A house with a 1.70 x 10 3 m 2 floor area and2.90 m high ceilings is to be heate

Question

A house with a 1.70 x 103 m2 floor area and2.90 m high ceilings is to be heated with propane. How manykilograms of propane would be needed to raise the temperature ofthe air inside the house from 14.4 oC to 23.1oC ? In this calculation your are ignoring thetemperature of the walls etc. The molar specific heat of the airinside the house is 30.2 J mol -1oC-1 The average density of air over this temperaturerange can be estimated to be 1.22 g/L. The average molar mass ofair is 29.1 g/mol. 1 L = 1 ×103 m3 The molar heat of combustion ofpropane is -2220.047 kJ/mol

Explanation / Answer

Volume = (1.7e3 m2)*(2.90 m) = 493000 m3 Q = moles*Cp*T moles = Volume*density/molar mass = 493000 m3 *(1000 L /1m3)*(1.22g/L)*(1 mol/29.1 g) = 20668728.5 moles air Q= 20668728.5 moles*(30.2 J/mol/C)*(23.1 C - 14.4 C) = 5.43050173e9Joules molar mass propane, C3H8 = 3*12 + 8 = 44 5.43050173e9 Joules *( 1mol propane/ 2220.047 kJ) *(1 kJ /1000 J)*(44 g/ mol )*(1 kg / 10000g) = 107.629287 kg propane ~ 108 kg propane

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