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水蒸气比潜热产生更多温室效应的时间到了吗?

我和一个气候变化否认者有一场争论。

他说，大气中水蒸气的增加通过从地面吸收蒸发热量而产生的冷却效应比额外的温室效应产生的热量“大得多”。

我说额外的温室效应与时间相结合，所以在一定的时间T(我怀疑“很小”)之后，这些热量将补偿从地面吸收的潜热。

有人能给出T的数量级吗?

注意，这个问题不是关于“正常的”水循环，即潜热在蒸发时被吸收，在凝结时被释放，而是关于对流层中水蒸气的(无可争议的)全球增加。

回答

这实际上在典型的地球Trenberth能量收支图中有所暗示:[!$86.4\ maththrm {\frac{W}{\，m^2}}$是由汽化向上移动到大气中的能量。$340.3\ mathm {\frac{W}{\，m^2}}$是总体温室效应。自然温室效应的绝大部分是水蒸气。水循环表明，平均而言，水蒸气在大气中停留两周([这篇文章][3]现在建议8-9天)。因此，从这些细节来看，这意味着我们可以认为水蒸气的温室回报大约是其相变能量的四倍。因此，四分之一的停留时间可以粗略地估算出，平均一个水分子需要2-4天的时间才能把它从地表带到大气中的能量送回地球，而这些能量是水分子通过吸收和最终的凝结/沉积从地表带到大气中的能量。我可以想象，你的朋友可能不太愿意只使用这个数字来考虑估算，所以也许我们可以尝试另一种方法，深入了解这些数字的来源....对于大气中稳定的水蒸气水平，水汽最终将能量从地表转移到大气的唯一主要方式是通过其产生的降水(其他水的转变往往对水汽没有净变化，也不会从地面转移净能量……蒸发然后在地面再凝结的水没有净能量变化，同样，水蒸气变成云，然后在那个高度再蒸发，在这个过程中没有净能量变化)。 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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