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).
1)How much heat is needed?

A) 27 kJ B) 35 kJ C) 49 kJ D) 57 kJ E) None of the above

2) How much octane needs to be burned?
A) 0.845 L B) 1.048 L C) 1.124 L D) 1.648 L E) None of the above

Answer 1

Given,

Cp for Air = 0.718 J / g K

Density of Air = 1.225 g / L

Volume of Air = 40 m^3 = 40,000 L

Mass = Density x Volume

=> Mass of Air = 1.225 x 40,000 = 49,000 g

Initial Temperature of Air (T1) = 10 degree C

Final Temperature of Air (T2) = 20 degree C

Heat absorbed by air = m Cp (T2 - T1)

=> Heat = 49000 x 0.718 x (20 - 10) = 351820 J = 351.82 kJ

E) None of the above

Given,

The enthalpy of combustion of octane is 5074.9 kJ/mol

=> 5074.9 kJ heat is released when 1 mole of Octane is burnt

=> 1 kJ heat is released when 1 / 5074.9 mole of Octane is burnt

=> 351.82 kJ heat is released when (1 / 5074.9) x 351.82 = 0.0693 moles of Octane is burnt

Molar Mass of Octane = 114.23 g / mol

=> Mass of octane = 0.0693 x 114.23 = 7.916 g

Density of octane = 703 kg / m^3 = 703 g / L

Volume = 7.916 / 703 = 0.0114 L = 11.4 mL

E) None of these