Assuming there is no heat loss from the water bottle, only the total number of fusion reactions is important; a lower rate of fusion will simply result in slower heating. If one fusion reaction releases 8.4 x 10^-14 calories, it will take about 1.2 x 10^13 reactions to produce 1 calorie, or 1.2 x 10^16 reactions to heat 1 liter of water by 1 degree C.

Assuming that the water started out at 14.5 C (to match the standard definition of the calorie) under standard atmospheric pressure, it would reach the boiling point of 100 C after just over 10^18 fusion reactions. Vaporizing 1 liter of boiling water at standard pressure takes about 539,000 calories, or about 6.5 x 10^18 fusion reactions' worth of heat. The specific heat of water vapor is about 2 kilojoules, or 477 calories or 5.7 x 10^15 fusion reactions, per kilogram per degree (ignoring effects of pressure and dissociation of water molecules), so heating 1 kg of water vapor from 100 degrees to 5000 degrees would take about 2.8 x 10^19 fusion reactions' worth of heat.

Therefore, the final answer is 10^18 + 6.5 x 10^18 + 2.8 x 10^19, or about 3.6 x 10^19 fusion reactions to heat the water in the bottle to 5000 degrees.
Source(s):
http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/phase.html
http://www.engineeringtoolbox.com/water-thermal-properties-d_162.html


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