The combustion of propane (C₃H₈) produces 2220 kJ of energy per mole of propane consumed. How many grams of propane will be required to heat 67.0 gal of bathtub water from 25.0°C to 35.0°C if the process is 80.0% efficient? (1 gal = 3.785 L, 1 cal = 4.184 J, the density of water is 1.00 g/mL, the specific heat of water is 1 cal/(g°C)

Explanation:

1. Heat required to heat 67.0 gal of bathtub water from 25.0°C to 35.0°C

a) If the process were 100% efficient the heat would be:

b) With 80% efficiency

2. Amount of propane required:

a) moles = 7,3224.3535kJ/(2,220kJ/mol) = 3.299 mol

b) Multiply by the molar mass to obtain the mass in grams:

The result must be reported with 3 significant figures.

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