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 × ΔTmass = 67.0 gal × 3.785 liter / gal × 1,000 g/liter = 140,045 gspecific heat: 4.184 J/gºC (from tables)ΔT = 35.0°C - 25.0°C = 10.0°CHeat = 140,045g × 4.184 g/JºC × 10.0ºC = 5,859,482.8 J

b) With 80% efficiency

Efficiency = 0.8 = heat out / heat inHeat in = 5,859,482.8J /  0.8 = 7,324,353.5JDivide 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.

This is so as the liquid you are heating is copper (II) sulfate (I think so) so when you heat it to saturation, there will still be some water molecules left behind, which will allow copper (II) sulfate crystals to be formed since there is water of crystallisation. so the formula is (CuSO4.5H2O).

Hence, if you heat it for a longer period of time when all the water has evaporated, you will obtain a white powder (CuSO4) as crystals cannot form without water of crystallisation