Question

The enthalpy of combustion of octane is 5074.9 kJ/mol and its density is 703 kg/m3 . On a mild winter day a room of volume 40 m3 needs to be warmed up from 10 to 20 OC? (Consider: cair = 0.718 J/g K; dair = 1.225 g/L).

Answer 1

Calculate heat required to raise the temperature of 40 m^3 of air from 10^oC to 20^o C using relation.

Q = m C_air T

... = 49000 x 0.718 x10   

... = 351820 J

(where , the given values for m = ( 40000 x 1.225 ) = 49000 gms. , C_air =0.718J K / gm ,

T = (293-283 = 10 K )

Now, given that the enthalpy of combustion of octane is 5074.9 kJ / ( or = 5074900 J ) / mol

So , 5074900 Joules of heat is given out by combustion of one mole or 114 gms of octane

therefore, 351820 J-----------------------------------------------------{ (114 x 351820) / 5074900 } gms-----

..............................................................................................................= 7.903 gms of octane

Given the density of octane as equal to 703 kg / m³  ( or = 703000gms / m³ )

therefore volume of octane required = 7.903 / 703000

......................................................... = 1.124 x 10^-5 m³

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Note - (i) The question doesn't specify what is required.

...........(ii) The volume of octane has been calculated as per data provided.