was a true hotter source.

If this trick works then it would be perverse that the filter would absorb (or reflect) energy yet allow its output to heat something to a higher temperature (but more slowly).

Additional detail

Note that this is a thought experiment. I am not suggesting that this is practical or potentially useful. It is just an exercise to help me understand. I am not expecting to obtain free energy.

I understand that filter will block lots of energy and hence object B will receive less energy, as if A was further away. We could partially or fully compensate for this using lenses and mirrors. Also, suppose that filter is not attempting to greatly increase the colour temperature e.g maybe 5500K to 5600K.

I am exploring the claim even if we concentrate the light from a black body, we cannot use it to achieve a temperature above the source.

Does color temperature limit how much a laser of a given wavelength can heat a target?

Answer 1

While a lot of the other answers touch on the various aspects of the problem, the real reason this can't be done is more subtle.

First, it's not true that you can't use sunlight to heat something to above the temperature of the sun! Trivial solution: Use a solar panel to convert the sunlight to electricity, then use an arc furnace to heat it up to whatever temperature you want, up to the limits of the arc furnace. You can easily exceed 6000 K.

Another, less trivial answer that avoids conversion to electricity and uses only optical elements: Use a solar-pumped laser. The coherent light output from the laser can be focused to a very intense point, achieving temperatures much higher than the surface of the sun.

It's also not true that you can't mimic the spectrum of a hotter body using a (more or less) passive filter. To mimic the spectrum of a hotter body, you need to dampen out the longer wavelengths, and add new shorter wavelengths. The former could be done with a filter, and the latter could be done using e.g. some kind of frequency doubling crystal. The actual engineering details would probably be complicated, but there's no fundamental reason why this wouldn't be possible in principle. Obviously, the filter would absorb much of the energy of the incoming light, but the spectrum of a hotter body could be emulated.

Another answer pointed out that you can't focus sunlight to an arbitrarily small point because of etendue. This is true in any optical system that uses perfect mirrors and lenses, but there's no reason to limit ourselves to such systems.

So going back to the question: Is it possible, via some clever arrangement of absorbers and emitters, to raise the temperature of some body to higher than the temperature of the sun using sunlight energy input alone? Of course it is. Nothing in physics prevents this.

However, the actual underlying question here is: Can we use blackbody radiation in a closed system to heat some other object in the system to higher than the temperature of the source, in a way that violates the 2nd law of thermodynamics? And the answer is no. In all of the examples I listed, the 2nd law is not violated. Solar panels have fundamental efficiency limits. Lasers have efficiency limits for much the same reasons. Any kind of clever arrangement you devise will ultimately result in no loss of entropy in a closed system.

Why is this the case? The trivial, but maybe non-satisfying answer is "because the 2nd law is true." The actual detailed answer would depend on the system in question. But for most systems you can reduce the answer to some manifestation of Liouville's theorem, for example as explained in this answer.

Answer 2

Thought experiment 1

A is a large black body at temperature $T_A$ in empty space. B is a small black body some distance away at the lower temperature $T_B$.

I expect that B will gradually be warmed but not to above the temperature of A.

If the system of the two bodies is isolated, e.g., if they are placed in a container with reflecting walls, then eventually the two bodies and the radiation in the container would have the same temperature. The body A would definitely cool down (as per the second law of thermodynamics), whereas whether body B heats up or cools down depends on how many energy will leak into radiation (which is neglected inn the formulation of the problem, but which is the essential third component of it.)

If the two bodies are in an open vacuum, they will both continue cooling down, since the emitted radiation carries away some energy and never returns.

Thought experiment 2

I place a filter between A and B. This attenuates the longer wavelengths so that the transmitted radiation has the same spectrum as a hotter object. Correct me if I am wrong but I expect that it would be possible to do this such that an observer could not tell whether or not it was a true hotter source.

If this trick works then it would be perverse that the filter would absorb (or reflect) energy yet allow its output to heat something to a higher temperature (but more slowly).

This is just a radiative reformulation of what is known as Maxwell demon (closely related to Brownian ratchet) - a clever device that is supposed to beat the laws of thermodynamics, but separating particles (in this case photons) by their energy. The catch is the same - the experiment requires an ideal filter (the "demon") to work. A real filter will be heated itself and start emitting and/or leaking some of the high energy radiation towards the cold body, until eventually the whole system comes to equilibrium, at a temperature that is lower than the temperature of the object that was initially the hottest.

Remark

There was some discussion of the need for the bodies to be in thermodynamic equilibrium.

When we characterize the body by a temperature, we assume that, to some approximation, it is in thermodynamic equilibrium - since temperature is defined in terms of such an equilibrium, and does not make any sense otherwise.

Answer 3

Filtering longer wavelengths will not make the spectrum equivalent to one of a hotter object. You'll simply be cutting out the region where the colder object peaks and leaving the tails that have a far lower spectral radiance.

If we assume object B is also a black body such that it absorbs all radiation equally, then an observer near object B could in fact assume the opposite: that object A is colder than what it should be. This could happen if, for example, there was no access to spectral data and only the total energy flux, $M$. If you integrate Planck's distribution for all wavelengths, you would get the Stefan-Boltzmann law $(M=\sigma T^4)$. What you're proposing is to effectively only integrate over a small region such that the total energy flux ends up being lower. If the observer near object B does not know of any filtering, they would simply assume that object A has a temperature $T_A<T_{A \ \mathrm{(real)}}$ since $M_{\mathrm{measured}}<M_{\mathrm{real}}$.

Answer 4

I expect that B will gradually be warmed but not to above the temperature of A.

Probably not even close to it. While B is receiving some radiation from A, B will be radiating to empty space in all other directions.

The temperature of B where the total energy received from A equals the total energy transmitted away from B is expected to be less than the temperature of A.

I place a filter between A and B. This attenuates the longer wavelengths so that the transmitted radiation has the same spectrum as a hotter object.

A passive filter cannot create the spectrum of a hotter object. A hotter object would produce

• some output at shorter wavelengths
• more total power per area

The result of the filter will be that B receives less energy from A than before and it can reach thermal equilibrium at a lower temperature than before.

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A specific power output regardless of its size? Does a 1kg object at 10000K have more power than a 1000kg object at 5000K?

It's not per mass, it's per surface area. A larger object at the same temperature will radiate more power than a smaller object.

Your filtered object will transmit (much) less power than an object with a higher temperature, since the power goes up as the fourth power of temperature.
$$ P = A \epsilon \sigma T^4$$

What my speculation is based on is that the blackbody spectrum never quite goes to zero for shorter wavelengths.

It may not mathematically go to zero, but the power that comes from that spectrum is useless for raising the temperature of another object.

An object at 500K technically has non-zero power in visible light, but it is not expected that even one visible photon would be released from such an object during the current lifetime of the universe.

Answer 5

The other answers came close but does not seem to have hammered this into you:

This attenuates the longer wavelengths so that the transmitted radiation has the same spectrum as a hotter object.

This cannot become real because it is not just the relative abundance of shorter v.s. longer wavelengths that matter. You have to add the short wavelengths that basically are not there in the original because of the exponential cut-off.

If you attenuate so much of the longer wavelengths that you tease out a spectrum that looks like a hotter body, then what the standard theorems mean, would be that such a filter is physically impossible, unless it starts from something that can create even higher temperatures, e.g. it uses a battery. And so in any case, your result is impossible to achieve.

Answer 6

To add to the existing answers, we can quantitatively show that $T_B$ will never exceed $T_A$ in your "Thought experiment 2" using Planck's law. The key idea is that the filter must be symmetric; though it does decrease the heat lost by $B$ to $A$, it also decreases the heat gained by $B$ from $A$.

For simplicity, suppose $B$ is a sphere of radius $R_B$ and $A$ is a thin concentric spherical shell of radius $R_A \gg R_B$. I will suppose that your filter lets in all radiation of wavelength $\lambda < \lambda'$ unaltered, but reduces the intensity of radiation of wavelength $\lambda \geq \lambda'$ by a factor of $r<1$. The heat flow from $B$ to $A$ is $$4\pi R_B^2 \int_0^{2\pi}\text{d}\theta \int_0^{\pi/2} \sin\theta\cos\theta\;\text{d}\theta\left(\int_0^{\lambda'}B(\lambda, T_B)\;\text{d}\lambda + r\int_{\lambda'}^{\infty}B(\lambda, T_B)\;\text{d}\lambda \right) = 4\pi^2 R_B^2\left(\int_0^{\lambda'}B(\lambda, T_B)\;\text{d}\lambda + r\int_{\lambda'}^{\infty}B(\lambda, T_B)\;\text{d}\lambda \right),$$ where $B$ is the Planck radiance $$B(\lambda, T) = \frac{2hc^2}{\lambda^5} \frac{1}{e^{hc/\lambda k_BT}-1}.$$ The heat flow from $A$ to $B$ is $$4\pi R_A^2 \int_0^{2\pi} \text{d}\theta\int_0^{\arcsin(R_B/R_A)}\sin\varphi\cos\varphi\;\text{d}\phi\left(\int_0^{\lambda'}B(\lambda, T_A)\;\text{d}\lambda + r\int_{\lambda'}^{\infty}B(\lambda, T_A)\;\text{d}\lambda \right) = 4\pi^2 R_B^2\left(\int_0^{\lambda'}B(\lambda, T_A)\;\text{d}\lambda + r\int_{\lambda'}^{\infty}B(\lambda, T_A)\;\text{d}\lambda \right),$$ so conveniently, we only need to analyze the value of $$I:=\left(\int_0^{\lambda'} + r\int_{\lambda'}^\infty \right) \frac{2hc^2}{\lambda^5} \frac{\text{d}\lambda}{e^{hc/\lambda k_BT}-1}.$$ But note that by dimensional analysis $I\propto T^4,$ so for some function $f(r)$ we have \begin{align*} \text{Heat flow from }A\text{ to }B &= (R_B^2 f(r))T_A^4\\ \text{Heat flow from }B\text{ to }A &= (R_B^2 f(r))T_B^4. \end{align*} That is to say, if $T_B<T_A,$ heat will flow from $A$ to $B$ and equilibrium will be reached when $T_A=T_B$, regardless of whether or not there is a filter (the rate at which equilibrium is reached may depend on the filter however).

Of course, all of this result is independent of the shape of the blackbodies or the type of filter - I just chose the easiest case to analyze. The miracle of the second law of thermodynamics is that it allows us to circumvent all calculations. No matter the combination of mirrors, lenses or filters, radiation from $A$ can never spontaneously heat $B$ to a temperature $T_B>T_A$ without any external work.