Mass of Propane to heat Water

Question

The combustion of propane (C₃H₈) produces 2220 kJ of energy per mole of propane consumed. What mass in grams of propane will be required to heat 63.0 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 1.00 cal/(g°C)?

Answer 1

So... this is a unit conversion. First determine how much energy you will need to put into the water to raise the temperature by 10 degrees Celsius.

63 gal of water * 3.785 g/L = 238.455 L of water...

Convert L of water to g so.... 238.455 L of water * 1000 grams/L = 238455 grams of water.

Proceed to then calculate the energy needed.

238455 grams of water * 1 cal/gC * 10 degree C difference = 2384550 calories needed.

Convert the calories to J. 2384550/ 4.184 cal/J = 569921J.

Now find how much propane you need.

2220J * 0.8 (because it is only 80 percent efficient to burn) = 1776kJ/mol of propane.

So total energy/ energy per mol of propane (accounting for inefficiency):

569921J / (1776 *1000) J/mol = 0.3209 moles of propane used.

Convert to grams of propane: 0.3209 moles * (12 (from each carbon) *3 +8(from hydrogen))grams/mol = 14.1196 grams of propane needed.

Convert to sig figs: 14.1 grams of propane needed assuming calculations are accurate.