Question

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)

Answer 1

Answer:

• 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.