Question of the Chapter:


"50 gal of water at room temperature (25oC) in a household water tank needs to be heated to 60oC (hot), how many grams of methane gas are required ?"
 

Given the standard enthalpy of combustion of methane, -802.2 kJ/mol.


 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 

Key:

(1) the heat needed:

    q = m s (T2-T1),     m = Vd

        (50gal)(3.785L/gal)(1000mL/L)(1g/mL) x (4.184J/caloC) x (60-25)oC

      = 2.77 x 107J = 27,700 kJ = 27.7 MJ
 

(2) (802.2 kJ/mol)(1mol/16g.04g) = 50.0 kJ/g
 

(3)  (27,700 kJ) / (50.0 kJ/g) = 554 g if all the heat is used.

(4) If the efficiency of the burner is 50%, half of the heat is wasted:

    2 x (27,700 kJ) / (50.0 kJ/g) = 1,108 g ~ 2.5 lbs