LP gas burns according to the following exothermic reaction: C3H8(g)+5O2(g)--gt;3CO2(g)+4H2O(g)delta H of reaction=-2004 kjWhat mass of LP gas is necessary to heat 1.3 L of water from room temperature 25 C to boiling 100. C? Assume that during heating, 14% of the heat emitted by the LP gas combustion goes to heat the water. The rest is lost as heat to the surroundings.
mass of water that needs to be put to boiling point Mass Water = density volume = (1000 kg/m^3) (1.3)(0.1)^3 = 1.3 kg = 1300 grs of water It takes 4.184 joules to elevate 1 degree of temperature per grm of water. So the heat you need to increase the temp from 25 to 100 is Heat = 4.184 (1300) = 5439.2 J The heat exchanger appears to have an efficiency of 14%, then the amount of heat required from the propane to heat up the water is Heat from propane = 5439 / 0.14 = 38850 J The heat released by the combustion of LP gas is 2004 kJ/mol of propane. Then, the number of moles that you need is Moles of Propane = 38850 / 2004E3 = 0.019 moles Multiply that by the molecular mass you get Mass of propane = 0.019 * 44.1 = 0.85 grs of LP gas
