What volume of propane (C3H8) (measured at 0.00°C and 1.00 atm) must be burned in a cooker to raise the temperature of 2.00 kg of water from 20.0°C to 80.0°C, assuming the process efficiency is 60.0%? (Consider as liquid the water generated by the combustion) (2 pts.)

Reaction: C3H8 + 5 O2 → 3 CO2 + 4 H2O
ΔcH° 3 * (-393.5) + 4 * (-285.9) - (-103.8) = 2,220 kJ/mol
qrequired 2,000 * 4.184 * (80.0 - 20.0) / 0.60 = 837 kJ
n (C3H8) 837 / 2,220 = 0.380 mol
V (C3H8) 0.380 * 0.0821 * 273 / 1 = 8.52 L

Can someone tell me why we are dividing the efficiency rather than multiplying it?