A house with a 17.3 x 103 ft2 floor area and 8.00 ft high ceilings is to be heated with propane. How many kilo

Question

A house with a 17.3 x 103 ft2 floor area and 8.00 ft high ceilings is to be heated with propane. How many kilograms of propane would be needed to raise the temperature of the air inside the house from 65.0 oF to
70.0 oF ?. In this calculation your are ignoring the temperature of the walls etc.

The moloar specific heat of the air inside the house is 30.2 J mol -1 oC -1
The average density of air over this temperature range can be estimated to be 1.22 g/L.
The molar mass of air is 29.1 g/mol.
There are 28.317 liters in a cubic foot.
The molar heat of combustion of propane is -2220.047 kJ/mol

Answer 1

The volume of air in the house is (17300 ft^2)*(8 ft) = 138400 ft^3, assuming 17.3 x 103 means 17.3 x 10^3.

The volume of air in liters is (138400 ft^3)(28.317 L / ft^3) = 3919073 L.

The mass of the air is (3919073 L)(1.22 g/L) = 4781269 g.

The amount of air in moles is (4781269 g) / (29.1 g/mol) = 164305 mol.

The difference between 65 F and 70 F is 5 degrees F, which is equal to 5/1.8 = 2.78 degrees C.

The amount of heat required to increase the temperature of this amount of air by this amount of temperature is (164305 mol)(2.78 C)(30.2 J mol^-1 C^-1) = 13783345 J.

The number of moles of propane needed to generate this heat is (13783345 J) / (2220047 J/mol) = 6.21 mol.

You haven't been given the molar mass of propane, but I found that it was 44.096 g/mol. So the mass of propane you need is (6.21 mol)(44.096 g/mol) = 274 g, which is 0.274 kg.