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)

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:

b) With 80% efficiency

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:

The result must be reported with 3 significant figures.

c. because step up transformer steps up the input voltage i.e increases the input voltage  

the number of turns are greater on the secondary side of step up transformer  

transformer does not handle dc current

it means there are 3 atoms of oxygen