Combustion of natural gas (primarily methane) occurs in most household heaters.
ID: 573231 • Letter: C
Question
Combustion of natural gas (primarily methane) occurs in most household heaters. The heat given off in this reaction is used to raise the temperature of the air in the house. Assuming that all the energy given off in the reaction goes to heating up only the air in the house, determine the mass of methane required to heat the air in a house by 15.0 C . Assume the following: house dimensions are 35.0 m × 35.0 m × 3.2 m ; molar heat capacity of air is 30 JK1mol1 ; 1.00 mol of air occupies 22.7 L for all temperatures concerned.
Explanation / Answer
The dimensions of the house are given; the volume of air in the house = (35.0 m)*(35.0 m)*(3.2 m) = 3920 m3.
We know that 1 m3 = 1000 L; hence, the volume of air in the house = (3920 m3)*(1000 L/1 m3) = 3.92*106 L.
Again, it is given that 1.00 mole of air occupies 22.7 L at all temperatures; hence, the number of moles of air in the house = (3.92*106 L)*(1.00 mole/22.7 L) = 172,687.225 mole.
Next, determine the heat energy involved in heating the air in the house by 15°C; the heat energy is given as
Heat energy involved = (moles of air)*(molar heat capacity of air)*(rise in temperature) = (172687.225 mole)*(30 JK-1mol-1)*(15°C)*(1 K/1°C) = 7.7709*107 J (1°C rise in temperature is the same as 1 K rise in temperature since both the scales have equal number of graduations).
The heat of combustion of methane is -890.8 kJ/mol. Ignore the negative sign and obtain the mole(s) of methane that must be combusted; therefore, we have mole(s) of methane = (7.7709*107 J)/(890.8 kJ/mol) = (7.7709*107 J)/[(890.8 kJ/mol)*(1000 J/1 kJ)] = 87.235 mole.
Molar mass of methane, CH4 = (1*12 + 4*1.008) g/mol = 16.032 g/mol.
Mass of methane combusted = (87.325 mole)*(16.032 g/mol) = 1399.9944 g = (1399.9944 g)*(1 kg/1000 g) = 1.3999944 kg 1.40 kg (ans).