## Monday, April 11, 2011

### How much rain would a raindrop drop...

Today in church Jonathan commented on how the arrival of a wave of rain was followed by the air becoming noticeably cooler. I idly replied that it shouldn't, because the gravitational energy released by the raindrops' falling should be manifested as an increase in heat, then immediately became intrigued as to whether or not that would be the case (these things always seem to come at the most inconvenient times!).

Just for fun, I did some simple calculations, which I thought I'd share with you all. I assumed a very simple model, with a raindrop with mass = 1 gram, falling a total height of 1 kilometer (which is not a bad assumption for the height of the clouds about Hilo, I think). Anyway, given those assumptions, a raindrop falling to Earth releases \begin{align}\Delta E&=mg\Delta h\\
&\approx0.001\,\text{kg}\cdot\left(-9.8\frac{\text{m}}{\text{s}^2}\right)(-1000\,\text{m})\\
&\approx10\,\text{J}\end{align}about 10 joules of energy (a joule is about as much energy as it would take to lift a small apple one meter straight up, or the amount of energy released if that same apple fell a meter downwards). If you think that seems like a lot of energy for a raindrop, so did I. If the raindrop were to retain all this energy as kinetic energy, it would impact the ground with a speed of about 100 meters per second, or almost 225 miles per hour. In practice, most of this energy is lost to friction with the air as the raindrop falls, which would indeed heat up the atmosphere as I thought.

However, there's another factor to take into consideration: absorption of heat from the atmosphere by the raindrop. It's relatively simple to calculate the amount of energy it would take to heat up a raindrop with a 1-gram mass by a degree Celsius: \begin{align} Q&=mc\Delta T\\
&\approx1\,\text{g}\cdot4.181\frac{\text{J}}{\text{g}\cdot\text{K}}\Delta T\\
&\approx4\frac{\text{J}}{\text{K}}\Delta T \end{align} Thus for every degree Celsius (or kelvin, both of which equal about 1.8 degrees Fahrenheit) the raindrop heats up it absorbs about 4 joules of energy. So if the raindrop starts off about 5 degrees Fahrenheit cooler than the ambient air temperature at ground level, it will absorb as much energy as it emitted and leave the atmosphere no different temperature-wise than at the beginning. Of course, if the raindrop is more than 5 degrees Fahrenheit cooler (as is quite possible), it will absorb more energy than it releases and cause a net cooling effect. And it wouldn't have to absorb much excess heat, either, considering the sheer number of raindrops in a typical shower.

Well, there's your geeky musing for today. It makes a nice opportunity to show off the beautiful output of $$\LaTeX$$, too. A hui hou!