Question

A new type of 38.0 W lamp is designed to produce the same illumination as a conventional 53.0 W lamp. How much energy does this lamp save during 370.0 h of use? Answer in units of J.

If electrical energy costs $0.080/kW·h, how much money is saved in 370.0 h? Answer in units of $.

Answer 1

Time period = T = 370 hours = 370 x (3600) sec = 1.332 x 10⁶ sec

Power usage of the new lamp = P₁ = 38 W = 0.038 kW

Power usage of the conventional lamp = P₂ = 53 W = 0.053 kW

To find the energy usage in Joules we will use the power in watts and the time in seconds.

Energy used by the new lamp in 370 hours in Joules = E₁

E₁ = P₁T

E₁ = (38)(1.332x10⁶)

E₁ = 50.616 x 10⁶ J

Energy used by the conventional lamp in 370 hours in Joules = E₂

E₂ = P₂T

E₂ = (53)(1.332x10⁶)

E₂ = 70.596 x 10⁶ J

Energy saved by the new lamp = E

E = E₂ - E₁

E = 70.596x10⁶ - 50.616x10⁶

E = 1.998 x 10⁷ J

Electrical energy cost = C = 0.08 $/(kW.h)

To find the energy usage in kilowatt-hours we will use the power in kilowatts and the time in hours.

Energy used by the new lamp in 370 hours in kW.h = E₃

E₃ = P₁T

E₃ = (0.038)(370)

E₃ = 14.06 kW.h

Cost to run the new lamp for 370 hours = C₁

C₁ = CE₃

C₁ = (0.08)(14.06)

C₁ = 1.1248 $

Energy used by the conventional lamp in 370 hours in kW.h = E₄

E₄ = P₂T

E₄ = (0.053)(370)

E₄ = 19.61 kW.h

Cost to run the conventional lamp for 370 hours = C₂

C₂ = CE₄

C₂ = (0.08)(19.61)

C₂ = 1.5688 $

Amount of money saved by the new lamp = C

C = C₂ - C₁

C = 1.5688 - 1.1248

C = 0.444 $

a) Energy the new lamp saves in 370 hours of use = 1.998 x 10⁷ J

b) Money saved by the new lamp in 370 hours of use = 0.444 $