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Black color substance is assumed to absorb all the light/radiation that falls on it. White color substance on the other hand, reflects all the radiation that falls on it. Since all the momentum of photons is absorbed by black substance, change in momentum is $\Delta p$ whereas for white substance, change in momentum is $2\Delta p$. From this analogy, shouldn't the white be hotter than black?

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As you say the white surface reflects all photons. It redirects the photons, which fly away in a different direction transporting their energy away again from the surface. The temperature of the white material is not impacted. The black surface however absorbs the photon energy, which will tend to increase its temperature.

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The photons after reflecting from the white surface have same energy as before. So, no energy is transferred to white surface by the photons. However, the photons falling on black surface lose all its energy (as no photons are reflected, we can say they all come to a halt). This energy raises the temperature of black surfaces.

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