Temperature of the surface of the sun is about 5750K. Can you heat an object to more than 6000K using magnifying glass and sunlight? According to second law heat cannot be transferred from colder to hotter object and therefore it should not be possible to heat anything to temperature higher than 5750K using only sunlight and magnifying glass.

Also, I know that its possible to achieve much higher temperatures using electricity from solar cells, but this is not the same case since solar cell is not 100% efficient (and works only when its temperature is lower than 5750K) while magnifying glass can be 100% efficient.


marked as duplicate by Emilio Pisanty, ACuriousMind, Kyle Kanos, Ryan Unger, John Rennie May 17 '15 at 11:11

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    $\begingroup$ Your second law argument is spot on. It is also true that one can achieve any desired temperature if one has a third temperature bath that is lower than the temperature of the sun. That is exactly how solar cells can extract net energy that can be used in any way we wish, including to power lasers, particle accelerators etc. which are all machines that can achieve much higher temperatures. The max. efficiency of those solar cells is basically given by the Carnot efficiency of the cell running between the temperature of the sun and 300K and is approx. 95%. $\endgroup$ – CuriousOne May 16 '15 at 20:20
  • $\begingroup$ Here's an existing answer to this question. $\endgroup$ – Brionius May 16 '15 at 20:25
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    $\begingroup$ Landau/Lifshitz V (Statistical Mechanics) discusses the case of the lens in detail at the end of §63 Thermal Radiation (at least in the geometrical optics limit). (I hope the section number is correct, got this from the most recent German edition of the Landau/Lifshitz series). $\endgroup$ – Sebastian Riese May 16 '15 at 20:41

You can pump heat from cold objects to hot objects if you pay some more energy (that's what your refrigerator is doing) and that doesn't violate second law of thermodynamics.

You should note as you heat object, its thermal radiation will increase. Intensity (that is power per unit surface area) of thermal radiation is proportional to $T^4$ so when the temperature of your object is higher than the sun its intensity will be higher but if your lens could focus rays coming from an area of the sun larger than area of your object you could still provide more power. But how big your object could be?

Suppose we have a lens with a radius of 10 cm that could focus all the lights incident on it on our object. The flux that our lens receives is much less that flux at the surface of sun ($R_{\odot}$ is radius of the sun, $d$ is the average sun-earth distance):

$$L_{lens}=L_{\odot}({R_{\odot}\over d})^2=2\times10^{-5}L_{\odot}\\ P_{lens}=A_{lens}L_{lens}=\pi r^2 L_{lens} $$

When your object is at the same temperature as the sun its intensity is the same as that of the sun (assuming they have the same emissivity):

$$L_{object}=L_{\odot}\\ P_{object}=A_{object}L_{\odot} $$

So in order to have the same input and output power:

$$A_{object}=2\times10^{-5}\pi r^2 = 0.66 \text{ mm}^2$$

That's less than one square millimetre and if you want to go to higher temperatures you should have even smaller surface area. What's happening here is that you're radiating a lot of the input energy back into surrounding to increase your temperature by just a small amount.

Of course you could do this by solar cells as well, but you need larger solar cells as their efficiency is typically about 20%.

  • $\begingroup$ Still, your explanation would allow to produce temperatures higher than the source. The issue is that you based your reasoning only on geometric optics, while at this point the wavy nature of light will play a key role (as I understand it). $\endgroup$ – Fabrice NEYRET Jan 6 '17 at 18:21

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