这实际上是暗示在典型的Trenberth地球能源预算图:[![在这里输入图像描述][1]][1]86.4美元\ mathrm{\压裂{W} {\, m ^ 2}} $是能量运动从蒸发到大气中。340.3美元\ mathrm{\压裂{W} {\, m ^ 2}} $是整个温室效应。很重的大多数自然温室效应(水蒸气)[2]。水循环表明,平均而言,大气中的水蒸气停留两周([文章][3]现在显示8 - 9天)。所以从这些细节,这意味着我们可以考虑水蒸气的温室回到其相变能源移动大约四倍。所以四分之一的实习时间是一个粗略的估计,大约需要2 - 4天平均水分子将能量发送回地球,因为它长大从表面到大气的吸收,最终凝结/沉积。我可以想象你的朋友可能不会感到舒服只有使用这个数字来考虑估计,也许可以尝试另一种方式的隧穿到这些数字是从哪里来的....稳定大气中水汽含量,最后唯一的主要方式,从表面大气水分传输能量是由其合成沉淀(其他水转换往往没有净变化对水分,不要将净能量从地上……水蒸发然后再浓缩在地面没有净能量变化,同样水蒸气变成了云,然后在这一水平没有reevaporates净能量变化过程)。 The average daily precipitation on Earth is [around 2 mm per area][4]. The energy for evaporation is [2256 kJ/kg water][5]. So for a square meter, that's $$0.2 \frac{\,\mathrm{cm}}{\,\mathrm{day}} \cdot 1\,\mathrm{m^2} \cdot \left(\frac{1\,\mathrm{g}}{\,\mathrm{cm^3}}\right) \cdot 2256 \frac{\mathrm{J}}{\mathrm{g}} \cdot \left(\frac{100\,\mathrm{cm}}{1\,\mathrm{m}}\right)^2 = 45120 \frac{\mathrm{J}}{\mathrm{m^2}}\;\mathrm{per}\;\mathrm{day}$$ (the third term is the density of water of $\frac{1\,\mathrm{g}}{\,\mathrm{cm^3}}$ and the fifth term matches units) Which, converting to per second to get in Watts (multiply by $\mathrm{\frac{1\,day}{86400\,sec}}$), that's about $.5\mathrm{\frac{W}{\,m^2}}$, which comes out to be on roughly the same scale of value as the figure's latent heat value. How long does it take to achieve that in greenhouse reradiation by that particular water vapor? In a sense it's not easy to calculate the energy directly by each molecule... we know the radiative properties of water, what [wavelengths water reflects/absorbs/radiates more at][6], but it's a cumulative distribution, each molecule affected by its location and interaction with the others that determines the overall amount. We can look how the greenhouse effect [roughly doubles the energy we receive in a day, meaning half of all surface energy is from previous surface radiation being returned back][7]... which works out to be that $340.3\mathrm{\frac{W}{\,m^2}}$ of greenhouse return). But either way, it all leads back to residence time meaning it's that same ratio that was given from the figure, 86.4/340.3, and so the same time. It may really help to think of it not so much as how long it stays and how much that water molecule contributes while it is in the atmosphere, but instead about the whole water cycle process being what *maintains* the amount of water vapor in the atmosphere to allow for that amount of continued greenhouse return, and so it's just one continuous amount versus the other. But if you really do want to think of it as the time for water vapor to generate its greenhouse effect matching its latent heat transfer, it looks like it must be **on the order of about 3 days**, or around a quarter of its time in the atmosphere, to result in the energy amounts we see. [1]: https://i.stack.imgur.com/XSsWy.jpg [2]: https://skepticalscience.com/water-vapor-greenhouse-gas.htm [3]: https://hess.copernicus.org/articles/21/779/2017/hess-21-779-2017.html#:~:text=Based%20on%20state-of-the-art%20data%2C%20we,atmosphere%20of%208%E2%80%9310%20days [4]: https://www.smithsonianmag.com/arts-culture/what-is-the-daily-rainfall-on-earth-and-more-questions-from-our-readers-37945925/ [5]: https://www.engineeringtoolbox.com/fluids-evaporation-latent-heat-d_147.html [6]: https://en.wikipedia.org/wiki/Water_vapor_windows [7]: https://atmos.washington.edu/academics/classes/2001Q4/211/notes_greenhouse.html
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