# How much energy holds the full visible light spectrum? Same to heat absorbed by back color?

I would love to know how much energy a black body absorbs from the visible light.

In the following site we can calculate the eV of a given wavelength, but if I'd like to know the sum of eV for the whole range, hence for color black. How much would it be? .

I understand it'd be the integral of the function of the solar spectrum between those ranges. But looking for the solar curve in visible light range it seems a bit tricky.

The following seems to be the solar function at sea level : https://sunwindsolar.com/wp-content/uploads/2013/09/insolation-curve-adjusted.jpg

Is it possible then to calculate how much energy is in the whole visible light range? Would it be the same to energy absorbed by a 100% black item?

And could that calculation be transformed to watts per second per square meter?

Thanks

• This is more of a physics question than an earth science question. I would suggest reposting the question on SE Physics.
– Fred
Commented Sep 15, 2022 at 18:10
• I think it can be moved by moderators, less work and confusion overall, stays coherent, rather than having to close one, open another Commented Sep 15, 2022 at 19:45
• And considering its focus on the sun's spectrum, and then its isolation curve at Earth's surface, I think it's strayed very much into being an Earth Science Stack Exchange question, though agree the main idea is more of a general physics questions. Commented Sep 15, 2022 at 19:47
• Should I do something then? Commented Sep 15, 2022 at 22:00
• Are you just asking how much power/area there is in the visible solar spectrum, as seen from Earth, in the visible waveband? Sun at zenith? From which location? How precise does the answer need to be? Commented Sep 19, 2022 at 6:15

The accepted value of the solar constant for extraterrestrial solar radiation at the edge of the atmosphere, is 1.94 cal/cm$$^2$$ min, or 1353 W/m$$^2$$. Without making further assumptions, we now consider the question, how much sunlight in the visible light range would be available, in watts/square-meter per second, at the base of the atmosphere given perfect atmospheric conditions? A close estimate can be made if we know W/m$$^2$$ for extraterrestrial solar radiation in the visible light range and the estimated attenuation to reach the base of the atmosphere. The essential information is found by reference to J. A. Duffie and W. M. Beckman in their first edition, Solar Energy Thermal Processes (John Wiley and Sons, New York. 1974), p. 5ff. They provide an estimate of extraterrestrial solar radiation in the visible-light range based on data summarized by M. P. Thekaekara (Supplement to Proc. 20th Ann. Meeting, Inst. For Environ. Sci., 21 (1974), Data on Incident Solar Radiation).
Within the visible-light solar-radiation emission range of 0.38$$\mu$$m to 0.78$$\mu$$m, essentially the visible edge of the near ultraviolet to the visible edge of the near infrared, the extraterrestrial solar-emission fraction is about 47.3%, or 640 W/m$$^2$$. That portion reaching the base of the atmosphere with the sun directly overhead, would be reduced to about 70% of this value, or, thereby, to approximately 448 W/m$$^2$$, more or less. The 30% loss is due to atmospheric attenuation. These numbers are based on an accepted value of the solar constant for extraterrestrial radiation of 1353 W/m$$^2$$ and atmospheric transmission ratio of 0.7 for a given air-mass ratio of 1.0. Perfect sea-level atmospheric conditions for an air-mass ratio of one are approximated by dry, desert air, with the sun directly overhead, and horizontal visibility greater than 200 km.
But what exactly is the impact of the atmosphere? What if the sun is not directly overhead? We can still make very good estimates for clear-air conditions. As indicated above, solar radiation incident at the top of the atmosphere is termed extraterrestrial solar radiation. In the atmosphere, water vapor, carbon dioxide, dust, aerosols and particulates, and optical properties related to density of the atmosphere, attenuate the transmission of solar radiation passing through the atmosphere. And as noted previously, even under perfect conditions, with dry, desert air, the sun directly overhead, and horizontal visibility greater than 200 km, we could only expect about 70 percent of the extraterrestrial solar radiation, more or less, to reach the base of the atmosphere. The estimated net transmission of sunlight through the atmosphere is most easily related to something called the atmospheric air-mass ratio. At sea level, the air-mass ratio is a direct analogue of the secant of the angle to the sun from the zenith. In this case, with perfect conditions and the sun directly overhead, the secant angle is 0 degrees. Consequently, the air-mass ratio would be very close to 1.0. If the sun were 60 degrees from being overhead (secant(60$$^\circ$$) = 2.0), the slant path of sunlight to the base of the atmosphere would be coming through an atmospheric air mass of about 2.0. The fraction of solar radiation at the base of the atmosphere is easily determined by correcting for the air-mass ratio, as the unit air-mass transmission fraction is 70%. Consequently, for the sun 60 degrees from overhead (secant = 2.0), only 0.7$$^2$$, or about 49 percent of the extraterrestrial solar radiation would reach the base of the atmosphere. Note, the secant is used as an exponent to correct the fraction of sunlight available. The intensity of solar radiation at the base of the atmosphere can be calculated for any time of day if we know the slant path angle to the sun at the desired time. The secant correction, however, will not work for much more extreme angles.