Question

An 11.0-W energy-efficient fluorescent lightbulb is designed to produce the same illumination as a conventional 40.0-W incandescent lightbulb. Assuming a cost of $0.116/kWh for energy from the electric company, how much money does the user of the energy-efficient bulb save during 140 h of use? (Give your answer to the nearest cent.)
$

Answer 1

Rate of electricity = R = 0.116 $/kW.h

Time period the bulb runs for = T = 140 hours

Power used by an incandescent lightbulb = P₁ = 40 W = 0.04 kW

Energy used by the incandescent lightbulb in 140 hours = E₁

E₁ = P₁T

E₁ = (0.04)(140)

E₁ = 5.6 kW.h

Cost of electricity for 140 hours of incandescent lightbulb = C₁

C₁ = E₁R

C₁ = (5.6)(0.116)

C₁ = 0.65 $

Power used by a fluorescent lightbulb = P₂ = 11 W = 0.011 kW

Energy used by the fluorescent lightbulb in 140 hours = E₂

E₂ = P₂T

E₂ = (0.011)(140)

E₂ = 1.54 kW.h

Cost of electricity for 140 hours of fluorescent lightbulb = C₂

C₂ = E₂R

C₂ = (1.54)(0.116)

C₂ = 0.18 $

Money saved by the energy efficient bulb = C

C = C₁ - C₂

C = 0.65 - 0.18

C = 0.47 $

C = 47 cents

Money the user of the energy efficient bulb saves during 140 hours of use = 0.47 $ = 47 cents