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30 June, 07:16

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)

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  1. 30 June, 08:38
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    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.
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