14. Combustion of methane gas is the process through which most household heater

Question

14. Combustion of methane gas is the process through which most household heaters warm homes. The heat given off by the combustion is used to heat the air in the home. Assuming that 85.0% of the heat given off by the combustion is used to heat the home (15.0% is lost to the surroundings), what is the mass of methane (CH4) needed to warm the home by 10°C. Assume the house dimensions are 30.0 m x 30.0 m x 30.0 m, the specific heat capacity of air is 30 J/K mol, and 1.00 mol of air occupies 22.4 L at 25.0 °c.

Explanation / Answer

CH4 + O2= CO2 + H2O

balance

CH4 + 2O2= CO2 + 2H2O

now..

85% is used to heat...

find mass of CH4 required to incrase dT = 10°C

size of house:

V = 30*30*30 = 27000 m3 = 27000*10^3 L

note that 22.4 L of any mol will be at 0°C, so the data of 25°C is FALSE, anyways, we will use it since it is stated as a fact

mol of Air = V / D = ( 27000*10^3 L ) / (22.4L/mol) = 1205357.14 mol of air

then

Q = n*C*(Tf-Ti) = 1205357.14 mol * 30 J/Kmol * 10 K = 36160714.2 J

Q = 36160714.2 J = 36160.71 kJ

now..

Hcombustion methane = -882.0 kJ/mol

Qrequired (85%)= 36160.71 /0.85 = 42542.01 kJ

mol = Qrequired / Hcomb = 42542.01/882.0  

mol of CH4 = 48.23

mass = mol*MW = 48.23*16

mass of CH4 = 771.68 g

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