Weather Banter - View Single Post - Could we use endothermic(heat absorbing) reactions to reduce hurricane strength?

[email protected] · October 20th 05, 06:29 AM posted to sci.chem,sci.geo.meteorology,sci.physics

wrote:
...
The heat flux is given by Q = (k/Delta-x)*A*DeltaT = (0.6/0.01)*1*5 =
300 J/s. (you can check it yourself, the units work out propoerly - i
was lazy and didnt want to type them again).

Alright so we have heat flowing at 300 J/s. As the temperature
difference drops (i.e. the temp of the surface rises), Q also drops
linearly with the drop in DeltaT. Let us use 150 J/s as an average
over a large part of this temperature rise at the surface.

The amount of heat it takes to raise the temp of that water by 5C is
given by (specific heat)*(mass)*(deltaT)

energy needed = 4.19 J/(g-C)*(.01m*1m^2*1000000g/m^3)*(5C) = 209500 J

Now at 150 J/s this says our water will be roughly heated in 209500/150
s =~ 1400 s =~ 23.5 min.
.
.
.
I'd say plan on cooling
down to a rather deeper depth than 1 cm - like maybe a couple hundred
feet and pray that they are no currents down there to sweep your cool
water a couple hundred miles away.

You see that the heat flux has the thickness in the denominator while
the amount of heat is linearly dependent on mass which is density *
area * thickness. So the amount of heat has thickness in the numerator.
Then when you calculate the amount of time by dividing the amount of
heat by the heat flux you see this is dependent on the square of the
thickness. So a layer only 10 cm, 4 inches, thick would last 10^2 = 100
times as long as a 1 cm thick layer or 23.5*100 = 2350 minutes, about
40 hours.
Another surprising thing is that the temperature difference DeltaT
cancels out when you calculate the length of time. So it wouldn't get
better by making the layer cooler.

- Bob