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)

146 g

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:

Heat = mass * specific heat * ΔT

mass = 67.0 gal * 3.785 liter / gal * 1,000 g/liter = 140,045 g

specific heat: 4.184 J/gºC (from tables)

ΔT = 35.0°C - 25.0°C = 10.0°C

Heat = 140,045g * 4.184 g/JºC * 10.0ºC = 5,859,482.8 J

b) With 80% efficiency

Efficiency = 0.8 = heat out / heat in

Heat in = 5,859,482.8J / 0.8 = 7,324,353.5J

Divide by 1,000 to convert to kJ: 7,324.3535kJ

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:

mass = 3.299 mol * 44.097g/mol = 145.5 g ≈ 146 g

The result must be reported with 3 significant figures.