Why do spiders' webs stay wet longer than the environment? I have read this answer:

The reason you get droplets is due to a phenomenon called Rayleigh Instability. This causes a smooth film of water on a fibre to break up into droplets.
The droplets stick to the fibres of the web due to capillary forces.

What makes water-droplet/dew stick to spider's web and what keeps them there?

After irrigation, I noticed that the lawn and bushes dry very fast, and the spider's webs stay wet much longer (both when the web has direct sunshine or is in the shades, it dries much slower than the leaves around). I am specifically asking why this happens, why does the water stuck in the spider's web not dry up like water droplets from the grass and leaves dry up. My only thought was that the leaves and grass are green and this might heat up a little on the sunshine and this heat could make the water dry up faster than on the spider's web.
Question:

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*Why do spider's webs stay wet longer than the environment?

 A: A very nice observation. I think your thought is correct.
I would add to that the fact that when light hits the leaves it bounces back also as IR. Then IR might have a good chance to be trapped in the droplet. On the other hand when light goes into a sphere droplet (like the ones on spider web) it has a good chance to get out soon enough so that it is not absorbed by water.
There is also the differences of the exposed surface areas. You have to compare the droplets on the leaves with ones on the spider web in which has the larger ratio of exposed area to volume. Since water evaporates from the surface, that can be an important factor in how long it takes for all the volume of the water to evaporate. For example, if we approximate the droplets on leaves with a cylinder with radius $r_c$ and and small height $h$ and droplets of the spider web with a sphere with radius $r_s$, then the ratio of surface areas of these to their volumes are compared with each other as,
$$ \frac{\pi r_c^2}{h \pi r_c^2} =\frac{1}{h} \quad \mbox{compared to} \quad \frac{4\pi r_s^2}{\frac{4}{3}\pi r_s^3}=\frac{3}{r_s} $$
Now we should see if it is typical of $h$ to be smaller than $\frac{r_s}{3}$. If it is so then droplets on leaves probably evaporate faster.
