I recently got a curious question about the sun rays. Is there a material that can absorbs most of them and turn it into pure heat?

For example, we all know that dark material (black t-shirts) get much hot than white material, Although, I've heard that there is some metals that can get much hotter from direct sunlight.

I would like to know, which one can get much hotter from the sun?


The sun emitts a whole spectrum of wavelengths, so what we'd like in this case would be a material that has a low reflectivity for a large bandwidth. Black t-shirts absorp more heat because they simply absorp the more wavelengths, including most of the visual ones, and that's why they appear black. As we know, white light is a combination of the wavelengths in the visual spectrum so therefore we can conclude that a white t-shirt is reflecting alot of wavelengths (in the visual spectrum). The infrared spectrum (typically heat) is close to the visual spectrum.

The ideal situation for this would be the model of a blackbody. A blackbody is a body that absorbs incoming radiation at a factor $\alpha$ that's between 0 (no absorption) and 1 (full absorption) and then emitts radiation as well with a factor $\epsilon$ between 0 and 1. For you to have a material that retains alot of heat from incoming radiation you'd need a material with a high absorption factor. You can look up example values for a number of materials in the following link https://www.engineeringtoolbox.com/solar-radiation-absorbed-materials-d_1568.html

I hope this answered your question.

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  • $\begingroup$ A blackbody absorbs all radiation incident upon it. $\endgroup$ – Rob Jeffries Sep 18 '18 at 22:59
  • $\begingroup$ " and that's why they appear white" u meant black $\endgroup$ – Manu de Hanoi Sep 19 '18 at 2:27
  • $\begingroup$ Of course i meant black, thank you @ManudeHanoi . Rob, a perfect blackbody absorps all incoming radiation, however, real materials and coatings can be modeled with the absorption and emission coefficient. $\endgroup$ – DakkVader Sep 19 '18 at 8:28
  • $\begingroup$ @RobJeffries I could not tag you in the above comment, hence this one as well $\endgroup$ – DakkVader Sep 19 '18 at 8:29
  • $\begingroup$ Adding the word "perfect" to "blackbody" is unnecessary. A blackbody is an unattainable ideal. A "grey body" is one with a wavelength-independent emissivity that is less than 1 (also an unattainable ideal). $\endgroup$ – Rob Jeffries Sep 19 '18 at 12:24

The quick answer is ... any material with a high emissivity. According to Wikipedia, Ventablack holds the record of $\varepsilon = 0.99955$ in the visible range. That material would have a higher temperature than any other material, if put under the Sun and after a little while so steady-state is reached. Note that this could be not quite exact, because even though most of the Sun's energy is in the visible spectrum, some of it is in the UV and some of it in the IR range. But roughly, what I said holds.

The longer answer is that the heat equation of the material is $\nabla \cdot (\kappa \nabla T) + A\varepsilon(T^4-T_\text{Sun}^4) + \text{convection term} = C_p \frac{\partial T}{\partial t}$. So while it is true that, at the beginning of the moment when you place the objects under the sunlight there might be a transient period in which the temperature varies with time, after a while the temperature will reach a final temperature that is almost exclusively independent of $C_p$, unlike what is claimed in Árpád Szendrei's (wrong) answer. Indeed, at one moment $\frac{\partial T}{\partial t}\approx 0K/s$ so the $C_p$ terms drops out of the equation and thus there is no way that $T$ will depend on that coefficient.

Almost lastly, I would like to mention that $\kappa$ enters into play if the material is put into contact with another one, in which case the final temperature distribution will be affected by the heat sink (the material making contact) in such a way that $\nabla T\neq 0K/m$. So in order to get the final answer, you would have to specify exactly the setup of the experiment. In deep space, I expect the solution to depend only weakly on $\kappa$, because the whole material will roughly have a homogeneous temperature distribution after a certain time, and in that case the answer mostly depends on $\varepsilon$.

Also, I do not agree with DakkVader's sentence

The ideal situation for this would be the model of a blackbody.

A nice counter example of such a claim is a black hole (even though it cannot be said that it is a material per se, it still qualifies as a black body), which is either a perfect or very near perfect black body. The more light it absorbs, the colder it becomes (it has a negative specific heat). But do not misunderstand me, his answer is a very good one, I am just being overly picky here.

Árpád Szendrei's answer has another problem. It focuses on the real part of the refractive index, which plays no role whatsoever in the absorption of the EM waves (sunlight) into the material. In reality, the refractive index of metals is complex, it contains both a real and a complex part. And that's precisely the complex part that matters when it comes to absorption in a material. I.e. how the amplitude of the EM wave decays with distance from the surface, inside the material. So, unlike DakkVader's good answer, Árpád Szendrei's one contains gross inaccuracies making it simply wrong.

To sum up:

If the problem is not well specified, the answer depends on the geometry of the material and how it is connected with its surrounding. But as a rule of thumb, only $\kappa$ and $\varepsilon$ will matter (not $C_p$). You'd want to find a material with a low $\kappa$ and high $\varepsilon$. The exact value will depend on the exact setup in which the material finds itself.

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First, you are asking which metal heats up the fastest:

I understand you are asking about direct sunlight. Now we can check how direct heat energy will raise the temperature of the metals, and then correct this measurement with the refractive index of the metals.

Let's take 100g samples of aluminum, copper, silver.

100J of energy will raise the temperature of:

  1. Aluminum by 1.1C,

  2. silver 4.2C,

  3. copper 2.5C.

The refractive index of:

  1. Aluminum is 1.09

  2. silver is 0.15

  3. copper is 0.46

Now based on this information, to answer your question, silver will heat up the fastest because of direct sunlight.

Sunlight is white, not yellow, this is a common misconception. White light is made up of a combination of all wavelengths of photons, the whole spectrum. There is no physically white wavelength photon. Our eyes perceive colors by having receptors for RGB, red, green and blue wavelength photons. When we see white light, that is the combination of all these wavelength photons.

Now the reason that a white object appears white is not just because it reflects all light. It is because it reflects most of all visible wavelength photons, and because the sunlight is white, naturally a white object will appear white because white light, which is made up of a combination all visible wavelength photons, will all be reflected from the surface of this white object. Please see here why white objects are all white:


This will explain you too why black objects appear black. Black means our eyes do not receive any visible wavelength photons. A black object will absorb almost all visible light, and does not reflect nor emit any visible wavelength light.

You are correct that black objects will heat up faster (if same material), but that is because they:

  1. absorb all visible light and do not reflect any visible light

  2. inelastically scatter most of non-visible light, and that creates vibrational energy in the molecules (heat up)

  3. do not elastically scatter almost any light (this would be reflection)

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  • $\begingroup$ You considered only the real part of the refractive index. But it is the complex part that matters to determine how well EM waves are absorbed into the material. But eventually most of the EM wave will either be absorbed or reflected (very very little will just pass through the metal), so it is better to focus on the emissivity than on the refractive index, in order to answer the question. The heat equation has a term proportional the emissivity multiplied by ($T^4-T_\text{Sun}^4$). Also I do not think focusing on the specific heats is a good idea, because we can assume a steady-state due to t $\endgroup$ – thermomagnetic condensed boson Sep 20 '18 at 14:15
  • $\begingroup$ the nature of the problem (object below sunlight). What matters is the final temperature of the object, and this temperature is almost entirely independent of $c_p$. $\endgroup$ – thermomagnetic condensed boson Sep 20 '18 at 14:16
  • $\begingroup$ @coniferous_smellerULPBG-W8ZgjR the question of the OP was which metal heats up faster. I answered that question. My answer is at the level of QM. He mentions too white and black objects, I explained why they appear white and black. I explained how black objects heat up faster. I explained the real cause of heating up on the QM level. $\endgroup$ – Árpád Szendrei Sep 20 '18 at 17:12
  • $\begingroup$ @coniferous_smellerULPBG-W8ZgjR you say "But eventually most of the EM wave will either be absorbed or reflected" This is not correct. In case of metals, the ratio of reflection is the most. What is correct to say is that almost all visible light will be reflected from a metal. And that almost all non-visible wavelength light will be inelastically scattered. This inelastic scattering is what transforms non-visible wavelength photons' energy into the vibrational energy of the molecules of the metal, and this ia what heats up the metal mostly. $\endgroup$ – Árpád Szendrei Sep 20 '18 at 17:16
  • $\begingroup$ @coniferous_smellerULPBG-W8ZgjR you are confusing absorption with inelastic scattering. You seem to think that metals heat up mostly because of absorption. This is not correct. Most of the heat that is transferred to the metal from EM waves is by inelastic scattering. $\endgroup$ – Árpád Szendrei Sep 20 '18 at 17:18

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