# Falling water - hot or cold?

Hello everyone! Since this is my first post as one of the virtuosi, I
should probably introduce myself a little. I'm a first year graduate
student in physics at Cornell university. I did my undergraduate work at
Oberlin college (I know, you've never heard of it), and I'm currently
just trying to keep my head above water and take in as much physics as I
can. Additionally, I'm trying to find work for the summer, I might post
more on that later. Today, the question that is on my mind is: How much
does a water droplet heat up when it goes over niagara falls?
(image from http://grandcanyon.free.fr/)
Let's begin with a little motivation. Why would a water droplet heat up
when it falls? Well, the physical mechanism is that as it falls through
air, air resistance dissipates energy. This energy is dissipated mostly
as heat, so we expect to put some additional heat into our water droplet
as it falls. More quantitatively, assume the water is going to fall from
some height h to the ground. The gravitational potential energy of our
water droplet is given by $$ PE=mgh $$ where m is the mass of the
droplet and g is acceleration due to gravity. We assume that the
particle starts with no initial velocity. It is very easy to place an
upper bound on how much the droplet will warm. The maximum heating will
happen if all of the potential energy were converted to thermal energy.
The temperature change wound be $$ mc\Delta T = mgh $$ so $$ \Delta T
= \frac{gh}{c} $$ Where c is the specific heat of water. Niagara falls
is 51m tall, g is 9.8m/s*s and c is 4.1kJ/kg*K so this gives a maximum
temperature change of .12C=.22F, fractions of a degree. However, we can
do better than this. A small droplet will experience a linear drag
force, and a bouyant force from the air. The combination of these will
result in a terminal
velocity of $$
v_t=\frac{2}{9}\frac{(\rho_p-\rho_f)}{\mu} g R^2 $$ Where
$\rho_p \text{ and } \rho_f$ are the droplet and air density
respectively, $\mu$ is the dynamic viscosity and R is the radius. This
gives the kinetic energy the particle gains and we assume the rest of
the potential energy goes to thermal energy. So, $$mc\Delta T =
mgh-\tfrac{1}{2}mv_t^2$$ $$\Delta T = \frac{gh-v_t^2}{c} $$ Now
we can plug in some numbers. A little bit of searching will turn up
appropriate values for the density of water (\~1000kg/m^3), the density
of air (\~1.2kg/m^3) and the dynamic viscosity of air (\~.8mPa*s). We
estimate the radius of a small droplet of water is 1.5mm (\~1/16"),
which seems about a medium sized raindrop. This gives a terminal
velocity of 6m/s. Using the above equation this gives a temperature
change of $$ \Delta T = .0024C/m*h-.009C $$ It looks like our
calculation only makes sense after the after droplet has fallen more
than 4.5m (this is probably about the distance it takes to reach
terminal velocity)! Niagara falls is 51m tall, so this gives a change in
temperature of .11C=.2F. The water only heats up by a fraction of a
degree. This more realistic estimate is still rather close to our upper
bound. I leave it as a question to you, the reader, to estimate the
temperature change of a typical raindrop.

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