111 Alternating Current versus Direct Current
[latexpage]
Learning Objectives
- Explain the differences and similarities between AC and DC current.
- Calculate rms voltage, current, and average power.
- Explain why AC current is used for power transmission.
Alternating Current
Most of the examples dealt with so far, and particularly those utilizing batteries, have constant voltage sources. Once the current is established, it is thus also a constant. Direct current (DC) is the flow of electric charge in only one direction. It is the steady state of a constant-voltage circuit. Most well-known applications, however, use a time-varying voltage source. Alternating current (AC) is the flow of electric charge that periodically reverses direction. If the source varies periodically, particularly sinusoidally, the circuit is known as an alternating current circuit. Examples include the commercial and residential power that serves so many of our needs. (Figure) shows graphs of voltage and current versus time for typical DC and AC power. The AC voltages and frequencies commonly used in homes and businesses vary around the world.
(Figure) shows a schematic of a simple circuit with an AC voltage source. The voltage between the terminals fluctuates as shown, with the AC voltage given by
where \(V\) is the voltage at time \(t\), \({V}_{0}\) is the peak voltage, and \(f\) is the frequency in hertz. For this simple resistance circuit, \(I=\text{V/R}\), and so the AC current is
where \(I\) is the current at time \(t\), and \({I}_{0}={V}_{0}\text{/R}\) is the peak current. For this example, the voltage and current are said to be in phase, as seen in (Figure)(b).
Current in the resistor alternates back and forth just like the driving voltage, since \(I=\text{V/R}\). If the resistor is a fluorescent light bulb, for example, it brightens and dims 120 times per second as the current repeatedly goes through zero. A 120-Hz flicker is too rapid for your eyes to detect, but if you wave your hand back and forth between your face and a fluorescent light, you will see a stroboscopic effect evidencing AC. The fact that the light output fluctuates means that the power is fluctuating. The power supplied is \(P=\text{IV}\). Using the expressions for \(I\) and \(V\) above, we see that the time dependence of power is \(P={I}_{0}{V}_{0}\phantom{\rule{0.30em}{0ex}}{\text{sin}}^{2}\phantom{\rule{0.30em}{0ex}}\text{2}\pi \text{ft}\), as shown in (Figure).
 Do you observe the same thing with the headlights on your car? Explain what you observe. Warning: Do not look directly at very bright light.
We are most often concerned with average power rather than its fluctuations—that 60-W light bulb in your desk lamp has an average power consumption of 60 W, for example. As illustrated in (Figure), the average power \({P}_{\text{ave}}\) is
This is evident from the graph, since the areas above and below the \(\left(1/2\right){I}_{0}{V}_{0}\) line are equal, but it can also be proven using trigonometric identities. Similarly, we define an average or rms current \({I}_{\text{rms}}\) and average or rms voltage \({V}_{\text{rms}}\) to be, respectively,
and
where rms stands for root mean square, a particular kind of average. In general, to obtain a root mean square, the particular quantity is squared, its mean (or average) is found, and the square root is taken. This is useful for AC, since the average value is zero. Now,
which gives
as stated above. It is standard practice to quote \({I}_{\text{rms}}\), \({V}_{\text{rms}}\), and \({P}_{\text{ave}}\) rather than the peak values. For example, most household electricity is 120 V AC, which means that \({V}_{\text{rms}}\) is 120 V. The common 10-A circuit breaker will interrupt a sustained \({I}_{\text{rms}}\) greater than 10 A. Your 1.0-kW microwave oven consumes \({P}_{\text{ave}}=\text{1.0 kW}\), and so on. You can think of these rms and average values as the equivalent DC values for a simple resistive circuit.
To summarize, when dealing with AC, Ohm’s law and the equations for power are completely analogous to those for DC, but rms and average values are used for AC. Thus, for AC, Ohm’s law is written
The various expressions for AC power \({P}_{\text{ave}}\) are
and
(a) What is the value of the peak voltage for 120-V AC power? (b) What is the peak power consumption rate of a 60.0-W AC light bulb?
Strategy
We are told that \({V}_{\text{rms}}\) is 120 V and \({P}_{\text{ave}}\) is 60.0 W. We can use \({V}_{\text{rms}}=\frac{{V}_{0}}{\sqrt{2}}\) to find the peak voltage, and we can manipulate the definition of power to find the peak power from the given average power.
Solution for (a)
Solving the equation \({V}_{\text{rms}}=\frac{{V}_{0}}{\sqrt{2}}\) for the peak voltage \({V}_{0}\) and substituting the known value for \({V}_{\text{rms}}\) gives
Discussion for (a)
This means that the AC voltage swings from 170 V to \(\text{–170 V}\) and back 60 times every second. An equivalent DC voltage is a constant 120 V.
Solution for (b)
Peak power is peak current times peak voltage. Thus,
We know the average power is 60.0 W, and so
Discussion
So the power swings from zero to 120 W one hundred twenty times per second (twice each cycle), and the power averages 60 W.
Why Use AC for Power Distribution?
Most large power-distribution systems are AC. Moreover, the power is transmitted at much higher voltages than the 120-V AC (240 V in most parts of the world) we use in homes and on the job. Economies of scale make it cheaper to build a few very large electric power-generation plants than to build numerous small ones. This necessitates sending power long distances, and it is obviously important that energy losses en route be minimized. High voltages can be transmitted with much smaller power losses than low voltages, as we shall see. (See (Figure).) For safety reasons, the voltage at the user is reduced to familiar values. The crucial factor is that it is much easier to increase and decrease AC voltages than DC, so AC is used in most large power distribution systems.
(a) What current is needed to transmit 100 MW of power at 200 kV? (b) What is the power dissipated by the transmission lines if they have a resistance of \(1\text{.}\text{00}\phantom{\rule{0.25em}{0ex}}\Omega \)? (c) What percentage of the power is lost in the transmission lines?
Strategy
We are given \({P}_{\text{ave}}=\text{100 MW}\), \({V}_{\text{rms}}=\text{200 kV}\), and the resistance of the lines is \(R=1\text{.}\text{00}\phantom{\rule{0.25em}{0ex}}\Omega \). Using these givens, we can find the current flowing (from \(P=\text{IV}\)) and then the power dissipated in the lines (\(P={I}^{2}R\)), and we take the ratio to the total power transmitted.
Solution
To find the current, we rearrange the relationship \({P}_{\text{ave}}={I}_{\text{rms}}{V}_{\text{rms}}\) and substitute known values. This gives
Solution
Knowing the current and given the resistance of the lines, the power dissipated in them is found from \({P}_{\text{ave}}={I}_{\text{rms}}^{2}R\). Substituting the known values gives
Solution
The percent loss is the ratio of this lost power to the total or input power, multiplied by 100:
Discussion
One-fourth of a percent is an acceptable loss. Note that if 100 MW of power had been transmitted at 25 kV, then a current of 4000 A would have been needed. This would result in a power loss in the lines of 16.0 MW, or 16.0% rather than 0.250%. The lower the voltage, the more current is needed, and the greater the power loss in the fixed-resistance transmission lines. Of course, lower-resistance lines can be built, but this requires larger and more expensive wires. If superconducting lines could be economically produced, there would be no loss in the transmission lines at all. But, as we shall see in a later chapter, there is a limit to current in superconductors, too. In short, high voltages are more economical for transmitting power, and AC voltage is much easier to raise and lower, so that AC is used in most large-scale power distribution systems.
It is widely recognized that high voltages pose greater hazards than low voltages. But, in fact, some high voltages, such as those associated with common static electricity, can be harmless. So it is not voltage alone that determines a hazard. It is not so widely recognized that AC shocks are often more harmful than similar DC shocks. Thomas Edison thought that AC shocks were more harmful and set up a DC power-distribution system in New York City in the late 1800s. There were bitter fights, in particular between Edison and George Westinghouse and Nikola Tesla, who were advocating the use of AC in early power-distribution systems. AC has prevailed largely due to transformers and lower power losses with high-voltage transmission.
Generate electricity with a bar magnet! Discover the physics behind the phenomena by exploring magnets and how you can use them to make a bulb light.
Section Summary
- Direct current (DC) is the flow of electric current in only one direction. It refers to systems where the source voltage is constant.
- The voltage source of an alternating current (AC) system puts out \(V={V}_{0}\phantom{\rule{0.25em}{0ex}}\text{sin 2}\pi \text{ft}\), where \(V\) is the voltage at time \(t\), \({V}_{0}\) is the peak voltage, and \(f\) is the frequency in hertz.
- In a simple circuit, \(I=\text{V/R}\) and AC current is \(I={I}_{0}\phantom{\rule{0.25em}{0ex}}\text{sin 2}\pi \text{ft}\), where \(I\) is the current at time \(t\), and \({I}_{0}={V}_{0}\text{/R}\) is the peak current.
- The average AC power is \({P}_{\text{ave}}=\frac{1}{2}{I}_{0}{V}_{0}\).
- Average (rms) current \({I}_{\text{rms}}\) and average (rms) voltage \({V}_{\text{rms}}\) are \({I}_{\text{rms}}=\frac{{I}_{0}}{\sqrt{2}}\) and \({V}_{\text{rms}}=\frac{{V}_{0}}{\sqrt{2}}\), where rms stands for root mean square.
- Thus, \({P}_{\text{ave}}={I}_{\text{rms}}{V}_{\text{rms}}\).
- Ohm’s law for AC is \({I}_{\text{rms}}=\frac{{V}_{\text{rms}}}{R}\).
- Expressions for the average power of an AC circuit are \({P}_{\text{ave}}={I}_{\text{rms}}{V}_{\text{rms}}\), \({P}_{\text{ave}}=\frac{{V}_{\text{rms}}^{\phantom{\rule{0.25em}{0ex}}\phantom{\rule{0.25em}{0ex}}\phantom{\rule{0.25em}{0ex}}2}}{R}\), and \({P}_{\text{ave}}={I}_{\text{rms}}^{\phantom{\rule{0.25em}{0ex}}\phantom{\rule{0.25em}{0ex}}\phantom{\rule{0.25em}{0ex}}2}R\), analogous to the expressions for DC circuits.
Conceptual Questions
Give an example of a use of AC power other than in the household. Similarly, give an example of a use of DC power other than that supplied by batteries.
Why do voltage, current, and power go through zero 120 times per second for 60-Hz AC electricity?
You are riding in a train, gazing into the distance through its window. As close objects streak by, you notice that the nearby fluorescent lights make dashed streaks. Explain.
Problem Exercises
(a) What is the hot resistance of a 25-W light bulb that runs on 120-V AC? (b) If the bulb’s operating temperature is \(\text{2700º}\text{C}\), what is its resistance at \(\text{2600º}\text{C}\)?
Certain heavy industrial equipment uses AC power that has a peak voltage of 679 V. What is the rms voltage?
480 V
A certain circuit breaker trips when the rms current is 15.0 A. What is the corresponding peak current?
Military aircraft use 400-Hz AC power, because it is possible to design lighter-weight equipment at this higher frequency. What is the time for one complete cycle of this power?
2.50 ms
A North American tourist takes his 25.0-W, 120-V AC razor to Europe, finds a special adapter, and plugs it into 240 V AC. Assuming constant resistance, what power does the razor consume as it is ruined?
In this problem, you will verify statements made at the end of the power losses for (Figure). (a) What current is needed to transmit 100 MW of power at a voltage of 25.0 kV? (b) Find the power loss in a \(1\text{.}\text{00 -}\phantom{\rule{0.25em}{0ex}}\Omega \) transmission line. (c) What percent loss does this represent?
(a) 4.00 kA
(b) 16.0 MW
(c) 16.0%
A small office-building air conditioner operates on 408-V AC and consumes 50.0 kW. (a) What is its effective resistance? (b) What is the cost of running the air conditioner during a hot summer month when it is on 8.00 h per day for 30 days and electricity costs \(9.00 cents\text{/kW}\cdot \text{h}\)?
What is the peak power consumption of a 120-V AC microwave oven that draws 10.0 A?
2.40 kW
What is the peak current through a 500-W room heater that operates on 120-V AC power?
Two different electrical devices have the same power consumption, but one is meant to be operated on 120-V AC and the other on 240-V AC. (a) What is the ratio of their resistances? (b) What is the ratio of their currents? (c) Assuming its resistance is unaffected, by what factor will the power increase if a 120-V AC device is connected to 240-V AC?
(a) 4.0
(b) 0.50
(c) 4.0
Nichrome wire is used in some radiative heaters. (a) Find the resistance needed if the average power output is to be 1.00 kW utilizing 120-V AC. (b) What length of Nichrome wire, having a cross-sectional area of \(5.00{\text{mm}}^{2}\), is needed if the operating temperature is \(\text{500º C}\)? (c) What power will it draw when first switched on?
Find the time after \(t=0\) when the instantaneous voltage of 60-Hz AC first reaches the following values: (a) \({V}_{0}/2\) (b) \({V}_{0}\) (c) 0.
(a) 1.39 ms
(b) 4.17 ms
(c) 8.33 ms
(a) At what two times in the first period following \(t=0\) does the instantaneous voltage in 60-Hz AC equal \({V}_{\text{rms}}\)? (b) \(-{V}_{\text{rms}}\)?
Glossary
- direct current
- (DC) the flow of electric charge in only one direction
- alternating current
- (AC) the flow of electric charge that periodically reverses direction
- AC voltage
- voltage that fluctuates sinusoidally with time, expressed as V = V0 sin 2πft, where V is the voltage at time t, V0 is the peak voltage, and f is the frequency in hertz
- AC current
- current that fluctuates sinusoidally with time, expressed as I = I0 sin 2πft, where I is the current at time t, I0 is the peak current, and f is the frequency in hertz
- rms current
- the root mean square of the current, \({I}_{\text{rms}}={I}_{0}/\sqrt{2}\) , where I0 is the peak current, in an AC system
- rms voltage
- the root mean square of the voltage, \({V}_{\text{rms}}={V}_{0}/\sqrt{2}\) , where V0 is the peak voltage, in an AC system


Chapter 20 Electric Current, Resistance, and Ohm’s Law
20.4 Electric Power and Energy
Summary
- Calculate the power dissipated by a resistor and power supplied by a power supply.
- Calculate the cost of electricity under various circumstances.
Power in Electric Circuits
Power is associated by many people with electricity. Knowing that power is the rate of energy use or energy conversion, what is the expression for electric power? Power transmission lines might come to mind. We also think of lightbulbs in terms of their power ratings in watts. Let us compare a 25-W bulb with a 60-W bulb. (See Figure 1(a).) Since both operate on the same voltage, the 60-W bulb must draw more current to have a greater power rating. Thus the 60-W bulb’s resistance must be lower than that of a 25-W bulb. If we increase voltage, we also increase power. For example, when a 25-W bulb that is designed to operate on 120 V is connected to 240 V, it briefly glows very brightly and then burns out. Precisely how are voltage, current, and resistance related to electric power?
Electric energy depends on both the voltage involved and the charge moved. This is expressed most simply as [latex]{\text{PE} = qV}[/latex], where [latex]{q}[/latex] is the charge moved and [latex]{V}[/latex] is the voltage (or more precisely, the potential difference the charge moves through). Power is the rate at which energy is moved, and so electric power is
Recognizing that current is [latex]{I = q/t}[/latex] (note that [latex]{\Delta t=t}[/latex] here), the expression for power becomes
Electric power ([latex]{P}[/latex]) is simply the product of current times voltage. Power has familiar units of watts. Since the SI unit for potential energy (PE) is the joule, power has units of joules per second, or watts. Thus, [latex]{1 \;\text{A} \cdot \;\text{V} = 1 \;\text{W}}[/latex]. For example, cars often have one or more auxiliary power outlets with which you can charge a cell phone or other electronic devices. These outlets may be rated at 20 A, so that the circuit can deliver a maximum power [latex]{P = IV = (20 \;\text{A})(12 \;\text{V}) = 240 \;\text{W}}[/latex]. In some applications, electric power may be expressed as volt-amperes or even kilovolt-amperes (
[latex]{1 \;\text{kA} \cdot \;\text{V} = 1 \;\text{kW}}[/latex]).
To see the relationship of power to resistance, we combine Ohm’s law with [latex]{P = IV}[/latex]. Substituting [latex]{I = V/R}[/latex] gives [latex]{P = (V/R)V = V^2/R}[/latex]. Similarly, substituting [latex]{V = IR}[/latex] gives [latex]{P = I(IR) = I^2R}[/latex]. Three expressions for electric power are listed together here for convenience:
Note that the first equation is always valid, whereas the other two can be used only for resistors. In a simple circuit, with one voltage source and a single resistor, the power supplied by the voltage source and that dissipated by the resistor are identical. (In more complicated circuits, [latex]{P}[/latex] can be the power dissipated by a single device and not the total power in the circuit.)
Different insights can be gained from the three different expressions for electric power. For example, [latex]{P = V^2/R}[/latex] implies that the lower the resistance connected to a given voltage source, the greater the power delivered. Furthermore, since voltage is squared in [latex]{P = V^2/R}[/latex], the effect of applying a higher voltage is perhaps greater than expected. Thus, when the voltage is doubled to a 25-W bulb, its power nearly quadruples to about 100 W, burning it out. If the bulb’s resistance remained constant, its power would be exactly 100 W, but at the higher temperature its resistance is higher, too.
Example 1: Calculating Power Dissipation and Current: Hot and Cold Power
(a) Consider the examples given in Chapter 20.2 Ohm’s Law: Resistance and Simple Circuits and Chapter 20.3 Resistance and Resistivity. Then find the power dissipated by the car headlight in these examples, both when it is hot and when it is cold. (b) What current does it draw when cold?
Strategy for (a)
For the hot headlight, we know voltage and current, so we can use [latex]{P = IV}[/latex] to find the power. For the cold headlight, we know the voltage and resistance, so we can use [latex]{P = V^2/R}[/latex] to find the power.
Solution for (a)
Entering the known values of current and voltage for the hot headlight, we obtain
The cold resistance was [latex]{0.350 \;\Omega}[/latex], and so the power it uses when first switched on is
Discussion for (a)
The 30 W dissipated by the hot headlight is typical. But the 411 W when cold is surprisingly higher. The initial power quickly decreases as the bulb’s temperature increases and its resistance increases.
Strategy and Solution for (b)
The current when the bulb is cold can be found several different ways. We rearrange one of the power equations, [latex]{P = I^2R}[/latex], and enter known values, obtaining
Discussion for (b)
The cold current is remarkably higher than the steady-state value of 2.50 A, but the current will quickly decline to that value as the bulb’s temperature increases. Most fuses and circuit breakers (used to limit the current in a circuit) are designed to tolerate very high currents briefly as a device comes on. In some cases, such as with electric motors, the current remains high for several seconds, necessitating special “slow blow” fuses.
The Cost of Electricity
The more electric appliances you use and the longer they are left on, the higher your electric bill. This familiar fact is based on the relationship between energy and power. You pay for the energy used. Since [latex]{P=E/t}[/latex], we see that
is the energy used by a device using power [latex]{P}[/latex] for a time interval [latex]{t}[/latex]. For example, the more lightbulbs burning, the greater [latex]{P}[/latex] used; the longer they are on, the greater [latex]{t}[/latex] is. The energy unit on electric bills is the kilowatt-hour ([latex]{\text{kW} \cdot \;\text{h}}[/latex]), consistent with the relationship [latex]{E = Pt}[/latex]. It is easy to estimate the cost of operating electric appliances if you have some idea of their power consumption rate in watts or kilowatts, the time they are on in hours, and the cost per kilowatt-hour for your electric utility. Kilowatt-hours, like all other specialized energy units such as food calories, can be converted to joules. You can prove to yourself that [latex]{1 \;\text{kW} \cdot \;\text{h} = 3.6 \times 10^6 \;\text{J}}[/latex].
The electrical energy ([latex]{E}[/latex]) used can be reduced either by reducing the time of use or by reducing the power consumption of that appliance or fixture. This will not only reduce the cost, but it will also result in a reduced impact on the environment. Improvements to lighting are some of the fastest ways to reduce the electrical energy used in a home or business. About 20% of a home’s use of energy goes to lighting, while the number for commercial establishments is closer to 40%. Fluorescent lights are about four times more efficient than incandescent lights—this is true for both the long tubes and the compact fluorescent lights (CFL). (See Figure 1(b).) Thus, a 60-W incandescent bulb can be replaced by a 15-W CFL, which has the same brightness and color. CFLs have a bent tube inside a globe or a spiral-shaped tube, all connected to a standard screw-in base that fits standard incandescent light sockets. (Original problems with color, flicker, shape, and high initial investment for CFLs have been addressed in recent years.) The heat transfer from these CFLs is less, and they last up to 10 times longer. The significance of an investment in such bulbs is addressed in the next example. New white LED lights (which are clusters of small LED bulbs) are even more efficient (twice that of CFLs) and last 5 times longer than CFLs. However, their cost is still high.
Making Connections: Energy, Power, and Time
The relationship [latex]{E = Pt}[/latex] is one that you will find useful in many different contexts. The energy your body uses in exercise is related to the power level and duration of your activity, for example. The amount of heating by a power source is related to the power level and time it is applied. Even the radiation dose of an X-ray image is related to the power and time of exposure.
Example 2: Calculating the Cost Effectiveness of Compact Fluorescent Lights (CFL)
If the cost of electricity in your area is 12 cents per kWh, what is the total cost (capital plus operation) of using a 60-W incandescent bulb for 1000 hours (the lifetime of that bulb) if the bulb cost 25 cents? (b) If we replace this bulb with a compact fluorescent light that provides the same light output, but at one-quarter the wattage, and which costs $1.50 but lasts 10 times longer (10,000 hours), what will that total cost be?
Strategy
To find the operating cost, we first find the energy used in kilowatt-hours and then multiply by the cost per kilowatt-hour.
Solution for (a)
The energy used in kilowatt-hours is found by entering the power and time into the expression for energy:
In kilowatt-hours, this is
Now the electricity cost is
The total cost will be $7.20 for 1000 hours (about one-half year at 5 hours per day).
Solution for (b)
Since the CFL uses only 15 W and not 60 W, the electricity cost will be $7.20/4 = $1.80. The CFL will last 10 times longer than the incandescent, so that the investment cost will be 1/10 of the bulb cost for that time period of use, or 0.1($1.50) = $0.15. Therefore, the total cost will be $1.95 for 1000 hours.
Discussion
Therefore, it is much cheaper to use the CFLs, even though the initial investment is higher. The increased cost of labor that a business must include for replacing the incandescent bulbs more often has not been figured in here.
Making Connections: Take-Home Experiment—Electrical Energy Use Inventory
1) Make a list of the power ratings on a range of appliances in your home or room. Explain why something like a toaster has a higher rating than a digital clock. Estimate the energy consumed by these appliances in an average day (by estimating their time of use). Some appliances might only state the operating current. If the household voltage is 120 V, then use [latex]{P = IV}[/latex]. 2) Check out the total wattage used in the rest rooms of your school’s floor or building. (You might need to assume the long fluorescent lights in use are rated at 32 W.) Suppose that the building was closed all weekend and that these lights were left on from 6 p.m. Friday until 8 a.m. Monday. What would this oversight cost? How about for an entire year of weekends?
Section Summary
- Electric power [latex]{P}[/latex] is the rate (in watts) that energy is supplied by a source or dissipated by a device.
- Three expressions for electrical power are
[latex]{P = IV},[/latex][latex]{P =}[/latex] [latex]{\frac{V^2}{R}}[/latex],
and
[latex]{P = I^2 R}.[/latex] - The energy used by a device with a power [latex]{P}[/latex] over a time [latex]{t}[/latex] is [latex]{E = Pt}[/latex].
Conceptual Questions
1: Why do incandescent lightbulbs grow dim late in their lives, particularly just before their filaments break?
2: The power dissipated in a resistor is given by [latex]{P = V^2/R}[/latex], which means power decreases if resistance increases. Yet this power is also given by [latex]{P = I^2R}[/latex], which means power increases if resistance increases. Explain why there is no contradiction here.
Probles & Exercises
1: What is the power of a [latex]{1.00 \times 10^2 \;\text{MV}}[/latex] lightning bolt having a current of [latex]{2.00 \times 10^4 \;\text{A}}[/latex]?
2: What power is supplied to the starter motor of a large truck that draws 250 A of current from a 24.0-V battery hookup?
3: A charge of 4.00 C of charge passes through a pocket calculator’s solar cells in 4.00 h. What is the power output, given the calculator’s voltage output is 3.00 V? (See Figure 2.)
4: How many watts does a flashlight that has [latex]{6.00 \times 10^2 \;\text{C}}[/latex] pass through it in 0.500 h use if its voltage is 3.00 V?
5: Find the power dissipated in each of these extension cords: (a) an extension cord having a [latex]{0.0600 -\;\Omega}[/latex] resistance and through which 5.00 A is flowing; (b) a cheaper cord utilizing thinner wire and with a resistance of [latex]{0.300 \;\Omega}[/latex]
6: Verify that the units of a volt-ampere are watts, as implied by the equation [latex]{P = IV}[/latex].
7: Show that the units [latex]{1 \;\text{V}^2 / \;\Omega = 1 \;\text{W}}[/latex], as implied by the equation [latex]{P = V^2/R}[/latex].
8: Show that the units [latex]{1 \;\text{A}^2 \cdot \;\Omega = 1 \;\text{W}}[/latex], as implied by the equation [latex]{P = I^2 R}[/latex].
9: Verify the energy unit equivalence that [latex]{1 \;\text{kW} \cdot \;\text{h} = 3.60 \times 10^6 \;\text{J}}[/latex].
10: Electrons in an X-ray tube are accelerated through [latex]{1.00 \times 10^2 \;\text{kV}}[/latex] and directed toward a target to produce X-rays. Calculate the power of the electron beam in this tube if it has a current of 15.0 mA.
11: An electric water heater consumes 5.00 kW for 2.00 h per day. What is the cost of running it for one year if electricity costs [latex]{12.0 \;\text{cents/kW} \cdot \;\text{h}}[/latex]? See Figure 3.
12: With a 1200-W toaster, how much electrical energy is needed to make a slice of toast (cooking time = 1 minute)? At [latex]{9.0 \;\text{cents/kW} \cdot \;\text{h}}[/latex], how much does this cost?
13: What would be the maximum cost of a CFL such that the total cost (investment plus operating) would be the same for both CFL and incandescent 60-W bulbs? Assume the cost of the incandescent bulb is 25 cents and that electricity costs [latex]{10 \;\text{cents/kWh}}[/latex]. Calculate the cost for 1000 hours, as in the cost effectiveness of CFL example.
14: Some makes of older cars have 6.00-V electrical systems. (a) What is the hot resistance of a 30.0-W headlight in such a car? (b) What current flows through it?
15: Alkaline batteries have the advantage of putting out constant voltage until very nearly the end of their life. How long will an alkaline battery rated at [latex]{1.00 \;\text{A} \cdot \;\text{h}}[/latex] and 1.58 V keep a 1.00-W flashlight bulb burning?
16: A cauterizer, used to stop bleeding in surgery, puts out 2.00 mA at 15.0 kV. (a) What is its power output? (b) What is the resistance of the path?
17: The average television is said to be on 6 hours per day. Estimate the yearly cost of electricity to operate 100 million TVs, assuming their power consumption averages 150 W and the cost of electricity averages [latex]{12.0 \;\text{cents/kW} \cdot \;\text{h}}[/latex].
18: An old lightbulb draws only 50.0 W, rather than its original 60.0 W, due to evaporative thinning of its filament. By what factor is its diameter reduced, assuming uniform thinning along its length? Neglect any effects caused by temperature differences.
19: 00-gauge copper wire has a diameter of 9.266 mm. Calculate the power loss in a kilometer of such wire when it carries [latex]{1.00 \times 10^2 \;\text{A}}[/latex].
20: Integrated Concepts
Cold vaporizers pass a current through water, evaporating it with only a small increase in temperature. One such home device is rated at 3.50 A and utilizes 120 V AC with 95.0% efficiency. (a) What is the vaporization rate in grams per minute? (b) How much water must you put into the vaporizer for 8.00 h of overnight operation? (See Figure 4.)
21: Integrated Concepts
(a) What energy is dissipated by a lightning bolt having a 20,000-A current, a voltage of [latex]{1.00 \times 10^2 \;\text{MV}}[/latex], and a length of 1.00 ms? (b) What mass of tree sap could be raised from 18.0ºC to its boiling point and then evaporated by this energy, assuming sap has the same thermal characteristics as water?
22: Integrated Concepts
What current must be produced by a 12.0-V battery-operated bottle warmer in order to heat 75.0 g of glass, 250 g of baby formula, and [latex]{3.00 \times 10^2 \;\text{g}}[/latex] of aluminum from 20.0ºC to 90.0ºC in 5.00 min?
23: Integrated Concepts
How much time is needed for a surgical cauterizer to raise the temperature of 1.00 g of tissue from 37.0ºC to 100ºC and then boil away 0.500 g of water, if it puts out 2.00 mA at 15.0 kV? Ignore heat transfer to the surroundings.
24: Integrated Concepts
24: Hydroelectric generators (see Figure 5) at Hoover Dam produce a maximum current of [latex]{8.00 \times 10^3 \;\text{A}}[/latex] at 250 kV. (a) What is the power output? (b) The water that powers the generators enters and leaves the system at low speed (thus its kinetic energy does not change) but loses 160 m in altitude. How many cubic meters per second are needed, assuming 85.0% efficiency?
25: Integrated Concepts
(a) Assuming 95.0% efficiency for the conversion of electrical power by the motor, what current must the 12.0-V batteries of a 750-kg electric car be able to supply: (a) To accelerate from rest to 25.0 m/s in 1.00 min? (b) To climb a
[latex]{2.00 \times 10^2 \;\text{-m}}[/latex] -high hill in 2.00 min at a constant 25.0-m/s speed while exerting
[latex]{5.00 \times 10^2 \;\text{N}}[/latex] of force to overcome air resistance and friction? (c) To travel at a constant 25.0-m/s speed, exerting a [latex]{5.00 \times 10^2 \;\text{N}}[/latex] force to overcome air resistance and friction? See Figure 6.
26: Integrated Concepts
A light-rail commuter train draws 630 A of 650-V DC electricity when accelerating. (a) What is its power consumption rate in kilowatts? (b) How long does it take to reach 20.0 m/s starting from rest if its loaded mass is [latex]{5.30 \times 10^4 \;\text{kg}}[/latex], assuming 95.0% efficiency and constant power? (c) Find its average acceleration. (d) Discuss how the acceleration you found for the light-rail train compares to what might be typical for an automobile.
27: Integrated Concepts
(a) An aluminum power transmission line has a resistance of [latex]{0.0580 \;\Omega / \text{km}}[/latex]. What is its mass per kilometer? (b) What is the mass per kilometer of a copper line having the same resistance? A lower resistance would shorten the heating time. Discuss the practical limits to speeding the heating by lowering the resistance.
28: Integrated Concepts
(a) An immersion heater utilizing 120 V can raise the temperature of a [latex]{1.00 \times 10^2 \text{-g}}[/latex] aluminum cup containing 350 g of water from 20.0ºC to 95.0ºC in 2.00 min. Find its resistance, assuming it is constant during the process. (b) A lower resistance would shorten the heating time. Discuss the practical limits to speeding the heating by lowering the resistance.
29: Integrated Concepts
(a) What is the cost of heating a hot tub containing 1500 kg of water from 10.0ºC to 40.0ºC, assuming 75.0% efficiency to account for heat transfer to the surroundings? The cost of electricity is [latex]{9 \;\text{cents/kW} \cdot \text{h}}[/latex]. (b) What current was used by the 220-V AC electric heater, if this took 4.00 h?
30: Unreasonable Results
(a) What current is needed to transmit [latex]{1.00 \times 10^2 \;\text{MW}}[/latex] of power at 480 V? (b) What power is dissipated by the transmission lines if they have a [latex]{1.00 - \Omega}[/latex] resistance? (c) What is unreasonable about this result? (d) Which assumptions are unreasonable, or which premises are inconsistent?
31: Unreasonable Results
(a) What current is needed to transmit [latex]{1.00 \times 10^2 \;\text{MW}}[/latex] of power at 10.0 kV? (b) Find the resistance of 1.00 km of wire that would cause a 0.0100% power loss. (c) What is the diameter of a 1.00-km-long copper wire having this resistance? (d) What is unreasonable about these results? (e) Which assumptions are unreasonable, or which premises are inconsistent?
32: Construct Your Own Problem
Consider an electric immersion heater used to heat a cup of water to make tea. Construct a problem in which you calculate the needed resistance of the heater so that it increases the temperature of the water and cup in a reasonable amount of time. Also calculate the cost of the electrical energy used in your process. Among the things to be considered are the voltage used, the masses and heat capacities involved, heat losses, and the time over which the heating takes place. Your instructor may wish for you to consider a thermal safety switch (perhaps bimetallic) that will halt the process before damaging temperatures are reached in the immersion unit.
Glossary
- electric power
- the rate at which electrical energy is supplied by a source or dissipated by a device; it is the product of current times voltage
Solutions
Problems & Exercise
1: [latex]{2.00 \times 10^{12} \;\text{W}}[/latex]
5: (a) 1.50 W
(b) 7.50 W
7: [latex]{\frac{\textbf{V}^2}{\Omega}}[/latex] [latex]{=}[/latex] [latex]{\frac{\text{V}^2}{\text{V/A}}}[/latex] [latex]{ = \;\text{AV} =}[/latex] [latex]{(\frac{\text{C}}{\text{s}}) \; (\frac{\text{J}}{\text{C}})}[/latex]
9: [latex]{1 \;\text{kW} \cdot \;\text{h} =}[/latex] [latex]{(\frac{1 \times 10^3}{1 \;\text{s}})}[/latex] [latex]{(1 \;\text{h})}[/latex] [latex]{(\frac{3600 \;\text{s}}{1 \;\text{h}})}[/latex] [latex]{ = 3.60 \times 10^6 \;\text{J}}[/latex]
11: $438/y
13: $6.25
15: 1.58 h
17: $3.94 billion/year
19: 25.5 W
21: (a) [latex]{2.00 \times 10^9 \;\text{J}}[/latex]
(b) 769 kg
23: 45.0 s
25: (a) 343 A
(b) [latex]{2.17 \times 10^3 \;\text{A}}[/latex]
(c) [latex]{1.10 \times 10^3 \;\text{A}}[/latex]
27: (a) [latex]{1.23 \times 10^3 \;\text{kg}}[/latex]
(b) [latex]{2.64 \times 10^3 \;\text{kg}}[/latex]
3o: (a) [latex]{2.08 \times 10^5 \;\text{A}}[/latex]
(b) [latex]{4.33 \times 10^4 \;\text{MW}}[/latex]
(c) The transmission lines dissipate more power than they are supposed to transmit.
(d) A voltage of 480 V is unreasonably low for a transmission voltage. Long-distance transmission lines are kept at much higher voltages (often hundreds of kilovolts) to reduce power losses.


Effusion and Diffusion of Gases (8.4)
OpenStax
By the end of this section, you will be able to:
- Define and explain effusion and diffusion
- State Graham’s law and use it to compute relevant gas properties
If you have ever been in a room when a piping hot pizza was delivered, you have been made aware of the fact that gaseous molecules can quickly spread throughout a room, as evidenced by the pleasant aroma that soon reaches your nose. Although gaseous molecules travel at tremendous speeds (hundreds of meters per second), they collide with other gaseous molecules and travel in many different directions before reaching the desired target. At room temperature, a gaseous molecule will experience billions of collisions per second. The mean free path is the average distance a molecule travels between collisions. The mean free path increases with decreasing pressure; in general, the mean free path for a gaseous molecule will be hundreds of times the diameter of the molecule
In general, we know that when a sample of gas is introduced to one part of a closed container, its molecules very quickly disperse throughout the container; this process by which molecules disperse in space in response to differences in concentration is called diffusion (shown in Figure 8.31). The gaseous atoms or molecules are, of course, unaware of any concentration gradient; they simply move randomly—regions of higher concentration have more particles than regions of lower concentrations, and so a net movement of species from high to low concentration areas takes place. In a closed environment, diffusion will ultimately result in equal concentrations of gas throughout, as depicted in Figure 8.31. The gaseous atoms and molecules continue to move, but since their concentrations are the same in both bulbs, the rates of transfer between the bulbs are equal (no net transfer of molecules occurs).
We are often interested in the rate of diffusion, the amount of gas passing through some area per unit of time:
rate of diffusion = [latex]\frac{\text{amount of gas passing through an area}}{\text{unit of time}}[/latex]
The diffusion rate depends on several factors: the concentration gradient (the increase or decrease in concentration from one point to another), the amount of surface area available for diffusion, and the distance the gas particles must travel. Note also that the time required for diffusion to occur is inversely proportional to the rate of diffusion, as shown in the rate of diffusion equation.
A process involving the movement of gaseous species similar to diffusion is effusion, the escape of gas molecules through a tiny hole such as a pinhole in a balloon into a vacuum (Figure 8.32). Although diffusion and effusion rates both depend on the molar mass of the gas involved, their rates are not equal; however, the ratios of their rates are the same.
If a mixture of gases is placed in a container with porous walls, the gases effuse through the small openings in the walls. The lighter gases pass through the small openings more rapidly (at a higher rate) than the heavier ones (Figure 8.33). In 1832, Thomas Graham studied the rates of effusion of different gases and formulated Graham’s law of effusion: The rate of effusion of a gas is inversely proportional to the square root of the mass of its particles:
rate of effusion ∝ [latex]\frac{1}{\sqrt{ℳ}}[/latex]
This means that if two gases A and B are at the same temperature and pressure, the ratio of their effusion rates is inversely proportional to the ratio of the square roots of the masses of their particles:
[latex]\frac{\text{rate of effusion of A}}{\text{rate of effusion of B}} = \frac{\sqrt{ℳ_{B}}}{\sqrt{ℳ_{A}}}[/latex]
EXAMPLE 8.20
Applying Graham’s Law to Rates of Effusion
Calculate the ratio of the rate of effusion of hydrogen to the rate of effusion of oxygen.
Solution
From Graham’s law, we have:
[latex]\frac{\text{rate of effusion of hydrogen}}{\text{rate of effusion of oxygen}} = \frac{\sqrt{32 \cancel{\text{ g mol }^{-1}}}}{\sqrt{2 \cancel{\text{ g mol }^{-1}}}} = \frac{\sqrt{16}}{\sqrt{1}} = \frac{4}{1}[/latex]
Hydrogen effuses four times as rapidly as oxygen.
Check Your Learning
At a particular pressure and temperature, nitrogen gas effuses at the rate of 79 mL/s. Under the same conditions, at what rate will sulfur dioxide effuse?
Answer: 52 mL/s
EXAMPLE 8.21
Effusion Time Calculations
It takes 243 s for 4.46 × 10−5 mol Xe to effuse through a tiny hole. Under the same conditions, how long will it take 4.46 × 10−5 mol Ne to effuse?
Solution
It is important to resist the temptation to use the times directly and to remember how rate relates to time as well as how it relates to mass. Recall the definition of rate of effusion:
rate of effusion = [latex]\frac{\text{amount of gas transferred}}{\text{time}}[/latex]
and combine it with Graham’s law:
[latex]\frac{\text { rate of effusion of gas } \mathrm{Xe}}{\text { rate of effusion of gas}\mathrm{Ne}}=\frac{\sqrt{\mathscr{M}_{\mathrm{Ne}}}}{\sqrt{\mathscr{M}_{\mathrm{Xe}}}}[/latex]
to get:
[latex]\frac{\frac{\text { amount of } \mathrm{Xe} \text { transfered }}{\text { time for } \mathrm{Xe}}}{\frac{\text { amount of Ne transfered }}{\text { time for } \mathrm{Ne}}}=\frac{\sqrt{\mathscr{M}_{\mathrm{Ne}}}}{\sqrt{\mathscr{M}_{\mathrm{Xe}}}}[/latex]
Noting that amount of A = amount of B and solving for time for Ne:
[latex]\frac{\frac{\cancel{\text{ amount of } \mathrm{Xe}}}{\text { time for } \mathrm{Xe}}}{\frac{\cancel{\text { amount of Ne }}}{\text { time for } \mathrm{Ne}}} = \frac{\text { time for } \mathrm{Ne}}{\text { time for } \mathrm{Xe}}=\frac{\sqrt{\mathscr{M}_{\mathrm{Ne}}}}{\sqrt{\mathscr{M}_{\mathrm{Xe}}}}=\frac{\sqrt{\mathscr{M}_{\mathrm{Ne}}}}{\sqrt{\mathscr{M}_{\mathrm{Xe}}}}[/latex]
and substitute values:
[latex]\frac{\text { time for } \mathrm{Ne}}{243 \mathrm{s}}=\sqrt{\frac{20.2\cancel{\text{ g mol }}}{131.3\cancel{\text{ g mol }}}} = 0.392[/latex]
Finally, solve for the desired quantity:
time for Ne = 0.392 x 243 s = 95.3 s
Note that this answer is reasonable: Since Ne is lighter than Xe, the effusion rate for Ne will be larger than that for Xe, which means the time of effusion for Ne will be smaller than that for Xe.
Check Your Learning
A party balloon filled with helium deflates to [latex]\frac{2}{3}[/latex] of its original volume in 8.0 hours. How long will it take an identical balloon filled with the same number of moles of air (ℳ = 28.2 g/mol) to deflate to [latex]\frac{1}{2}[/latex] of its original volume?
Answer: 32 h
EXAMPLE 8.22
Determining Molar Mass Using Graham’s Law
An unknown gas effuses 1.66 times more rapidly than CO2. What is the molar mass of the unknown gas? Can you make a reasonable guess as to its identity?
Solution
From Graham’s law, we have:
[latex]\frac{\text { rate of effusion of Unknown }}{\text { rate of effusion of } \mathrm{CO}_{2}}=\frac{\sqrt{\mathscr{M}_{\mathrm{CO}_{2}}}}{\sqrt{\mathscr{M}_{Unknown}}}[/latex]
Plug in known data:
[latex]\frac{1.66}{1}=\frac{\sqrt{44.0 \mathrm{g} / \mathrm{mol}}}{\sqrt{\mathscr{M}_{Unknown}}}[/latex]
Solve:
[latex]\mathscr{M}_{U n k n o w n}=\frac{44.0 \mathrm{g} / \mathrm{mol}}{(1.66)^{2}}=16.0 \mathrm{g} / \mathrm{mol}[/latex]
The gas could well be CH4, the only gas with this molar mass.
Check Your Learning
Hydrogen gas effuses through a porous container 8.97 times faster than an unknown gas. Estimate the molar mass of the unknown gas.
Answer: 163 g/mol
How Sciences Interconnect
Use of Diffusion for Nuclear Energy Applications: Uranium Enrichment
Gaseous diffusion has been used to produce enriched uranium for use in nuclear power plants and weapons. Naturally occurring uranium contains only 0.72% of 235U, the kind of uranium that is “fissile,” that is, capable of sustaining a nuclear fission chain reaction. Nuclear reactors require fuel that is 2–5% 235U, and nuclear bombs need even higher concentrations. One way to enrich uranium to the desired levels is to take advantage of Graham’s law. In a gaseous diffusion enrichment plant, uranium hexafluoride (UF6, the only uranium compound that is volatile enough to work) is slowly pumped through large cylindrical vessels called diffusers, which contain porous barriers with microscopic openings. The process is one of diffusion because the other side of the barrier is not evacuated. The 235UF6 molecules have a higher average speed and diffuse through the barrier a little faster than the heavier 238UF6 molecules. The gas that has passed through the barrier is slightly enriched in 235UF6, and the residual gas is slightly depleted. The small difference in molecular weights between 235UF6 and 238UF6 only about 0.4% enrichment, is achieved in one diffuser (Figure 8.34). But by connecting many diffusers in a sequence of stages (called a cascade), the desired level of enrichment can be attained.
The large-scale separation of gaseous 235UF6 from 238UF6 was first done during World War II, at the atomic energy installation in Oak Ridge, Tennessee, as part of the Manhattan Project (the development of the first atomic bomb). Although the theory is simple, this required surmounting many daunting technical challenges to make it work in practice. The barrier must have tiny, uniform holes (about 10–6 cm in diameter) and be porous enough to produce high flow rates. All materials (the barrier, tubing, surface coatings, lubricants, and gaskets) need to be able to contain, but not react with, the highly reactive and corrosive UF6. Because gaseous diffusion plants require very large amounts of energy (to compress the gas to the high pressures required and drive it through the diffuser cascade, to remove the heat produced during compression, and so on), it is now being replaced by gas centrifuge technology, which requires far less energy. A current hot political issue is how to deny this technology to Iran, to prevent it from producing enough enriched uranium for them to use to make nuclear weapons.


197 Dissociative Fugue (300.13)
Dissociative Fugue is characterized by sudden, unplanned trips from the home or workplace without the ability to remember some or all of the individual’s past. Some of these patients take on new characteristics or aspects not related to their original identity. They tend to be running away from something of which they are unaware. After a fugue episode resolves, patients are unable to remember the events of the state. Although moving occurs in other disorders, in fugue it is purposeful and it is not enacted in a confused or dazed state. In a typical case, the fugue is brief, with purposeful travel, and limited contact with others. About 0.2% of the general population is afflicted with this type of dissociative disorder. (American Psychiatric Association, 2000).
Mental Status
- The mental status exam varies widely.
- Patient may present alert and oriented only to oneself.
- Eye contact and relatedness are limited to fair at best.
- Psychomotor activity is characterized by normal activity.
- Thought processes are intact, although thought content may vary widely from preoccupations to preservations to obsessive fixations to none.
- Reasoning and judgment are lacking, and insight is poor.
- An increased finding of violent or homocidal ideation is present, but suicidal ideation is lacking.
The DSM-IV-TR criteria according to the American Psychiatric Association (2000) includes the following:
- A. “The predominant disturbance is sudden, unexpected travel away from home or one’s customary place of work, with inability to recall one’s past” (p. 526).
- B. “Confusion about personal identity or assumption of a new identity (partial or complete)” (p. 526).
- C. “The disturbance does not occur exclusively during the course of Dissociative Identity Disorder and is not due to the direct physiological effects of a substance (e.g., a drug of abuse, a medication) or a general medical condition (e.g.., temporal lobe epilepsy)” (p. 526).
- D. “The symptoms cause clinically significant distress or impairment in social, occupational, or other important areas of functioning” (p. 526).
Associated features
Dissociative Fugue was formerly known as Psychogenic Fugue. It is comorbid with Bipolar Disorder, Major Depressive Disorder, and Schizophrenia, as well as PTSD, Substance Related disorders, Panic and Anxiety Disorders, Eating Disorders, and Somatoform Disorders. Note: Dissociative Fugue is often mistaken for malingering. This happens because the disorder enables people to escape their responsibilities or undesirable or dangerous situations; therefore it is seen as if a person is taking the ‘easy-way-out’. A person in the midst of a Dissociative Fugue episode may appear only slightly confused or they may appear to have no symptoms at all and attract no attention. Eventually, however, the person will begin to show significant signs of confusion or distress as they become aware of memory loss or confusion about their identity. This amnesia is characteristic of the disorder. When the fugue ends, the person may experience depression, grief, shame, guilt, dysphoria, psychological stress, conflict, and suicidal and aggressive impulses, and he may give approximate, though inaccurate, answers to questions. (American Psychiatric Association, 2000)
Child vs. adult presentation
Dissociative Fugue usually begins in adulthood. There is little information about the presentation of this disorder in children. When it does affect children, it is most commonly due to severe trauma such as sexual abuse, but even then it does not usually present until adulthood.
Gender and cultural differences in presentation
- Some research revealed that this condition most often occurs in females, but the reason is unknown. One source stated that females are at a rate six to nine times higher than males, and it increases as age increases. This pattern is most likely associated with the stresses on a woman to be both mother and a family provider and caretaker, in conjunction with the societal pressures and gender prejudices. Most studies however, believed that Dissociative Fugue is equally prevalent across genders.
- There is little information on the cultural differences in presentation of Dissociative Fugue. It is important to remember that what may be considered dissociative in one culture, may be seen as normal in another. Cultures prone to warfare are more likely to experience the distressing pressures of war, which is a common causal traumatic event of this disorder. Various cultures with defined “running” syndrome may have symptoms that meet diagnostic criteria for Dissociative Fugue, such as the amok in Western Pacific cultures.
Epidemiology
- This is a relatively rare disorder, actually the rarest of the dissociative disorders, affecting about only 2 in 1000 people in the United States. The prevalence rate is estimated at 0.2%. It is much more common however among people who have been in wars, accidents, natural disasters, or other highly traumatic or stressful events. (American Psychiatric Association, 2000)
- Single episodes are most commonly reported and vary in duration. Recovery is usually rapid, but Dissociative Amnesia may sometimes be present. (American Psychiatric Association, 2000)
Etiology
Episodes of Dissociative Fugue are usually triggered by traumatic, overwhelming, stressful events. Traumatic experiences such as war, natural disasters, accidents, and sexual abuse during childhood, often increase the incidence of the disorder. More personal types of stress, like the shocking death of a loved one or unbearable pressures at work or home, might also lead to the unplanned travel and amnesia that is characteristic of Dissociative Fugue. (American Psychiatric Association, 2000)
Differential diagnosis
Dissociative fugue includes other dissociative disorders, seizure disorder, amnestic disorder, schizophrenia, mania, dementia (often of the Alzheimer type), malingering, frontal lobe disorders, head trauma and injury, and factitious disorder. Fugue differs from other mental disorders in that the flight behavior is organized and purposeful. Patients with seizure disorder do not assume a new identity and usually have an altered state of conscious with abnormal findings on electroencephalogram testing. (American Psychiatric Association, 2000)
Indications for hospitalization
In making a primary diagnosis, observing the patient in a controlled setting is often necessary. Patients reveal their level of need through interactions with others, inappropriate behavior without remorse, or by verbalizing their symptoms when they are aware of their suffering. In general, hospitalization is indicated when medical or surgical treatment is required, when the diagnosis is unclear, when no safe alternative exist for housing the patient, or as a means of stopping the ongoing abuse. Additionally, any time a patient experiences severe confusion regarding his or her identity or chronic amnesia regarding the total fugue episode, hospitalization if indicated. Hospitalization is also a tool for assessing and administering social services and medication, developing behavior, and ensuring that a patient will respond to medication under the safety and care of medical professionals. And, hospitalization provides containment. Most patients with dissociative fugue symptoms receive acute treatment in general hospital settings and psychiatric departments because they have a tendency to be brought in during an episode. In this way, the hospital provides the safety and treatment mechanism needed for a disorder that, without intervention, remains undiagnosed. Hospitalization most often occurs in order to provide emergency crisis treatment that is best provided in an acute are setting.
Empirically supported treatments
- Most fugues last for only hours or days, and then often disappear on their own. The goal of treatment is to assist the person to come to terms with the trauma or stress that triggered the fugue in the first place. Another goal of treatment is to help develop new coping methods to prevent further fugue episodes. As with most disorders, the particular treatment approach depends on the individual and the severity of his or her symptoms. The most likely treatment however will include a combination of psychotherapy, cognitive therapy, medication, family therapy, creative therapy, and clinical hypnosis.
- Psychotherapy
- is the main treatment for dissociative disorders such as Dissociative Fugue. Such treatments aim to increase insight into problems.
- Cognitive therapy
- focuses on changing dysfunctional thinking patterns.
- Medication
- is useful when the person also suffers from depression or anxiety.
- Family therapy
- aims to teach the family more about the disorder and learn about the symptoms of recurrence.
- Creative therapies
- such as music therapy and art therapy, let the person express themselves in safe manners.
- Clinical hypnosis
- uses intense relaxation, concentration, and focuses attention to achieve an altered state of awareness. This is risky however because of the risk of creating false memories. The prognosis for Dissociative Fugue is often very good because the episodes do not usually last longer than a few months and people generally recover quickly. Efforts to restore the memories of what happened during the fugue are usually unsuccessful, or take a long time to be recovered.
Illustrative case
A case study was reported in Psychology Today (Drawing a Blank, October 2007) and was also reported in Maclean’s Magazine (The Man Who Lost Himself, May 2007) about a man named Jeff Ingram. A short summary of this case goes as follows: Ingram, 40, is a former mill worker in Olympia, Washington. He left his home one morning headed for Alberta to visit a terminally ill friend. A few days later he woke up on a street in Denver with no idea of who he was. Ingram became confused, angry, and worried when he was being questioned by the hospital’s receptionist because he had no knowledge of his identity. Even months after being reunited with his family, Ingram still had no pre-fugue memories, including that of his three year relationship with then-fiancée. In order to prevent such confusion in the future, Ingram ordered GPS shoes and had his identity information tattooed on him. He also wears a zip disk with medical information around his neck. It is believed that the possible trigger of Ingram’s fugue episode was the stress of his friend’s battle of cancer. A more detailed article can be found in Maclean’s magazine (May 2007). Here is a news video clip on this case.
- Visit msnbc.com for Breaking News, World News, and News about the Economy
Proposed Diagnostic and Statistical Manual of Mental Disorders, Fifth Edition (DSM-V)
Dissociative Fugue subtype:
- 1. Amnesia includes lack of control over how much of a person’s past they can recall, confusion about their individual identity, or assumption of a new identity (partial or complete)
- 2. Sudden, unplanned journey away from home or a job.


Heat and Heat Transfer Methods
101 Temperature Change and Heat Capacity
Learning Objectives
- Observe heat transfer and change in temperature and mass.
- Calculate final temperature after heat transfer between two objects.
One of the major effects of heat transfer is temperature change: heating increases the temperature while cooling decreases it. We assume that there is no phase change and that no work is done on or by the system. Experiments show that the transferred heat depends on three factors—the change in temperature, the mass of the system, and the substance and phase of the substance.
The dependence on temperature change and mass are easily understood. Owing to the fact that the (average) kinetic energy of an atom or molecule is proportional to the absolute temperature, the internal energy of a system is proportional to the absolute temperature and the number of atoms or molecules. Owing to the fact that the transferred heat is equal to the change in the internal energy, the heat is proportional to the mass of the substance and the temperature change. The transferred heat also depends on the substance so that, for example, the heat necessary to raise the temperature is less for alcohol than for water. For the same substance, the transferred heat also depends on the phase (gas, liquid, or solid).
Heat Transfer and Temperature Change
The quantitative relationship between heat transfer and temperature change contains all three factors:
where [latex]Q[/latex] is the symbol for heat transfer, [latex]m[/latex] is the mass of the substance, and [latex]\text{Δ}T[/latex] is the change in temperature. The symbol [latex]c[/latex] stands for specific heat and depends on the material and phase. The specific heat is the amount of heat necessary to change the temperature of 1.00 kg of mass by [latex]1\text{.00ºC}[/latex]. The specific heat [latex]c[/latex] is a property of the substance; its SI unit is [latex]\text{J/}\left(\text{kg}\cdot \text{K}\right)[/latex] or [latex]\text{J/}\left(\text{kg}\cdot ºC\right).[/latex] Recall that the temperature change [latex]\left(\text{Δ}T\right)[/latex] is the same in units of kelvin and degrees Celsius. If heat transfer is measured in kilocalories, then the unit of specific heat is [latex]\text{kcal/}\left(\text{kg}\cdot ºC\right).[/latex]
Values of specific heat must generally be looked up in tables, because there is no simple way to calculate them. In general, the specific heat also depends on the temperature. Table 101.1 lists representative values of specific heat for various substances. Except for gases, the temperature and volume dependence of the specific heat of most substances is weak. We see from this table that the specific heat of water is five times that of glass and ten times that of iron, which means that it takes five times as much heat to raise the temperature of water the same amount as for glass and ten times as much heat to raise the temperature of water as for iron. In fact, water has one of the largest specific heats of any material, which is important for sustaining life on Earth.
Example 101.1 Calculating the Required Heat: Heating Water in an Aluminum Pan
A 0.500 kg aluminum pan on a stove is used to heat 0.250 liters of water from [latex]\text{20.0ºC}[/latex] to [latex]\text{80.0ºC}[/latex]. (a) How much heat is required? What percentage of the heat is used to raise the temperature of (b) the pan and (c) the water?
Strategy
The pan and the water are always at the same temperature. When you put the pan on the stove, the temperature of the water and the pan is increased by the same amount. We use the equation for the heat transfer for the given temperature change and mass of water and aluminum. The specific heat values for water and aluminum are given in Table 101.1.
Solution
Because water is in thermal contact with the aluminum, the pan and the water are at the same temperature.
- Calculate the temperature difference:
[latex]\text{Δ}T={T}_{\text{f}}-{T}_{\text{i}}=\text{60.0ºC.}[/latex]
- Calculate the mass of water. Because the density of water is [latex]\text{1000}\phantom{\rule{0.25em}{0ex}}\phantom{\rule{0.25em}{0ex}}{\text{kg/m}}^{3}[/latex], one liter of water has a mass of 1 kg, and the mass of 0.250 liters of water is [latex]{m}_{w}=0\text{.}\text{250}\phantom{\rule{0.25em}{0ex}}\text{kg}[/latex].
- Calculate the heat transferred to the water. Use the specific heat of water in Table 101.1:
[latex]{Q}_{w}={m}_{w}{c}_{w}\text{Δ}T=\left(0\text{.}\text{250}\phantom{\rule{0.25em}{0ex}}\text{kg}\right)\left(\text{4186}\phantom{\rule{0.25em}{0ex}}\text{J/kg}\text{ºC}\right)\left(\text{60.0}\text{ºC}\right)=\text{62}.8 kJ.[/latex]
- Calculate the heat transferred to the aluminum. Use the specific heat for aluminum in Table 101.1:
[latex]{Q}_{\text{Al}}={m}_{\text{Al}}{c}_{\text{Al}}\text{Δ}T=\left(\text{0.500 kg}\right)\left(\text{900 J/kgºC}\right)\left(\text{60.0ºC}\right){\text{= 27.0 × 10}}^{4}\text{J = 27.0 kJ.}[/latex]
- Compare the percentage of heat going into the pan versus that going into the water. First, find the total transferred heat:
[latex]{Q}_{\text{Total}}={Q}_{\text{W}}+{Q}_{\text{Al}}=\text{62}\text{.}8\phantom{\rule{0.25em}{0ex}}\text{kJ}+\text{27}\text{.}0\phantom{\rule{0.25em}{0ex}}\phantom{\rule{0.25em}{0ex}}\text{kJ = 89}\text{.}8\phantom{\rule{0.25em}{0ex}}\text{kJ.}[/latex]
Thus, the amount of heat going into heating the pan is
and the amount going into heating the water is
Discussion
In this example, the heat transferred to the container is a significant fraction of the total transferred heat. Although the mass of the pan is twice that of the water, the specific heat of water is over four times greater than that of aluminum. Therefore, it takes a bit more than twice the heat to achieve the given temperature change for the water as compared to the aluminum pan.
Example 101.2 Calculating the Temperature Increase from the Work Done on a Substance: Truck Brakes Overheat on Downhill Runs
Truck brakes used to control speed on a downhill run do work, converting gravitational potential energy into increased internal energy (higher temperature) of the brake material. This conversion prevents the gravitational potential energy from being converted into kinetic energy of the truck. The problem is that the mass of the truck is large compared with that of the brake material absorbing the energy, and the temperature increase may occur too fast for sufficient heat to transfer from the brakes to the environment.
Calculate the temperature increase of 100 kg of brake material with an average specific heat of [latex]\text{800 J/kg}\cdot \text{ºC}[/latex] if the material retains 10% of the energy from a 10,000-kg truck descending 75.0 m (in vertical displacement) at a constant speed.
Strategy
If the brakes are not applied, gravitational potential energy is converted into kinetic energy. When brakes are applied, gravitational potential energy is converted into internal energy of the brake material. We first calculate the gravitational potential energy [latex]\left(\text{Mgh}\right)[/latex] that the entire truck loses in its descent and then find the temperature increase produced in the brake material alone.
Solution
- Calculate the change in gravitational potential energy as the truck goes downhill
[latex]\text{Mgh}=\left(\text{10,000 kg}\right)\left(9\text{.}\text{80 m}{\text{/s}}^{2}\right)\left(\text{75.0 m}\right)=\text{7.}\text{35}×{\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}\text{J.}[/latex]
- Calculate the temperature from the heat transferred using [latex]Q\text{=}\text{Mgh}[/latex] and
[latex]\text{Δ}T=\frac{Q}{\text{mc}}\text{,}[/latex]
where [latex]m[/latex] is the mass of the brake material. Insert the values [latex]m=\text{100 kg}[/latex] and [latex]c=\text{800 J/kg}\cdot \text{ºC}[/latex] to find
[latex]\text{Δ}T=\frac{\left(7\text{.35}×{\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}\text{J}\right)}{\left(\text{100 kg}\right)\left(\text{800 J/kgºC}\right)}=\text{92ºC.}[/latex]
Discussion
This temperature is close to the boiling point of water. If the truck had been traveling for some time, then just before the descent, the brake temperature would likely be higher than the ambient temperature. The temperature increase in the descent would likely raise the temperature of the brake material above the boiling point of water, so this technique is not practical. However, the same idea underlies the recent hybrid technology of cars, where mechanical energy (gravitational potential energy) is converted by the brakes into electrical energy (battery).
| Substances | Specific heat (c) | |
|---|---|---|
| Solids | J/kg⋅ºC | kcal/kg⋅ºC2 |
| Aluminum | 900 | 0.215 |
| Asbestos | 800 | 0.19 |
| Concrete, granite (average) | 840 | 0.20 |
| Copper | 387 | 0.0924 |
| Glass | 840 | 0.20 |
| Gold | 129 | 0.0308 |
| Human body (average at 37 °C) | 3500 | 0.83 |
| Ice (average, -50°C to 0°C) | 2090 | 0.50 |
| Iron, steel | 452 | 0.108 |
| Lead | 128 | 0.0305 |
| Silver | 235 | 0.0562 |
| Wood | 1700 | 0.4 |
| Liquids | ||
| Benzene | 1740 | 0.415 |
| Ethanol | 2450 | 0.586 |
| Glycerin | 2410 | 0.576 |
| Mercury | 139 | 0.0333 |
| Water (15.0 °C) | 4186 | 1.000 |
| Gases3 | ||
| Air (dry) | 721 (1015) | 0.172 (0.242) |
| Ammonia | 1670 (2190) | 0.399 (0.523) |
| Carbon dioxide | 638 (833) | 0.152 (0.199) |
| Nitrogen | 739 (1040) | 0.177 (0.248) |
| Oxygen | 651 (913) | 0.156 (0.218) |
| Steam (100°C) | 1520 (2020) | 0.363 (0.482) |
Note that Example 101.2 is an illustration of the mechanical equivalent of heat. Alternatively, the temperature increase could be produced by a blow torch instead of mechanically.
Example 101.3 Calculating the Final Temperature When Heat Is Transferred Between Two Bodies: Pouring Cold Water in a Hot Pan
Suppose you pour 0.250 kg of [latex]\text{20}\text{.0ºC}[/latex] water (about a cup) into a 0.500-kg aluminum pan off the stove with a temperature of [latex]\text{150ºC}[/latex]. Assume that the pan is placed on an insulated pad and that a negligible amount of water boils off. What is the temperature when the water and pan reach thermal equilibrium a short time later?
Strategy
The pan is placed on an insulated pad so that little heat transfer occurs with the surroundings. Originally the pan and water are not in thermal equilibrium: the pan is at a higher temperature than the water. Heat transfer then restores thermal equilibrium once the water and pan are in contact. Because heat transfer between the pan and water takes place rapidly, the mass of evaporated water is negligible and the magnitude of the heat lost by the pan is equal to the heat gained by the water. The exchange of heat stops once a thermal equilibrium between the pan and the water is achieved. The heat exchange can be written as [latex]\mid {Q}_{\text{hot}}\mid ={Q}_{\text{cold}}[/latex].
Solution
- Use the equation for heat transfer [latex]Q=\text{mc}\text{Δ}T[/latex] to express the heat lost by the aluminum pan in terms of the mass of the pan, the specific heat of aluminum, the initial temperature of the pan, and the final temperature:
[latex]{Q}_{\text{hot}}={m}_{\text{Al}}{c}_{\text{Al}}\left({T}_{\text{f}}-\text{150ºC}\right)\text{.}[/latex]
- Express the heat gained by the water in terms of the mass of the water, the specific heat of water, the initial temperature of the water and the final temperature:
[latex]{Q}_{\text{cold}}={m}_{W}{c}_{W}\left({T}_{\text{f}}-\text{20.0ºC}\right)\text{.}[/latex]
- Note that [latex]{Q}_{\text{hot}}0[/latex] and [latex]{Q}_{\text{cold}}>0[/latex] and that they must sum to zero because the heat lost by the hot pan must be the same as the heat gained by the cold water:
[latex]\begin{array}{lll}\hfill {Q}_{\text{cold}}\text{+}{Q}_{\text{hot}}& \text{=}& \text{0,}\\ \hfill {Q}_{\text{cold}}& =& {\text{–Q}}_{\text{hot}},\\ {m}_{W}{c}_{W}\left({T}_{\text{f}}-\text{20.0ºC}\right)& =& {\mathrm{-m}}_{\mathrm{Al}}{c}_{\mathrm{Al}}\left({T}_{\text{f}}-\text{150ºC.}\right)\end{array}[/latex]
- This an equation for the unknown final temperature, [latex]{T}_{\text{f}}[/latex]
- Bring all terms involving [latex]{T}_{\text{f}}[/latex] on the left hand side and all other terms on the right hand side. Solve for [latex]{T}_{\text{f}}[/latex],
[latex]{T}_{\text{f}}=\frac{{m}_{\text{Al}}{c}_{\text{Al}}\left(\text{150ºC}\right)+{m}_{W}{c}_{W}\left(\text{20}\text{.0ºC}\right)}{{m}_{\text{Al}}{c}_{\text{Al}}+{m}_{W}{c}_{W}}\text{,}[/latex]
and insert the numerical values:
[latex]\begin{array}{lll}{T}_{\text{f}}& =& \frac{\left(\text{0.500 kg}\right)\left(\text{900 J/kgºC}\right)\left(\text{150ºC}\right)\text{+}\left(\text{0.250 kg}\right)\left(\text{4186 J/kgºC}\right)\left(\text{20.0ºC}\right)}{\left(\text{0.500 kg}\right)\left(\text{900 J/kgºC}\right)+\left(\text{0.250 kg}\right)\left(\text{4186 J/kgºC}\right)}\\ & =& \frac{\text{88430 J}}{\text{1496.5 J/ºC}}\\ & =& \text{59}\text{.1ºC.}\end{array}[/latex]
Discussion
This is a typical calorimetry problem—two bodies at different temperatures are brought in contact with each other and exchange heat until a common temperature is reached. Why is the final temperature so much closer to [latex]\text{20.0ºC}[/latex] than [latex]\text{150ºC}[/latex]? The reason is that water has a greater specific heat than most common substances and thus undergoes a small temperature change for a given heat transfer. A large body of water, such as a lake, requires a large amount of heat to increase its temperature appreciably. This explains why the temperature of a lake stays relatively constant during a day even when the temperature change of the air is large. However, the water temperature does change over longer times (e.g., summer to winter).
Take-Home Experiment: Temperature Change of Land and Water
What heats faster, land or water?
To study differences in heat capacity:
- Place equal masses of dry sand (or soil) and water at the same temperature into two small jars. (The average density of soil or sand is about 1.6 times that of water, so you can achieve approximately equal masses by using [latex]\text{50%}[/latex] more water by volume.)
- Heat both (using an oven or a heat lamp) for the same amount of time.
- Record the final temperature of the two masses.
- Now bring both jars to the same temperature by heating for a longer period of time.
- Remove the jars from the heat source and measure their temperature every 5 minutes for about 30 minutes.
Which sample cools off the fastest? This activity replicates the phenomena responsible for land breezes and sea breezes.
Check Your Understanding
If 25 kJ is necessary to raise the temperature of a block from [latex]\text{25ºC}[/latex] to [latex]\text{30ºC}[/latex], how much heat is necessary to heat the block from [latex]\text{45ºC}[/latex] to [latex]\text{50ºC}[/latex]?
The heat transfer depends only on the temperature difference. Since the temperature differences are the same in both cases, the same 25 kJ is necessary in the second case.
Summary
- The transfer of heat [latex]Q[/latex] that leads to a change [latex]\text{Δ}T[/latex] in the temperature of a body with mass [latex]m[/latex] is [latex]Q=\text{mc}\text{Δ}T[/latex], where [latex]c[/latex] is the specific heat of the material. This relationship can also be considered as the definition of specific heat.
Conceptual Questions
- What three factors affect the heat transfer that is necessary to change an object’s temperature?
- The brakes in a car increase in temperature by [latex]\text{Δ}T[/latex] when bringing the car to rest from a speed [latex]v[/latex]. How much greater would [latex]\text{Δ}T[/latex] be if the car initially had twice the speed? You may assume the car to stop sufficiently fast so that no heat transfers out of the brakes.
Problems & Exercises
- On a hot day, the temperature of an 80,000-L swimming pool increases by [latex]1\text{.}\text{50ºC}[/latex]. What is the net heat transfer during this heating? Ignore any complications, such as loss of water by evaporation.
- Show that [latex]1\phantom{\rule{0.25em}{0ex}}\text{cal/g}\cdot \text{ºC}=1\phantom{\rule{0.25em}{0ex}}\text{kcal/kg}\cdot \text{ºC}[/latex].
- To sterilize a 50.0-g glass baby bottle, we must raise its temperature from [latex]\text{22}\text{.}0\text{ºC}[/latex] to [latex]\text{95}\text{.}0º\text{C}[/latex]. How much heat transfer is required?
- The same heat transfer into identical masses of different substances produces different temperature changes. Calculate the final temperature when 1.00 kcal of heat transfers into 1.00 kg of the following, originally at [latex]\text{20}\text{.}0º\text{C}[/latex]: (a) water; (b) concrete; (c) steel; and (d) mercury.
- Rubbing your hands together warms them by converting work into thermal energy. If a woman rubs her hands back and forth for a total of 20 rubs, at a distance of 7.50 cm per rub, and with an average frictional force of 40.0 N, what is the temperature increase? The mass of tissues warmed is only 0.100 kg, mostly in the palms and fingers.
- A 0.250-kg block of a pure material is heated from [latex]\text{20}\text{.}0º\text{C}[/latex] to [latex]\text{65}\text{.}0º\text{C}[/latex] by the addition of 4.35 kJ of energy. Calculate its specific heat and identify the substance of which it is most likely composed.
- Suppose identical amounts of heat transfer into different masses of copper and water, causing identical changes in temperature. What is the ratio of the mass of copper to water?
- (a) The number of kilocalories in food is determined by calorimetry techniques in which the food is burned and the amount of heat transfer is measured. How many kilocalories per gram are there in a 5.00-g peanut if the energy from burning it is transferred to 0.500 kg of water held in a 0.100-kg aluminum cup, causing a [latex]\text{54}\text{.}9º\text{C}[/latex] temperature increase? (b) Compare your answer to labeling information found on a package of peanuts and comment on whether the values are consistent.
- Following vigorous exercise, the body temperature of an 80.0-kg person is [latex]\text{40}\text{.}0º\text{C}[/latex]. At what rate in watts must the person transfer thermal energy to reduce the the body temperature to [latex]\text{37}\text{.}0º\text{C}[/latex] in 30.0 min, assuming the body continues to produce energy at the rate of 150 W? [latex]\left(\text{1 watt = 1 joule/second or 1 W = 1 J/s}\phantom{\rule{0.25em}{0ex}}\right)[/latex].
- Even when shut down after a period of normal use, a large commercial nuclear reactor transfers thermal energy at the rate of 150 MW by the radioactive decay of fission products. This heat transfer causes a rapid increase in temperature if the cooling system fails [latex]\left(\text{1 watt = 1 joule/second or 1 W = 1 J/s and 1 MW = 1 megawatt}\right)[/latex]. (a) Calculate the rate of temperature increase in degrees Celsius per second
- [latex]\left(\text{ºC/s}\right)[/latex] if the mass of the reactor core is [latex]1\text{.}\text{60}×{\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{kg}[/latex] and it has an average specific heat of [latex]\text{0.3349 kJ/kgº}\cdot \text{C}[/latex]. (b) How long would it take to obtain a temperature increase of [latex]\text{2000º}\text{C}[/latex], which could cause some metals holding the radioactive materials to melt? (The initial rate of temperature increase would be greater than that calculated here because the heat transfer is concentrated in a smaller mass. Later, however, the temperature increase would slow down because the [latex]5×{\text{10}}^{5}\text{-kg}[/latex] steel containment vessel would also begin to heat up.)
Footnotes
1 The values for solids and liquids are at constant volume and at [latex]\text{25ºC}[/latex], except as noted.
2 These values are identical in units of [latex]\text{cal/g}\cdot ºC[/latex].
3[latex]{c}_{\text{v}}[/latex] at constant volume and at [latex]\text{20}\text{.}0ºC[/latex], except as noted, and at 1.00 atm average pressure. Values in parentheses are [latex]{c}_{\text{p}}[/latex] at a constant pressure of 1.00 atm.
Glossary
- specific heat
- the amount of heat necessary to change the temperature of 1.00 kg of a substance by 1.00 ºC


1.4 Managing Information for Online Learning
As you develop your identity as an online learner, you will want to consider the role of information management in your learning process. Strong independent learners actively read, evaluate, and use information for current course tasks, but more importantly, to develop a resource file of information that will support professional growth.
Many online learners feel overwhelmed at the volume of reading and the wealth of online resources available to them. Managing information well requires you to develop skills in identifying the purpose for your reading (What do you need to learn from this reading? Why is it important to you personally and professionally?), and the strategy that will help you achieve your purpose (skimming, reading key portions, a close reading). In the Strategic Reading module of this workshop, you will discover additional ways to manage the reading process.
Early in your learning journey, you will want to select a system for managing information. As you will frequently work with electronic texts and articles, selecting a system that allows you to store, search, and retrieve readings and notes from current and past courses. Online note systems, such as OneNote or Evernote, are highly effective for this purpose.
Benefits of Using an Online Notebook
Online notebook platforms allow you to do the following:
- Create individual notebooks for courses or assignments, creating additional sections or pages for weekly module topics
- Create flexible notes that include text, images, audio, and video files
- Store PDFs and other course documents for your own records (note: you will eventually lose access to the college’s LMS when courses are complete)
- Clip and store relevant information from the web
- Search stored content to quickly find relevant information, making connections between courses and your previous learning
- Sync notes across all of your devices
Choose a Notebook
Two fully-featured and common software platforms are Onenote and Evernote
| Advantages | Disadvantages | |
| OneNote |
|
|
| Evernote |
|
|
To evaluate which platform best serves your needs, you may wish to investigate by trying out each on your preferred devices. You may also wish to explore the following linked video tutorials for further information, and to begin developing your organizational system.
OneNote:
Evernote:
Comparison of Notebook Apps
Try it!
After investigating and installing your preferred digital notebook, create a notebook for each of your semester courses. Save your course presentation as a file in each of the notebooks.


15 What is Anti-Racist Pedagogy?
Now that we have explored what it means to be anti-racist, we will explore anti-racist pedagogy, a theoretical framework we will apply to our classrooms and instruction. Anti-racist pedagogy can be applied to any academic discipline as a way to explain and address the pervasiveness of racism. This framework calls on actively acknowledging and opposing racism by leveraging tools and strategies to design inclusive courses with more representation.
Just as being anti-racist is an iterative process, so too is the process of employing anti-racist pedagogy. It requires:
- Thoughtful consideration of authorial voices in learning materials,
- Conscientiously realigning the power structure within the learning environment to include learners as active participants in the learning process,
- Engagement in discussion to actively combat racism in our fields,
- Constant evaluation of the course design.
“Anti-racist pedagogy is not about simply incorporating racial content into courses, curriculum, and discipline. It is also about how one teaches, even in courses where race is not the subject matter.” (Kishimoto, 2018, p.540). Anti-racist pedagogy requires students and instructors to engage in critical analysis and self-reflection to reveal structural inequities.
How do we apply anti-racist pedagogy to all disciplines?
- We can do so by sharing with students how various disciplines were shaped by systemic racism, for example, disparities in medical care or contributions to the field made by BIPOC.
- Giving voice to and sharing the perspective of voices most often left out; teaching about events from the perspective of the oppressed, not just the oppressor (for example, the Tulsa Race Riots vs. Tulsa Race Massacre).
- Education is an institution of society shaped by racial disparity. Everything from lack of representation in research, health, and clinical trials, to inequitable access to quality education, we must acknowledge that while our educational systems were not built for all students, we have the opportunity to actively build our courses to be inclusive.
- Engaging in discussion with our students about how our disciplines were shaped by racism.
Becoming an Anti-Racist Educator
The Anti-Racist Discussion Pedagogy guide by Dr. Selfa Chew, Dr. Akil Houston, and Dr. Alisa Cooper recommends the work of an anti-racist instructor begin with oneself, and shares a number self-reflective questions to consider:
- Am I truly ready to do this work, even when it becomes uncomfortable?
- Talking about race can be uncomfortable, but we must lean in to the discomfort to have difficult conversations. Our discomfort also informs what conversations and discussions we are ready to engage with our students.
- Have I learned enough to confidently support this conversation?
- We do not have be experts on race and it is okay not to know everything! What is important is that we have prepared to engage in these discussions before asking students to engage in this work.
- How has my own experience shaped my perspective and biases?
- Our identities shape our experiences, especially as they relate to race and power. Reflecting on our identity helps to examine our biases and may shape how we approach conversations with our students. It is important to consider how our identity may impact how we engage in discussions about race based on the identities of our students.
- Am I ready to facilitate, rather than dictate?
- This student-centered framework requires us come along side to facilitate students’ learning, rather than disseminate information “down” to students. Anti-racist pedagogy enables us to learn from students as well as with them.
- How will I handle potentially challenging scenarios?
- Planning ahead will help us to deescalate difficult situations and empowers us to commit to leaning into uncomfortable conversations.
Optional Resources
- The Anti-Racist Discussion Pedagogy by Dr. Selfa Chew, Dr. Akil Houston, and Dr. Alisa Cooper
- Anti-Racist Pedagogy in Action: First Steps by Columbia Center for Teaching and Learning
- Anti-racist pedagogy: from faculty’s self-reflection to organizing within and beyond the classroom by Kyoko Kishimoto
- 1921 Tulsa Race Massacre by Tulsa Historical Society and Museum
- Inequality in Teaching and Schooling: How Opportunity Is Rationed to Students of Color in America by Linda Darling-Hammond
- Why Are Health Studies So White? by Natalie Jacewicz
- Developing Community Agreements by the National Equity Project


16.1 – Introduction to the Nervous System
Learning Objectives
- Identify the anatomy of the nervous system and describe the main functions of the nervous system
- Analyse, translate, and define medical terms and common abbreviations of the nervous system and use correct abbreviations
- Practice the spelling and pronunciation of nervous system terminology
- Identify the medical specialties associated with the nervous system and explore common diseases, disorders, and procedures related to the nervous system
Nervous System Word Parts
Click on prefixes, combining forms, and suffixes to reveal a list of word parts to memorize for the nervous system.
Prefix
- poly- (many, much)
- a- (absence of, without, no, not)
- an- (absence of, without, no, not)
- pan- (all, total)
- post- (after)
- para- (beside, beyond, around, abnormal)
- pre- (before)
- sub- (below, under)
- dys- (painful, abnormal, difficult, laboured)
- intra- (within
- hemi- (half)
- hyper- (above, excessive)
- inter- (between)
Combining Form
- cephal/o (head)
- cerebell/o (cerebellum)
- cerebr/o (cerebrum)
- dur/o (dura mater, hard)
- encephal/o (brain)
- esthesi/o (sensation, sensitivity, feeling)
- gangli/o (ganglion)
- ganglion/o (ganglion)
- gli/o (glia, gluey substance)
- mening/o (meninges)
- meningi/o (meninges)
- ment/o (mind)
- mon/o (one, single)
- myel/o (spinal cord)
- neur/o (nerve)
- phas/o (speech)
- poli/o (gray matter)
- psych/o (mind)
- quadr/i (four)
- radic/o (nerve root)
- radicul/o (nerve root)
- rhiz/o (nerve root)
Suffix
- -al (pertaining to)
- -algia (pain)
- -cele (hernia, protrusion)
- -cyte (cell)
- -ectomy (excision, surgical removal, cutting out)
- -genic (producing, originating, causing)
- -gram (the record, radiographic image)
- -graph (instrument used to record)
- -graphy (process of recording, radiographic imaging)
- -ia (condition of, diseased state or abnormal state)
- -ic (pertaining to)
- -iatrist (specialist, physician)
- -atry (specialty, treatment)
- -ictal (seizure, attack)
- -itis (inflammation)
- -logist (specialist, physician)
- -logy (study of)
- -lysis (loosening, dissolution, separating)
- -malacia (softening)
- -oid (resembling)
- -oma (tumour)
- -osis (abnormal condition)
- -paresis (slight paralysis)
- -pathy (disease)
- -plasty (surgical repair)
- -plegia (paralysis)
- -sclerosis (hardening)
- -rrhaphy (suturing, repairing)
- -tomy (incision, cut into)
Activity source: Nervous System Word Parts by Kimberlee Carter, from Building a Medical Terminology Foundation by Kimberlee Carter and Marie Rutherford, licensed under CC BY- 4.0. /Text version added.
Introduction to the Nervous System
The picture you have in your mind of the nervous system probably includes the brain, the nervous tissue contained within the cranium, and the spinal cord, the extension of nervous tissue within the vertebral column. That suggests it is made of two organs—and you may not even think of the spinal cord as an organ—but the nervous system is a very complex structure. Within the brain, many different and separate regions are responsible for many different and separate functions. It is as if the nervous system is composed of many organs that all look similar and can only be differentiated using tools such as the microscope or electrophysiology.
Watch The Nervous System, Part 1: Crash Course Anatomy & Physiology #8 (11 min) on YouTube
Media 16.1: CrashCourse. (2015, February 23). The Nervous System, Part 1: Crash Course Anatomy & Physiology #8 [Video]. YouTube. https://youtu.be/qPix_X-9t7E
Nervous System Medical Terms
Nervous System Medical Terms (Text Version)
Practice the following nervous system words by breaking into word parts and pronouncing.
- quadriplegia
- quadr/i/plegia
- paralysis of four (limbs)
- neurectomy
- neur/ectomy
- excision of nerves
- poliomyelitis
- poli/o/myel/itis
- inflammation of gray matter of the spinal cord
- intracerebral
- intra/cerebr/al
- pertaining to within the cerebrum
- encephalosclerosis
- encephal/o/sclerosis
- hardening of the brain
- interictal
- inter/ictal
- occurs between seizures or attacks
- rhizotomy
- rhiz/o/tomy
- incision into a nerve root
- hyperesthesia
- hyper/esthesi/a
- excessive sensitivity to stimuli
- mental
- ment/al
- pertaining to the brain
- neuralgia
- neur/algia
- pain of the nerves
- radiculopathy
- radicul/o/pathy
- disease of the nerve roots
- panplegia
- pan/plegia
- total paralysis
- monoparesis
- mon/o/paresis
- slight paralysis of one (limb)
- duritis
- dur/itis
- inflammation of the dura mater
- anesthesia
- an/esthesi/a
- without (loss of) feeling or sensation
- cerebellitis
- cerebell/itis
- inflammation of the cerebellum
- encephalomalacia
- encephal/o/malacia
- softening of the brain
- cerebral thrombosis
- cerebr/al thromb/osis
- Abnormal condition of blood clot pertaining to the cerebrum (blood clot in a blood vessel in the brain)
- craniocerebral
- crani/o/cerebr/al
- pertaining to the cranium and cerebrum
- glioma
- gli/oma
- tumour composed of the glial tissue
- psychiatrist
- psych/iatrist
- physician who studies and treats disorders of the mind
- cephalgia
- ceph/algia
- *rebel does not follow the rules*
- Pain in the head (headache)
- neuroma
- neur/oma
- tumour made up of nerve (cells)
- psychogenic
- psych/o/genic
- originating in the mind
- polyneuropathy
- poly/neur/o/pathy
- disease of many nerves
- psychosis
- psych/osis
- abnormal condition of the mind
- mononeuropathy
- mon/o/neur/o/pathy
- disease affecting a single nerve
- electroencephalograph
- electr/o/encephal/o/graph
- instrument used to record electrical activity of the brain
- radicotomy
- radic/o/tomy
- incision into a nerve root
- aphasia
- a/phas/ia
- condition of without speaking
- poliomyelitis
- poli/o/myel/itis
- inflammation of the gray matter of the spinal cord
- gangliitis
- gangli/itis
- inflammation of the ganglion
- hemiplegia
- hemi/plegia
- paralysis of half (right or left) side of the body
- subdural hematoma
- sub/dur/al hemat/oma
- tumour filled with blood pertaining to below the dura mater
- dysphasia
- dys/phas/ia
- condition of difficulty speaking
- encephalitis
- encephal/itis
- inflammation of the brain
- subdural
- sub/dur/al
- pertaining to below the dura mater
- neurology
- neur/o/logy
- study of nerves
- CT myelography
- CT myel/o/graphy
- process of recording the spinal cord with computed tomography
- glioblastoma
- gli/o/blast/oma
- tumour composed of developing glial tissue
- neuroid
- neur/oid
- resembling a nerve
- neuroarthropathy
- neur/o/arthr/o/pathy
- disease of the nerves and joints
- meningitis
- mening/itis
- inflammation of the meninges
- hemiparesis
- hemi/paresis
- slight paralysis of half (right or left) side of the body
- neurorrhaphy
- neur/o/rrhaphy
- suturing of a nerve
- psychology
- psych/o/logy
- study of the mind
- neurologist
- neur/o/logist
- specialist who studies and treats the nervous system
- neuroplasty
- neur/o/plasty
- surgical repair of a nerve
- psychopathy
- psych/o/pathy
- disease of the mind
- preictal
- pre/ictal
- occurs before a seizure or attack
- radiculitis
- radicul/itis
- inflammation of the nerve roots
- electroencephalography
- electr/o/encephal/o/graphy
- process of recording the electrical activity of the brain
- cerebral angiography
- cerebr/al angi/o/graphy
- process of recording (blood) vessel pertaining to the cerebrum
- psychiatry
- psych/iatry
- specialty of the mind
- psychosomatic
- psych/o/somat/ic
- pertaining to the mind and body
- neurotomy
- neur/o/tomy
- incision into a nerve
- cerebral
- cerebr/al
- pertaining to the brain
- neuropathy
- neur/o/pathy
- disease of (peripheral) nerves
- myelomalacia
- myel/o/malacia
- softening of the spinal cord
- encephalomyeloradiculitis
- encephal/o/myel/itis
- inflammation of brain, spinal cord and nerve roots
- monoplegia
- mon/o/plegia
- paralysis of one (limb)
- meningomyelocele
- mening/o/myel/o/cele
- protrusion of the meninges and spinal cord
- paresthesia
- par/esthesi/a
- * rebel does not follow the rules*
- abnormal sensation in the extremeties
- psychologist
- psych/o/logist
- specialist of the mind
- postictal
- post/ictal
- occurs after a seizure or attack
- meningioma
- meningi/oma
- tumour of the meninges
- neurolysis
- neur/o/lysis
- dissolution of nerve (for pain mangement)
- electroencephalogram
- electr/o/encephal/o/gram
- the record of electrical activity of the brain
- gliocyte
- gli/o/cyte
- glial cell
- polyneuritis
- poly/neur/itis
- inflammation of many nerves
- neuritis
- neur/itis
- inflammation of the nerves
- ganglionectomy
- ganglion/ectomy
- excision of a ganglion
- meningocele
- mening/o/cele
- protrusion of the meninges
- rhizomeningomyelitis
- rhiz/o/mening/o/myel/itis
- inflammation of the nerve root, meninges and spinal cord
Activity source: Nervous System Medical Terms by Kimberlee Carter, from Building a Medical Terminology Foundation by Kimberlee Carter and Marie Rutherford, licensed under CC BY- 4.0. /Text version added.
Attribution
Except where otherwise noted, this chapter is adapted from “Nervous System” in Building a Medical Terminology Foundation by Kimberlee Carter and Marie Rutherford, licensed under CC BY 4.0. / A derivative of Betts et al., which can be accessed for free from Anatomy and Physiology (OpenStax). Adaptations: dividing Nervous System chapter content into sub-chapters.
instrument to view small particles by enlarging the particles
The study of electrical properties of cells and tissues


Summary
13.1 An Awakening of Religion and Individualism
Evangelical Protestantism pervaded American culture in the antebellum era and fueled a belief in the possibility of changing society for the better. Leaders of the Second Great Awakening like Charles G. Finney urged listeners to take charge of their own salvation. This religious message dovetailed with the new economic possibilities created by the market and Industrial Revolution, making the Protestantism of the Second Great Awakening, with its emphasis on individual spiritual success, a reflection of the individualistic, capitalist spirit of the age. Transcendentalists took a different approach, but like their religiously oriented brethren, they too looked to create a better existence. These authors, most notably Emerson, identified a major tension in American life between the effort to be part of the democratic majority and the need to remain true to oneself as an individual.
13.2 Antebellum Communal Experiments
Reformers who engaged in communal experiments aimed to recast economic and social relationships by introducing innovations designed to create a more stable and equitable society. Their ideas found many expressions, from early socialist experiments (such as by the Fourierists and the Owenites) to the dreams of the New England intellectual elite (such as Brook Farm). The Second Great Awakening also prompted many religious utopias, like those of the Rappites and Shakers. By any measure, the Mormons emerged as the most successful of these.
13.3 Reforms to Human Health
Reformers targeted vices that corrupted the human body and society: the individual and the national soul. For many, alcohol appeared to be the most destructive and widespread. Indeed, in the years before the Civil War, the United States appeared to be a republic of drunkenness to many. To combat this national substance abuse problem, reformers created a host of temperance organizations that first targeted the middle and upper classes, and then the working classes. Thanks to Sylvester Graham and other health reformers, exercise and fresh air, combined with a good diet, became fashionable. Phrenologists focused on revealing the secrets of the mind and personality. In a fast-paced world, phrenology offered the possibility of knowing different human characteristics.
13.4 Addressing Slavery
Contrasting proposals were put forth to deal with slavery. Reformers in the antebellum United States addressed the thorny issue of slavery through contrasting proposals that offered profoundly different solutions to the dilemma of the institution. Many leading American statesmen, including slaveholders, favored colonization, relocating Black Americans to Africa, which abolitionists scorned. Slave rebellions sought the end of the institution through its violent overthrow, a tactic that horrified many in the North and the South. Abolitionists, especially those who followed William Lloyd Garrison, provoked equally strong reactions by envisioning a new United States without slavery, where Black people and White people stood on equal footing.  Slaveholders, in particular, saw slavery as a positive aspect of American society, one that reformed the lives of enslaved people by exposing them to civilization and religion.
13.5 Women’s Rights
The spirit of religious awakening and reform in the antebellum era impacted women’s lives by allowing them to think about their lives and their society in new and empowering ways. Of all the various antebellum reforms, however, abolition played a significant role in generating the early feminist movement in the United States. Although this early phase of American feminism did not lead to political rights for women, it began the long process of overcoming gender inequalities in the republic.