# Arrhenius' calculation of the temperature of Venus

In this answer, we learn that in the early 1900s, Nobel Prize winner Svante Arrhenius believed that Venus was covered in lush swamps. His book on this matter, "the destinies of the stars", is archived here. A few quotes from this book

The average temperature there [on Venus] is calculated to be about 47℃ assuming the sun constant to two calories per cubic centimeter per minute.

[about albedos] the figure 49.6 (Russell calculates the figure 45) for the entire Earth naturally falls closer - almost 3.6 times - to 59, the figure of Venus, than to 15.4, the figure for Mars

So. Arrhenius believed the albedo of Venus to be 0.59, and the solar constant to be two calories per square centimeter per minute, close to today's accepted value. (I'm assuming the "per cubic centimeter" in the original is an error). How did he arrive at a surface temperature on Venus of 47℃, from this data? (I understand that Venus is today known to be much hotter than that)

My initial thought was, let $$S$$ be the solar constant, $$R_E$$ and $$R_V$$ the radii of the Earth's and Venus' orbit respectively, and $$r_V$$ the radius of the planet Venus, and the albedo $$a=0.59$$, then Arrhenius may have believed the total solar power absorbed by Venus to be $$P_{in}=(1-a)S\left(\frac{R_E}{R_V}\right)^2 \pi r_V^2$$ The Stefan–Boltzmann law was known at the time, so he may have calculated the blackbody radiation emitted by Venus at temperature $$T$$ $$P_{out}=4\pi r_V^2 \sigma T^4$$ Then $$P_{in}=P_{out}$$ $$(1-a)S\left(\frac{R_E}{R_V}\right)^2 =4 \sigma T^4$$ $$\sqrt[4]{\frac{(1-a)S\left(\frac{R_E}{R_V}\right)^2}{4 \sigma}} = T$$ but that's not even remotely right, it gives 263K, or -10℃.