Question

You designed a camping stove that uses liquid propane as fuel and is so efficient that combustion takes place at 120 oC (temperature of the exhaust gas). How much useful heat will this stove produce per 100 g of liquid propane in winter conditions when temperature of the fuel and air are -30 oC? You can assume that air contains pure oxygen and that heat capacities are independent of temperature. C3H8 (g) + 5O2 --> 3CO2 + 4H2O Note: Vaporization of propane takes place at -30 oC.

Answer 1

enthalpy of vapourization of propane is 15.7 kJ/mol

heat capacity of propane = 0.39cal/g or 1.63 J/g

heat of combustion of propane is 2044kJ/mol

heat capacity of air is 1 kJ/Kg

100 g of propane is 100/44g/mol = 2.27 moles

So you will need 2.27 x 5 = 11.4 moles of oxygen which is 11.4/0.21 = 54 moles of air assuming a MW of 29 this is 54 x 29g/mol = 1569 g or 1.57 Kg

So to heat all of this we will need

Q = 2.27 x 15.7 kJ + 100 x 1.63 x 150 + 1.57 x 1 kJ x 150

Q = 295589 Joules of heat is needed

Heat liberated by propane burning is 2044 kJ x 2.27 = 4639880 J

So usefule heat is 4639880 - 295589 = 4344291 J or 4344.3kJ