You heat your house with liquid propane (C3H8) and you want to determine how muc

Question

You heat your house with liquid propane (C3H8) and you want to determine how
much propane it will take to change the temperature from 15 EC to 23 EC. Assume
you have an average house that is 186 m2 with ceiling height of 2.45 m. The air in
the house has a molar heat capacity of 29.1 J/mol@K. Because air is composed
primarily of nitrogen (78.084%) and oxygen (20.946%), these values can be used to
determine the

Explanation / Answer

volume of air in house = 186*2.45 = 455.7 m^3 = 455700 liters mass of air in house = volume*density = 555954 gms take 100 gms of air mass of nitrogen = 78.084 moles of nitrogen = 78.084/28.01 = 2.79 mass of oxygen = 20.946 moles of oxygen = 20.946/32 = 0.655 molar mass of air = 100/(2.79+0.655) = 29.03 gm/mol moles of air in house = 555954/29.03 = 19150.2 heat needed to increase the temp of the air by 8 degrees = 19150.2*29.1*8 = 4458.163 KJ propane burns as C3H8 + 5O2 -------> 3CO2 + 4H2O so heat released by combustion of 1 mole propane = 3(-393.5)+4(-241.826)-(-105) = -2042.94 KJ so moles of propane needed = 4458.163/2042.94 = 2.182 moles mass of propane needed = (2.182*44.1) = 96.24 gms milliliters of propane needed = 96.24/0.5853 = 164.42 liters of propane = 0.16442 L

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