Question

An 11.0 W energy-efficient fluorescent lamp is designed to produce the same illumination as a conventional 40.0 W lamp.
a. How much energy does this lamp save during 100.0 h of use?
b. If electrical energy costs $0.080/kW •h, how Much money is saved in 100.0 h?

Answer 1

a. The energy-efficient lamp saves (40.0 W - 11.0 W) = 29.0 W of power usage.

To calculate the energy saved during 100.0 h, we need to convert watts to kilowatt-hours (kWh):

29.0 W = 0.029 kW

Energy saved = power x time

Energy saved = 0.029 kW x 100.0 h

Energy saved = 2.9 kWh

Therefore, the energy-efficient lamp saves 2.9 kWh of energy during 100.0 h of use.

b. To calculate the cost savings, we need to multiply the energy savings (in kWh) by the cost of electricity per kWh:

Cost savings = energy saved x cost per kWh

Cost savings = 2.9 kWh x $0.080/kWh

Cost savings = $0.232

Therefore, the energy-efficient lamp saves $0.232 in electricity costs during 100.0 h of use.