We usually have to pay for the energy we use. It is interesting and easy to estimate the cost of energy for an electrical appliance if its power consumption rate and time used are known. The higher the power consumption rate and the longer the appliance is used, the greater the cost of that appliance. The power consumption rate is P=W/t=E/tP=W/t=E/t size 12{P= {W} slash {t} = {E} slash {t} } {}, where EE size 12{E} {} is the energy supplied by the electricity company. So the energy consumed over a time tt size 12{t} {} is

E=Pt.E=Pt. size 12{E= ital "Pt"} {}

(4)Electricity bills state the energy used in units of kilowatt-hours (kW⋅h),(kW⋅h), size 12{ \( "kW" cdot h \) ,} {} which is the product of power in kilowatts and time in hours. This unit is convenient because electrical power consumption at the kilowatt level for hours at a time is typical.

What is the cost of running a 0.200-kW computer 6.00 h per day for 30.0 d if the cost of electricity is $0.120 per kW⋅hkW⋅h size 12{"kW" cdot h} {}?

*Strategy*

Cost is based on energy consumed; thus, we must find EE size 12{E} {} from E=PtE=Pt size 12{E= ital "Pt"} {} and then calculate the cost. Because electrical energy is expressed in kW⋅hkW⋅h size 12{"kW" cdot h} {}, at the start of a problem such as this it is convenient to convert the units into kWkW size 12{"kW"} {} and hours.

*Solution*

The energy consumed in kW⋅hkW⋅h size 12{"kW" cdot h} {} is

E
=
Pt=(0.200kW)(6.00h/d)(30.0d)
=
36.0 kW⋅h,E
=
Pt=(0.200kW)(6.00h/d)(30.0d)
=
36.0 kW⋅h,alignl { stack {
size 12{E= ital "Pt"= \( 0 "." "200"" kW" \) \( 6 "." "00"" h/d" \) \( "30" "." 0" d" \) } {} #
size 12{" "="36" "." "0 kW" cdot "h,"} {}
} } {}

(5)and the cost is simply given by

cost=(36.0 kW⋅h)($0.120 per kW⋅h)=$4.32 per month.cost=(36.0 kW⋅h)($0.120 per kW⋅h)=$4.32 per month. size 12{"cost"= \( "36" "." 0" kW" cdot h \) \( $0 "." "120"" per kW" cdot h \) =$4 "." "32"" per month"} {}

(6)*Discussion*

The cost of using the computer in this example is neither exorbitant nor negligible. It is clear that the cost is a combination of power and time. When both are high, such as for an air conditioner in the summer, the cost is high.

The motivation to save energy has become more compelling with its ever-increasing price. Armed with the knowledge that energy consumed is the product of power and time, you can estimate costs for yourself and make the necessary value judgments about where to save energy. Either power or time must be reduced. It is most cost-effective to limit the use of high-power devices that normally operate for long periods of time, such as water heaters and air conditioners. This would not include relatively high power devices like toasters, because they are on only a few minutes per day. It would also not include electric clocks, in spite of their 24-hour-per-day usage, because they are very low power devices. It is sometimes possible to use devices that have greater efficiencies—that is, devices that consume less power to accomplish the same task. One example is the compact fluorescent light bulb, which produces over four times more light per watt of power consumed than its incandescent cousin.

Modern civilization depends on energy, but current levels of energy consumption and production are not sustainable. The likelihood of a link between global warming and fossil fuel use (with its concomitant production of carbon dioxide), has made reduction in energy use as well as a shift to non-fossil fuels of the utmost importance. Even though energy in an isolated system is a conserved quantity, the final result of most energy transformations is waste heat transfer to the environment, which is no longer useful for doing work. As we will discuss in more detail in Thermodynamics, the potential for energy to produce useful work has been “degraded” in the energy transformation.

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