
The process of making everyday items often requires the transformation of energy through many different forms, which can be followed as a flow of energy. Power is a basic quantity that measures the change in energy per unit time and has units of watts (W). Power is measured for many technologies all around us, such as light bulbs, batteries, automobile engines, washing machines, or stereo speakers. Voltage (measured in volts), electric current (measured in amperes), and power (measured in watts) are the most useful quantities for understanding the energy consumption of electric devices. Radiant energy spans the full electromagnetic spectrum, but our eyes can only see one small part of it called visible light.

power, watt (W), ampere (A), volt (V), electric current, radiant energy, intensity, light, horsepower (hp), renewable energy


$$P=\frac{\text{\Delta}E}{\text{\Delta}t}=\frac{W}{\text{\Delta}t}$$

 

Review problems and questions 

 Describe the energy flow that goes into producing a gold ring. Which step in the process do you think requires the most energy?

Many kinds of energy go into producing a gold ring. Mechanical energy is needed to dig and extract the ore, and electrical energy is needed for lights for the miners. Chemical energy in the form of fuel transports the gold to a factory or jeweler. Electrical or chemical energy converted to thermal energy melts the gold so that it can be deformed. Mechanical energy twists the metal into the desired shape. The mechanical energy needed to find and extract the ore is likely the most energyintensive step in the process.

 A 40 kg wheeled cart needs to be moved to the top of a platform that is 1.0 m high.
 How much power is required to pick up the cart and place it on the platform in 3.0 s?
 How much power is required to roll it 20 ft up a ramp, a process taking 20 s?
 How much power is required to roll it 40 ft up a shallower ramp, a process that also takes 20 s?

Answer:  130 W
 20 W
 20 W
Solution: Asked: power P needed to lift the cart
Given: mass of the cart m = 40 kg; height of the platform h = 1.0 m; time taken to lift cart t = 3.0 s; acceleration due to gravity g = 9.8 m/s^{2} Relationships: E_{p} = mgh, P = E/t Solve: $$\begin{array}{l}\begin{array}{lll}{E}_{p}\hfill & =\hfill & mgh\hfill \\ \hfill & =\hfill & \left(40\text{\hspace{0.17em}}\text{kg}\right)\times \left(9.8\text{\hspace{0.17em}}\text{m}/{\text{s}}^{\text{2}}\right)\times \left(1.0\text{\hspace{0.17em}}\text{m}\right)=392\text{\hspace{0.17em}}\text{J}\hfill \end{array}\\ \begin{array}{lll}P\hfill & =\hfill & \frac{E}{t}\hfill \\ \hfill & =\hfill & \frac{392\text{\hspace{0.17em}}\text{J}}{3\text{\hspace{0.17em}}\text{s}}=130\text{\hspace{0.17em}}\text{W}\hfill \end{array}\end{array}$$Answer: It will take 130 W of power to place the cart on the platform in 3.0 s.  Asked: power P required to roll the cart up the ramp
Given: potential energy E_{p} = 392 J at top of platform; time t = 20 s taken to reach top of platform Relationships: P = E/t Solve: $$\begin{array}{lll}P\hfill & =\hfill & \frac{E}{t}\hfill \\ \hfill & =\hfill & \frac{392\text{\hspace{0.17em}}\text{J}}{20\text{\hspace{0.17em}}\text{s}}=19.6\text{\hspace{0.17em}}\text{W}\hfill \end{array}$$Answer: It will take 20 W of power to roll the cart up the ramp in 20 s.  Power required only depends on the energy expended and the time taken. Since in this scenario the cart also takes 20 s to get to the top of the platform, it will also require 20 W of power.

 An immersion water heater is rated at 1,000 W. If it is plugged into a 120 volt outlet, how much current does it draw?

Answer: The water heater will draw 8.33 A of current.
Asked: current I drawn by the water heater Given: power P = 1,000 W drawn by heater; voltage V = 120 V of power source Relationships: P = IV Solve: $$\begin{array}{l}P=IV\\ I=\frac{P}{V}=\frac{1,000\text{\hspace{0.17em}}\text{W}}{120\text{\hspace{0.17em}}\text{V}}=8.33\text{\hspace{0.17em}}\text{A}\end{array}$$

 The cost of electricity in Mount Pleasant is 5 cents per kilowatthour.
 How much does it cost to operate a 100Wrated incandescent light bulb for 24 hours?
 How much does it cost to operate a 100Wrated fluorescent light bulb for 24 hours?
 How much does it cost to operate each for an entire year?
 Which one generates more light?

Answer:  $0.12
 $0.028
 It costs $43.80 to run a 100Wrated incandescent light bulb for a year and $10.07 to run a 100Wrated fluorescent light bulb for a year.
 Both generate the same amount of light (about 2 W of radiant energy).
Solution: Asked: cost of operating an incandescent light bulb
Given: cost of electricity = $0.05/kWh, power drawn by light bulb P = 100 W, time t = 24 hours Relationships: (total cost) = (cost of electricity) × (power) × (time) Solve: Using this relationship (and converting from kW to W) results in:$$\left(\frac{\$0.05}{\text{kW}\text{\hspace{0.17em}}\text{hr}}\right)\times \left(\frac{1\text{\hspace{0.17em}kW}}{\mathrm{1,000}\text{\hspace{0.17em}}\text{W}}\right)\times \left(100\text{\hspace{0.17em}}\text{W}\right)\times \left(24\text{\hspace{0.17em}}\text{hr}\right)=\$0.12$$Answer: It costs $0.12 to run a 100Wrated incandescent light bulb for 24 hr.  Asked: cost of operating a fluorescent light bulb
Given: cost of electricity = $0.05/kWh, power drawn by light bulb P = 23 W, time t = 24 hours Relationships: (total cost) = (cost of electricity) × (power) × (time) Solve: $$\left(\frac{\$0.05}{\text{kW}\text{\hspace{0.17em}}\text{hr}}\right)\times \left(\frac{1\text{\hspace{0.17em}kW}}{\mathrm{1,000}\text{\hspace{0.17em}}\text{W}}\right)\times \left(23\text{\hspace{0.17em}}\text{W}\right)\times \left(24\text{\hspace{0.17em}}\text{hr}\right)=\$0.028$$Answer: It costs $0.028 to run a 100Wrated fluorescent light bulb for 24 hr.  Asked: cost of operating incandescent and fluorescent light bulbs
Given: cost of electricity = $0.05/kWh, power drawn by light bulb = 100 W for incandescent and 23 W for fluorescent, time = 1 year Relationships: (total cost) = (cost of electricity) × (power) × (time) Solve: Incandescent:$$\begin{array}{l}\left(\frac{\$0.05}{\text{kW}\text{\hspace{0.17em}}\text{hr}}\right)\times \left(\frac{1\text{\hspace{0.17em}kW}}{\mathrm{1,000}\text{\hspace{0.17em}}\text{W}}\right)\times \left(100\text{\hspace{0.17em}}\text{W}\right)\times \left(1\text{\hspace{0.17em}}\text{hr}\right)\times \left(\frac{24\text{\hspace{0.17em}}\text{hr}}{1\text{\hspace{0.17em}}\text{day}}\right)\times \left(\frac{365\text{\hspace{0.17em}}\text{days}}{1\text{\hspace{0.17em}}\text{year}}\right)=\$43.80\end{array}$$Fluorescent:$$\begin{array}{l}\left(\frac{\$0.05}{\text{kW}\text{\hspace{0.17em}}\text{hr}}\right)\times \left(\frac{1\text{\hspace{0.17em}kW}}{\mathrm{1,000}\text{\hspace{0.17em}}\text{W}}\right)\times \left(23\text{\hspace{0.17em}}\text{W}\right)\times \left(1\text{\hspace{0.17em}}\text{hr}\right)\times \left(\frac{24\text{\hspace{0.17em}}\text{hr}}{1\text{\hspace{0.17em}}\text{day}}\right)\times \left(\frac{365\text{\hspace{0.17em}}\text{days}}{1\text{\hspace{0.17em}}\text{year}}\right)=\$10.07\end{array}$$Answer: It costs $43.80 to run a 100Wrated incandescent light bulb for a year and $10.07 to run a 100Wrated fluorescent light bulb for a year.  Both generate the same amount of light (about 2 W of radiant energy).

 From the standpoint of energy and power, what are the advantages for a car with high horsepower? What are the disadvantages?

More power means doing more work in less time, this means that a car with more horsepower can travel faster and accelerate faster. The major disadvantage is that it uses more energy—and hence costs more to run—so better gas mileage is usually obtained with a smaller and less powerful engine.

 Why does ultraviolet light cause sunburn but infrared light does not?

Ultraviolet light has higher energy than infrared light; the higher energy has the capacity to damage skin cells.

Take a Quiz 