Unit D Lesson 13: The Cost of Electricity
Completion requirements
Unit D Lesson 13: The Cost of Electricity
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Learning Targets |
Big Question: What is the cost of electricity?
You may have heard people complaining about the cost of electricity. The power ratings of our appliances, and how often the appliances are on and running, determines the amount of electrical energy we use.
You may have heard people complaining about the cost of electricity. The power ratings of our appliances, and how often the appliances are on and running, determines the amount of electrical energy we use.
At the end of this inquiry, you should be able to answer the following questions:
- How do I manipulate variables to calculate power use?
Pages 332 and 333 in your textbook will help you with the concept of calculating electricity cost.
Introduction
Power... Not the Same as Energy
The power company checks your home meter that keeps track of the amount of energy you use – and then it sends you a monthly bill How do you calculate energy used? What does power have to do with energy? Is there more than one way to calculate power? To answer these questions, we must first understand that power and energy are two different things. We know about energy from previous lessons, now what about power?
Mr. James Watt, a Scottish inventor, did a lot of research on steam engines and how to save power. Mr. Watt did such great work that the unit of power (Watt) was named after him.
The power company checks your home meter that keeps track of the amount of energy you use – and then it sends you a monthly bill How do you calculate energy used? What does power have to do with energy? Is there more than one way to calculate power? To answer these questions, we must first understand that power and energy are two different things. We know about energy from previous lessons, now what about power?
Mr. James Watt, a Scottish inventor, did a lot of research on steam engines and how to save power. Mr. Watt did such great work that the unit of power (Watt) was named after him.
Skill Builder
Power
The rate at which energy is used and transformed in a device is power. The unit of power is a watt (W). One watt is equal to 1 joule of energy per second. The more watts a device uses, the faster it converts energy from one form into another form. Two formulas are used to calculate power. The example that follows focuses on formula one.
The rate at which energy is used and transformed in a device is power. The unit of power is a watt (W). One watt is equal to 1 joule of energy per second. The more watts a device uses, the faster it converts energy from one form into another form. Two formulas are used to calculate power. The example that follows focuses on formula one.
Formula One
«math»«mstyle indentalign=¨center¨»«mtext»Power§#160;(P)§#160;=§#160;current§#160;(I)§#160;§#215;§#160;voltage§#160;(V)«/mtext»«mspace linebreak=¨newline¨»«/mspace»«mi mathvariant=¨bold-italic¨»P«/mi»«mo mathvariant=¨bold¨»=«/mo»«mi mathvariant=¨bold-italic¨»I«/mi»«mi mathvariant=¨bold-italic¨»V«/mi»«/mstyle»«/math»
«math»«mstyle indentalign=¨center¨»«mtext»Power§#160;(P)§#160;=§#160;current§#160;(I)§#160;§#215;§#160;voltage§#160;(V)«/mtext»«mspace linebreak=¨newline¨»«/mspace»«mi mathvariant=¨bold-italic¨»P«/mi»«mo mathvariant=¨bold¨»=«/mo»«mi mathvariant=¨bold-italic¨»I«/mi»«mi mathvariant=¨bold-italic¨»V«/mi»«/mstyle»«/math»
Formula Two
«math»«mstyle indentalign=¨center¨»«mtext»Power§#160;(«/mtext»«mi»P«/mi»«mtext»)§#160;=§#160;«/mtext»«mfrac»«mrow»«mtext»Energy§#160;(«/mtext»«mi»E«/mi»«mtext»)«/mtext»«/mrow»«mrow»«mtext»time§#160;(«/mtext»«mi»t«/mi»«mtext»)«/mtext»«/mrow»«/mfrac»«mspace linebreak=¨newline¨»«/mspace»«mi mathvariant=¨bold-italic¨»P«/mi»«mo mathvariant=¨bold¨»=«/mo»«mfrac»«mi mathvariant=¨bold¨»E«/mi»«mi mathvariant=¨bold¨»t«/mi»«/mfrac»«/mstyle»«/math»
«math»«mstyle indentalign=¨center¨»«mtext»Power§#160;(«/mtext»«mi»P«/mi»«mtext»)§#160;=§#160;«/mtext»«mfrac»«mrow»«mtext»Energy§#160;(«/mtext»«mi»E«/mi»«mtext»)«/mtext»«/mrow»«mrow»«mtext»time§#160;(«/mtext»«mi»t«/mi»«mtext»)«/mtext»«/mrow»«/mfrac»«mspace linebreak=¨newline¨»«/mspace»«mi mathvariant=¨bold-italic¨»P«/mi»«mo mathvariant=¨bold¨»=«/mo»«mfrac»«mi mathvariant=¨bold¨»E«/mi»«mi mathvariant=¨bold¨»t«/mi»«/mfrac»«/mstyle»«/math»
What is the power of a hair dryer that requires 10 A of current to operate on a 120-V circuit?
A. Write what you know.
A. Write what you know.
B. Select and choose formula.I = 10A
V = 120 V
P = ? W
C. Substitute into formula.P = I x V
D. Write answer in a sentence.P = (10A) x (120V)
= 1200 W
The power of the hair dryer is 1200 W. (Note that the number with the least number of significant digits in the question has two significant digits, so the answer to the question should also have only two significant digits, therefore the best answer to this question is: "The power of the hair dryer is 1.2 kW" – kW, or kilowatts, is obtained by dividing watts by 1000, because there are 1000 watts in a kilowatt.)
Practice
– Determining Power Using Electricity
Read the following questions and answer each question in the same format as the example above.
Read the following questions and answer each question in the same format as the example above.
A. Write what you know.
B. Select and choose formula.
C. Substitute into formula.
D. Write answer in a sentence.
Question 1
The average current that a television draws while it operates is 2.00 A. If the television is connected to a 120-V circuit, what is the power rating of the television?
A. Write what you know.
B. Select and choose formula.
C. Substitute into formula.
D. Write answer in a sentence.
The average current that a television draws while it operates is 2.00 A. If the television is connected to a 120-V circuit, what is the power rating of the television?
A. Write what you know.
I = 2A
V = 120 V
P = ? W
P = I x V
P = (2.00 A)(120 V)
= 240 W
The power rating of the TV is 240 W.
Question 2
What is the power rating of a 80-A toaster connected to a 120-V circuit?
A. Write what you know.
B. Select and choose formula.
C. Substitute into formula.
D. Write answer in a sentence.
What is the power rating of a 80-A toaster connected to a 120-V circuit?
A. Write what you know.
I = 80 A
V = 120 V
P = I x V
P = I x V
P = (80A)(120V)
= 9600 W
To record your answer with correct number of significant digits, convert W into kW by dividing by 1000
9600 W x 1 kW/1000 W
9.6 kW
The power rating of the toaster is 9.6 kW.
Determining Power Using Energy and Time
The practice you just completed used electricity variables (current, voltage) to determine power. But power can also be determined by calculating how many joules of energy are used every second. This is a fairly easy calculation, all you need to do is take the total amount of energy consumed and divide it by the number of seconds the device was in use. This is an important calculation in modern society, mostly because energy costs money. Utility companies would probably prefer that you use lots of high-power devices, because that means you will consume more energy, meaning more money for them. As a consumer, it might be in your best interest to select low-power devices as part if your strategy is to use less energy.
The practice you just completed used electricity variables (current, voltage) to determine power. But power can also be determined by calculating how many joules of energy are used every second. This is a fairly easy calculation, all you need to do is take the total amount of energy consumed and divide it by the number of seconds the device was in use. This is an important calculation in modern society, mostly because energy costs money. Utility companies would probably prefer that you use lots of high-power devices, because that means you will consume more energy, meaning more money for them. As a consumer, it might be in your best interest to select low-power devices as part if your strategy is to use less energy.
The second way to calculate
power
is to use the formula «math»«mi»P«/mi»«mo»=«/mo»«mfrac»«mi»E«/mi»«mi»t«/mi»«/mfrac»«/math»
.
Energy is measured in joules and time is measured in seconds. All electrical devices are labelled with the power usage and voltage. From these two values, you could calculate the energy consumption. Look at the following chart and notice the energy consumption rates.
Energy is measured in joules and time is measured in seconds. All electrical devices are labelled with the power usage and voltage. From these two values, you could calculate the energy consumption. Look at the following chart and notice the energy consumption rates.
Figure 1 – Note that power and energy calculations must use seconds, so in the cases above, hours and minutes are converted to seconds.
Converting Minutes to Seconds
Note that, when calculating power using energy and time, you need to be familiar with how many seconds are in a minute and hour. Often, the questions tell you the time the device is used in minutes. Then, you would multiply by 60 seconds to convert the time to seconds.
15 minutes x 60sec/min = 900 seconds
Note that, when calculating power using energy and time, you need to be familiar with how many seconds are in a minute and hour. Often, the questions tell you the time the device is used in minutes. Then, you would multiply by 60 seconds to convert the time to seconds.
15 minutes x 60sec/min = 900 seconds
Question 1
A microwave uses 1 440 000 joules of energy in 30 minutes. What is the power rating of the microwave?
A. Write what you know.
A microwave uses 1 440 000 joules of energy in 30 minutes. What is the power rating of the microwave?
A. Write what you know.
B. Select and choose formula.E = 1 440 000 J
t = 30 minutes x 60 seconds in a minute = 1800 s
P = ? W
P = E/tC. Substitute into formula.
P = 1 440 000 J / 1800 sD. Write answer in a sentence.
= 800 W
The microwave has a power rating of 800 W.
Question 2
An iron has a power rating of 500 watts. The iron is used for 20 minutes. How many joules of energy are used by the iron?
A. Write what you know.
An iron has a power rating of 500 watts. The iron is used for 20 minutes. How many joules of energy are used by the iron?
A. Write what you know.
B. Select and choose formula.P = 500 W
t = 20 minutes x 60 seconds per minute = 1200 seconds
E =? J
E = P x tC. Substitute into formula.
E = (500 W) x (1200s)
= 600 000J
D. Write answer in a sentence.The answer needs be displayed with 2 significant digits. You can either use scientific notation, or convert to megajoules.
600 000 J = 6.0 x \({{10}^{5}}\) J
The iron uses 6.0 x \({{10}^{5}}\) joules of energy.
Practice – Determining Power Using Energy and Time
Read the following questions and answer each question in the same format as the examples above.
Read the following questions and answer each question in the same format as the examples above.
A. Write what you know.
B. Select and choose formula.
C. substitute into formula
D. Write answer in a sentence.
Question 1
You bake a cake in a 7000 watt oven for 20 minutes, how many joules of electricity would the oven use?
You bake a cake in a 7000 watt oven for 20 minutes, how many joules of electricity would the oven use?
A. Write what you know.
P = 7000WB. Select and choose formula.
t = 20 minutes x 60 seconds per minute = 1200 seconds
E = ? J
E= P x tC. Substitute into formula.
E = (7000W) x (1200 s)D. Write answer in a sentence.
8 400 000J
The oven uses 8 400 000 J of energy.
The oven uses much energy so sometimes we say kilojoules instead of joules. 1 kilojoule is 1000 J.
Thus, 8 400 000 J is 8400 kJ.
For proper significant digits this answer needs to be converted into megajoules (MJ)
8400 kJ x 1 MJ/1000 kJ = 8.4 MJ
Question 2
What is the power rating of a microwave used to bake a potato? The microwave uses 1 600 000 J of energy in 35 minutes to bake the potato.
What is the power rating of a microwave used to bake a potato? The microwave uses 1 600 000 J of energy in 35 minutes to bake the potato.
A. Write what you know.
E = 1 600 000 JB. Select and choose formula.
t = 35 minutes x 60 seconds per minute = 2100 s
P = ? W
P = E/tC. Substitute into formula.
P = 1 600 000 J/ 2100 sD. Write answer in a sentence.
= 761.9 J/s OR
=761.9 W (because 1 J/s = 1W)
convert to kW
761.9 W x 1 kW/1000W
0.76 kW
The power used by the microwave is 0.76 kW.
Cost of Power
Electrical devices consume a lot of power. For example, a stove may have a power rating of 7000 Watts. A conversion to kilowatts is often used to make the numbers smaller and make calculating the cost of power easier. This works better for electric bills as well, as smaller numbers fit easier onto a page, and seeing smaller numbers probably makes consumers feel better about their electric bills (even though it's the same amount energy!)
Remember though, when we calculate how much energy is consumed, we usually use calculate the number in joules. But the joule is a very, very tiny unit of energy. As a practical alternative, using kilowatts (instead of watts) and hours (instead of seconds) we get a unit of energy known as the kilowatt-hour (kWh). To calculate kWh, simply use power in kilowatts, and multiply by the number of hours of energy consumption.
To change watts to kilowatts, simply divide the number of watts by 1000.
Electrical devices consume a lot of power. For example, a stove may have a power rating of 7000 Watts. A conversion to kilowatts is often used to make the numbers smaller and make calculating the cost of power easier. This works better for electric bills as well, as smaller numbers fit easier onto a page, and seeing smaller numbers probably makes consumers feel better about their electric bills (even though it's the same amount energy!)
Remember though, when we calculate how much energy is consumed, we usually use calculate the number in joules. But the joule is a very, very tiny unit of energy. As a practical alternative, using kilowatts (instead of watts) and hours (instead of seconds) we get a unit of energy known as the kilowatt-hour (kWh). To calculate kWh, simply use power in kilowatts, and multiply by the number of hours of energy consumption.
To change watts to kilowatts, simply divide the number of watts by 1000.
Example
1000 watts = 1 kilowatt
1000/1000 = 1 kW
130 watts = 0.130 kilowatts
130/1000 = 0.130 kW
1000 watts = 1 kilowatt
1000/1000 = 1 kW
130 watts = 0.130 kilowatts
130/1000 = 0.130 kW
Practice
Make the following conversions between watts and kilowatts:
1. 10 000 watts = ____kilowatts
2. 30 watts = ________kilowatts
3. ___________ = 25 kilowatts
Make the following conversions between watts and kilowatts:
1. 10 000 watts = ____kilowatts
2. 30 watts = ________kilowatts
3. ___________ = 25 kilowatts
1. 10 kW
2. 0.03 kW
3. 25 000 W
2. 0.03 kW
3. 25 000 W
Figure 2 – Every month, most individuals who own homes receive an electric bill that states how much money they owe for the electric energy they have used.
The Refrigerator – An Energy Consumption Example
The refrigerator is an excellent choice for an energy consumption example because almost everyone has one. Look at the following example carefully, and also realize the results might be different for different refrigerators. In fact, the next time you are in a store that sells appliances, open the doors on the refrigerators and see if you can find energy consumption information.
The refrigerator is an excellent choice for an energy consumption example because almost everyone has one. Look at the following example carefully, and also realize the results might be different for different refrigerators. In fact, the next time you are in a store that sells appliances, open the doors on the refrigerators and see if you can find energy consumption information.
A Two-part Refrigerator Question: How much energy, and how much will it cost?
i) If a refrigerator has a 700 W power rating, how many kilowatt-hours of energy will it require during a 30-day period?
ii) If electricity costs 11 cents per kilowatt-hour, how much would the refrigerator cost to operate during that period?
A. Write what you know.
*Change watts to kilowatts and time to hours.
P = 700 watt or 0.7 kW
t = 30 days x 24 hours in a day = 720 hours
A. Write what you know.
Cost = 11 cents/kWh = $0.11
Energy = 504 kWh
B. Select and chose formula.
Power x time = Energy in kWh
B. Select and chose formula.
Energy (kWh) x cost = Cost of electricity bill
C. Substitute into formula.
0.7 kW x 720 h = 504 kWh
C. Substitute into formula.
504kWh x 0.11 = $55.44
D. Write answer in a sentence.
The power required is 504 kWh during a 30-day period.
D. Write answer in a sentence.
It will cost $55.44 to operate the refrigerator for 30 days.
Watch
Do you wonder how much it costs to run the appliances in your home? To learn about cost of using appliances, watch the videos below: "Calculating the Operating Cost of Appliances in Your Home", and "Calculating the Cost of Electricity Usage in Physics" and if you need one more calculation tutorial, watch the 3rd video "Cost of Electricity", it's an example of the math involved in calculating electricity cost.
Practice – Determining The Cost of Energy
Read the questions and answer each question in the same format as the examples above.
Read the questions and answer each question in the same format as the examples above.
A. Write what you know.
B. Select and choose formula.
C. Substitute into formula.
D. Write answer in a sentence.
Question 1
A homeowner finds that she has 42 light bulbs (each are 100 W) in use in her home.
i) If all the bulbs are on for an average of 5 hours per day, how many kilowatt-hours of electricity will be consumed in a 30-day period?
ii) At 8 cents per kilowatt-hour, how much will operating the lights cost the homeowner during that 30-day period? (8 cents = $0.08)
A homeowner finds that she has 42 light bulbs (each are 100 W) in use in her home.
i) If all the bulbs are on for an average of 5 hours per day, how many kilowatt-hours of electricity will be consumed in a 30-day period?
ii) At 8 cents per kilowatt-hour, how much will operating the lights cost the homeowner during that 30-day period? (8 cents = $0.08)
i) If all the bulbs are on for an average of 5 hours per day, how many kilowatt-hours of electricity will be consumed in a 30-day period?
A. Write what you know.
ii) At 8 cents per kilowatt-hour, how much will operating the light cost the homeowner during that 30-day period? (8 cents= $0.08)
A. Write what you know.
A. Write what you know.
B. Select and chose formula.P= 42 light bulbs x 100 watts each = 4200 watts
4200 watts = 4.2 kW
t= 30 days x 5 hours a day = 150 h
C. Substitute into formula.Power (kW) x hours = kWh
4.2 kW x 150 h = 630 kWhD. Write answer in a sentence.
630 kWh will be consumed in a 30-day period.
ii) At 8 cents per kilowatt-hour, how much will operating the light cost the homeowner during that 30-day period? (8 cents= $0.08)
A. Write what you know.
B. Select and chose formula.Power = 630 kWh
Cost for each kWh = 8 cents
C. Substitute into formula.Cost of bill = Power (kWh) x Cost (per kWh)
D. Write answer in a sentence.630 kWh x $0.08 = $50.40
It will cost the homeowner $50.40 to use the light bulbs.
Question 2
Bob has a stereo that operates at 120 V, using 2.5 A of current.
i) What is the power rating of Bob's stereo? (Hint: Recall the previous power-calculation problems where P = I x V.)
ii) If Bob plays his stereo for an average of 5 hours each day, how much electricity (in kWh) will he use in a 30-day period?
Bob has a stereo that operates at 120 V, using 2.5 A of current.
i) What is the power rating of Bob's stereo? (Hint: Recall the previous power-calculation problems where P = I x V.)
ii) If Bob plays his stereo for an average of 5 hours each day, how much electricity (in kWh) will he use in a 30-day period?
i) How much power does Bob’s stereo need to operate?
Need a Hint:
A. Write what you know.
ii) If Bob plays his stereo for an average of 5 hours each day, how much electricity (in kWh) will he use in a 30-day period?
Need a Hint: Change 300 watts to kilowatts (300/1000 = 0.3 kW).
A. Write what you know.
Need a Hint:
Recall the previous power calculation problems where P= I x V
A. Write what you know.
B. Select and choose formula.I = 2.50 A
V = 120 V
P = ?
C. Substitute into formula.P = I x V
D. Write answer in a sentence.P = (2.50 A)(120 V)
= 300 W
Bob’s stereo uses 300 watts of power.
ii) If Bob plays his stereo for an average of 5 hours each day, how much electricity (in kWh) will he use in a 30-day period?
Need a Hint: Change 300 watts to kilowatts (300/1000 = 0.3 kW).
A. Write what you know.
B. Select and choose formula.P = 300 watts = 0.3 kW
t = 5 hours x 30 days = 150 h
C. Substitute into formula.P x t = power in kWh
D. Write answer in a sentence.0.3 kW x 150 h = 45 kWh
Bob will use 45 kWh in a 30-day period.
Interactive
Power
-
The BrainPOP video, reviews the concept of Acids and Bases.
Click here
to watch "Power".
You will need a username and password to access the video.
- Username: 0099
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Password: students
-
Click here
to complete the quiz on Power. Remember to check your answers at the end!