thermodynamicswater
Lets sat that I have a 1000 gallon pool sitting pretty at 82.4 degrees F (28 C). How much energy would it lose on a cold day (50 F or 10 C) and only the top was above ground (25 Sq Feet.)
Would this be greatly effected if the day was colder (0 degrees C) or if the sides if the pool were exposed to (100 Sq Feel)?
updated calculations - based on neutrino energy escaping and vapor inhalation risk
Your math is close but not quite right.
First - the number of tritium atoms.
There are 1000/(16+3+3) = 45 moles (as you said)
This means there are 45*2*$N_A$ = $5.5 \cdot 10^{25}$ atoms of Tritium
Now the half life is 12.3 years or 4500 days, that is $3.9\cdot 10^8 $s.
This means the 1/e time $\tau=t_{1/2}/\ln{2} = 5.6\cdot 10^8 \mathrm{s}$
The number of decays per second is $\frac{5.5\cdot 10^{25}}{5.6\cdot 10^8} = 9.8\cdot 10^{16} s^{-1}$
The energy given off in one decay is 19 keV (source: LBL) - but the mean energy imparted to the electron is less: only 5.7 keV (Wikipedia. The remainder goes to the anti-neutrino that is also produced in the decay. Since the interaction cross section between the neutrino and the water is extremely small, that energy can be considered "lost" in terms of heating the water.
This means the total energy deposited in the water (all of it, since the beta decay has a very short range) is about 90 W
Clearly this is a significant source of heating - about 22 calories per second, so it will heat a liter of water by one degree C in about 50 seconds. In an insulated container, liquid that started at room temperature would boil in about a little over an hour. During that time, it would produce about 14 ml of $^3He$ gas as well (0.6 mmol).
This liquid would probably not glow - the blue glow usually associated with radioactivity is due to Cerenkov radiation (roughly the optical equivalent of a sonic boom) and that requires particles to travel faster than the speed of light in the medium.
An electron with 19 keV is not really relativistic (rest mass 511 keV), so the speed is given near enough by
$$v = \sqrt{\frac{2E}{m}} = \sqrt{\frac{2\cdot 19\cdot 10^3 \cdot 1.6 \cdot 10^{-19}}{9.1\cdot 10^{-31}}} \approx 8\cdot 10^7 \mathrm{m/s}$$
This is sufficiently below the Cerenkov limit (which for water with a refractive index of about 1.3 would be about $2\cdot 10^8 \mathrm{m/s}$ - so there will be no blue glow.
But just because it doesn't glow does not mean it's not dangerous.
According to the CDC, a skin dose of 550 Gy would require amputation. Now with half the water near the skin radiating "away" from the skin, we expect the deposited dose per unit mass in the skin to be half of that deposited in the water; this means the dose rate is 45 Gy per second. You would reach the "amputate everything" dose in about 12 seconds. But because the beta has such low energy, it is said to be absorbed in the "dead upper layer of the skin". This might protect you... If it weren't for the fact that if you have tritiated liquid, you have tritiated vapor. Inhalation of that vapor will be really bad for you.
How bad? Assuming that the swimming pool is a comfortable 22 °C, the saturated vapor pressure of water at that temperature is about 20 Torr or 260 Pa. At 60 % relative humidity (not uncommon near a pool) that would be 150 Pa, or about 1/600th of the air. Assuming that all that inhaled tritium is exchanged with the water in the body, and that an average human exchanges 1 m3 of air with the environment per hour (source), this means 1.6 liter of tritium gas per hour - about 70 mmol. This would deposit 0.07 Gy per hour into the body - where there is nothing to protect you. It would take a bit longer to kill you... but kill you it would. Of course swallowing a bit of pool water would really speed up the process, as would any cuts or abrasions. And did you ever notice how crinkly your skin gets when you spend too much time in the water? That's water penetrating the skin. I don't have a good way to estimate that - the internet is full of stories on the subject, but I found no hard data. I suspect that it will be a significant source of tritium entering the body if you swim in your pool.
So even a very short exposure is likely to be most unpleasant. I recommend against using this as a method to heat your swimming pool.
As an aside, there are plenty of instances where Tritium is used as a good "permanent" light source for watch hands, sighting compasses, and other instruments. The short range of the beta particles and the practical half life make it quite good in this role; mixed with an efficient scintillator that turns the energy into visible light, you don't need a lot of tritium to light up a dial at night. You are allowed to use 25 mCi of tritium for this application without needing an NRC license.
Now 25 mCi is 925 MBq - million decays per second. Your liter of tritiated water exceeds this by about 9 billion; and it is in direct contact with the skin (instead of isolated between a few mm of glass which effectively stops the beta radiation).
This is why tritium watches are safe, and your swimming pool is not.
Building off of akhmeteli's excellent answer, I implemented the differential equations from the asteroid webpage in Mathematica. I then tried to tweak the numbers, within realistic bounds, to get the thing into space. In no realistic case was I able to get the thing more than a few hundred meters up before it completely burned away.
To maximize the distance travelled, we want $\Lambda$ and $A$ to be as small as possible; respectively, these correspond to the rate at which heat is transferred to the "asteroid" and the effective cross-sectional area of the object (taking into account turbulence). In addition, we want the heat of ablation $\zeta$ (the amount of heat require to vaporize a certain mass of the substance) to be relatively high, since this will reduce the rate at which mass is lost.
The parameter $\Gamma$ also has an effect; it describes the amount of drag experienced with the atmosphere. Interestingly, one can actually get the projectile higher by increasing the drag: a higher drag means the projectile slows down faster, but that means that the projectile can slow down enough enough that it doesn't burn up immediately.
My optimistic estimates are $\Lambda \approx 0.15$ (note that this number is used in the code example on the page) and $A = 1$ (which would be more streamlined than a sphere). I also used $\zeta = 10\times 10^6$ J/kg, since it was the highest "typical" value in the table.1 Finally, I used $\Gamma = 0.5$, an estimate given on that webpage for the lower atmosphere.
Here's the result of the simulation, with the parameters given above. The vaporization of the plug is complete at a height of 312 meters.
And here is the simulation for akhmeteli's parameters, with $\Gamma = 0.5$. The plug does not significantly change its velocity before it burns up; the final height is a little over 6 meters. As would be expected, this is within an order of magnitude of akhmeteli's back-of-the-envelope estimate.
If you tweak the unknown parameters of my "optimistic" case above, you can attain a height of 1 km if:
All in all, it seems unlikely that the plug got anywhere near space.
Feel free to tweak this code as you see fit. The code stops integration when either the mass of the steel falls below 1 gram, or the speed falls below 1 m/s. The code does implement a height-dependent atmospheric density via a simple exponential model, though it turns out not to be all that relevant for realistic parameters. The acceleration due to gravity is assumed to be constant.
Needs["DifferentialEquations`InterpolatingFunctionAnatomy`"];
Λ = 0.15;(*heat transfer *)
A = 1;(*shape factor *)
Γ = 0.5; (*drag coefficient*)
ρa0 = 1.25 ;(*atmo. density *)
v0 = 66000; (*initial velocity*)
m0 = 900 ;(*initial mass*)
ζ = 7*10^6;(*heat of ablation*)
ρm = 7800;(*steel density*)
h = 7000; (*atmospheric "height"*)
a = Γ A ρa0 / ρm^(2/3);
b = Λ A ρa0/(2 ζ ρm^(2/3));
soln = NDSolve[{x''[t] == - a Exp[-x[t]/h] x'[t]^2/m[t]^(1/3) - 9.8 m[t],
m'[t] == -b Exp[-x[t]/h] x'[t]^3 m[t]^(2/3), x[0] == 0,
x'[0] == v0, m[0] == m0,
WhenEvent[{m[t] < 0.001, x'[t] < 1}, "StopIntegration"]},
{x, m}, {t, 0, 1000}]
{ti, tf} = First[InterpolatingFunctionDomain[x /. First[soln]]]
Plot[x'[t] /. First[soln], {t, ti, tf}, PlotRange -> {0, 66000},
AxesLabel -> {"Time (s)", "Velocity (m/s)"}]
Plot[m[t] /. First[soln], {t, ti, tf}, PlotRange -> {0, 900},
AxesLabel -> {"Time (s)", "Mass (kg)"}]
x[tf] /. First[soln]
1 It is not clear to me whether these are the appropriate units for $\zeta$; the page is unclear. They're dimensionally correct, though.
Best Answer
As I mentioned in my comment, the heat loss will be largely driven by evaporation. To illustrate, here is a chart that relates surface temperature, relative humidity and wind velocity to rate of evaporation. This is for concrete, but the same basic physics applies:
If we assume that the "surface temperature of the concrete" and temperature of the pool can be equated, then we find the rate of evaporation per unit area for the conditions you gave as "somewhere between 1 and 2 $kg/m^2/hr$". With the latent heat of evaporation of water being around 80 cal/g, and the surface area of your pool at 5 square meters (pretty shallow pool... 1000 gallons isn't very much), the lower end of that range would give you a heat loss of
$$80 \cdot 4.2 \cdot 1000 \cdot 5 J/hr$$
which would require a heat source of about 460 W to maintain. Of course on a sunny day you get about $ 1 kW/m^2$ of heat from the sun (normal incidence), much of which will be reflected. These things are all of roughly the same order of magnitude, so your pool will stay warm on a sunny day with not too much wind. When the wind picks up, and especially when the relative humidity is lower, this gets much worse. As you can see from the graph, a wind of 25 mph will cause 10x more evaporation (for the same conditions) compared to a windless day.
A more authoritative (believable) calculation can be found at http://smud.apogee.net/comsuite/content/ces/?id=1023 - these people deal with energy of pools for a living.
Their page shows (among others) the following table for an indoor pool (so no wind) of 25 yards by 8 lanes:
I didn't do the math, but the numbers look comparable.