These two very simple to use formulas will probably allow easy order of magnitude scoping.
Time to heat with given heating power
T = V x 1000 cc/l x 4.17 x K / W seconds
or T = 4170 V.K/W
Power required to heat in time T
Power = W = = V x 1000 cc/l x 4.17 x K / T Watts
or W = 4170 V.K/T
T = seconds
V = litres
K = degrees K (or C) rise
W = Watts heating power
4.17 = Specific heat of water over range 20 C to 30 C adjusted for mean water density so units are J/cc/K
This is increased by any losses
eg electric kettle uninsulated is 85% - 95% efficient
Pumping energy adds partially to heating.
The key factor is the amount of heat stored in water per cc per degree K (or degree C) rise in temperature. Once this is known all else can be calculated.
A more useful than many table of water thermal properties may be found here on the well-worth-bookmarking "Engineering Toolbox" site. Rather than giving just a few figures at a few selected temperatures this gives properties at a wide range of temperatures, so it can be established if temperature has a significant effect on the relevant properties in the context of the current problem.
The specific heat of water is an average of about 4.18 Joule per gram* per degree K across the 20 C to 30C temperature range, or 4.17 Joule per cc per degree K. The 0.01 J/unit/K is not going to make much difference to your result.
- The per cc and per gram figures differ slightly because the density of water is very slightly less than 1 g/cc across this range. The specific heat will be about 4.18 Joule per gram and 4.17 J/cc per degree K. The usual value given is 4.182 J/g/K, often without reference to temperature range.
Other useful (and necessary) facts:
One Joule is provided by 1 Watt of heating per second or
1 J = 1W/s
1 litre = 1000 cc
1 metre^3 = 1000 litre
And 1 litre = 10 x 10 x 10 cm or 100 x 100 x 100 mm
Total_energy = SH x cc x delta-C = SH x litres x 1000 x delta-C
Power = energy /second = Total_energy / seconds.
Pressure matters little here except as it may affect water density.
Look up compressibility of water to see how much this matters.
Doubling water head affects volume (of course) but has an insignificant effect on pressure.
Energy in Volume of V litres =
V x 1000 cc/l x SG g/cc x SH x delta_T
V = 14,786 (user figure)
delta_T = 10 K (user figure)
SG = 1 (actual figiure rolled into SH - see text)
SH = 4.17 J/cc/K (SG is included in this - see text)
Energy = 14,786 x 1000 x 1 x 4.17 x 10 = 616,576,000 J or W.s
So you can heat it in 1 second using a 617,000! kW heater.
This may be hard.
or in one hour with 616,576/3600 seconds = 171,000 Watts of heat (+ losses)
Or using a 3kW element as used in domestic hot water heaters.
(Max size available is usually 3 kW - larger for special uses).
616576000/3000 =~ 57 hours!
To heat it in and '8 hour day' = 616.6 MJ/(8 x 3600) = 21.4 kW.
Heat capacity of tank may be significant.
Insulation of tank will be significant.
Other losses may occur.
But that gives you an idea.