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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.

Azad
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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%.

Azad
  • 841