How many kilograms of propane would be needed here?

Question

A house with a 18.1 × 103 ft2 floor area and 9.00 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 59.0 oF to
74.0 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

V(air) = 18.1 * 103 * 9 = 16800 cf * 28.3 lit/cf = 475000 liters

moles(air) = (475000 lit * 1.22 g/lit)/29.1 g/mol = 19900 mol

E = 30.2 J/mol*C * (74-59) C * 19900 mol = 9010 kJ

moles(propane) = 9010 kJ/2220 kJ/mol = 4.06 mol

molar mass of propane = 44.1 g/mol

mass(propane) = 44.1 g/mol * 4.06 mol = 179 g = 0.179 kg