In: Chemistry
If all the energy obtained from burning 0.840 pounds of propane is used to heat water, how many kilograms of water can be heated from 22.0 °C to 46.8 °C?
Balanced chemical reaction of burning of propane and energy evolved are as given below
C3H8 + 5O2 3CO2 + 4H2O H0rxn = -2218.6 KJ
molar mass of propane = 44.1 gm/mole
0.840 pound of propane = 381.0176 gm
mole of propane in 381.0176 gm = 381.0176 / 44.1 = 8.64 mole
According to reaction 1 mole propane produce heat = -2218.6 KJ then 8.64 mole porpane produce heat =
-2218.6 X 8.64 = -19168.4 KJ
heat released by 0.840 pound of propane = 19168.4 KJ = 19168400 J
Temperature change of water 220C to 46.80C
T = 46.8 - 22 = 24.80C
specific of water = 4.184 J/ g K
mass of water = ?
q = mass of water X specific heat of H2O(l) X T
19168400 = mass of water X 4.184 X24.8
mass of water = 19168400 / (4.184 X24.8)
mass of water = 184732 gm = 184.732 kg of water heated.
184.732 kg of water heated.from 220C to 46.80C