1. Reviewing and Practicing the Energy Math Tuesday, December 13th, 2016
Math for APES
Reviewing and Practicing the Energy Math
Tuesday, December 13th, 2016

2. Dimensional Analysis
Dimensional analysis is a strategy for performing conversions by multiplying by ratios that equal one.
Basic Principles of Dimensional Analysis
You can multiply a quantity by 1 without changing its value.
2 equal quantities create a ratio that equals one.

3. The Process of Dimensional Analysis
Set up conversion ratios so that the units being converted cancel out.

4. Calculate Usain Bolt’s speed in miles per hour
Calculate Usain Bolt’s speed in miles per hour. He ran a 100 meter dash in 9.58 seconds (10.4 m/s). Conversion factors: 1000 m = 1 km, 1.6 km = 1 mile,
10.4 meters x
seconds
1 km x
1000 m
1 mile x
1.6 km
60 seconds x
1 min
60 min =.
1 hr
23.4 miles
hours

5. Tips for Dimensional Analysis
Always include your units and cross off the ones that cancel out so that you can double check your work, making sure you did not miss a step or invert any ratios.
For some problems, the same unit may be used for more than one type of measurement. For these cases, you will need to include both the unit and what is being measured. =

6. Some ratios are normally given in a conventional orientation, such as miles per gallon, so that the information can be more easily interpreted. For example, a higher miles per gallon ratio means better fuel efficiency.
However for problem solving purposes these ratios can and may need to be inverted.
If a car gets 25 miles per gallon, it also uses 1 gallon per 25 miles.

7. Solve the problem with your table partner:
A car has a fuel efficiency of 20 miles per gallon and is used to drive 15,000 miles per year. Determine how many pounds of carbon are released into the atmosphere each year by the car.
Each gallon of gasoline burned releases 5 lbs of carbon dioxide.
Every lb of carbon dioxide contains 0.27 lbs of carbon
15,000 miles x 1 gallon x 5 lbs CO2 x lbs carbon
year miles gallon 1 lb CO2
=1, 012 lb carbon

8. The basic unit of energy
… is the joule (J). It is a very small unit, so when we are talking about a lot of energy, we use kilojoules (kJ).
Remember your metric conversions…
1000 J = 1 kJ (kilo = 1000)

9. Power Power (P) is the rate at which energy is used.
When determining the amount of energy (usually in J or kJ) you must also include a time component.
Power x Time = Energy (P x t = E)
This can be rearranged to determine power as well
P = E/t (Power = Energy /time)

10. Efficiency The unit for power is the watt
1 W = 1 J/sec (1 watt = 1 joule per second)
Therefore a 100 watt light bulb uses 100 J/sec of electrical energy.
If it is 20% efficient (typical for an average light bulb) then the bulb converts 20% of the electrical energy into light, and 80% is lost as waste heat.

11. Physics Laws
Notice that in the previous example, we can see the operation of both the First and Second Laws of Thermodynamics.
The First Law: energy can be converted from one form to another, but none is lost. We have accounted for all the energy, but most of the electrical energy (high quality) was converted to low quality energy (heat).
Therefore we also see the Second Law: in any energy conversion, some energy is converted into lower quality energy (usually heat) and is unable to perform useful work (in this case, light).

12. Find energy of appliances
Knowing the relationship between energy and power allows us to find the energy used when an appliance of known power (in watts) operates for a known amount of time (in seconds).
Example: How much energy (in kJ) does a 75 watt light bulb use when it is turned on for 25 minutes?
Equation: E = P x t
Conversion: Power (in watts) 1 watt = 1 J/sec

13. Solution
How much energy (in kJ) does a 75 watt light bulb use when it is turned on for 25 minutes? E = 75 J 60 sec 25 min sec min
110,000 J
(or 110 kJ)

14. If the wattage is not given
Some of the information about the current can usually be found.
To find the power (in watts) of any electrical appliance, use the equation P = V x I
P = power, V = voltage, I is current in amps (A).
American household voltage is 100 V (Air conditioners, electric stoves and dryers are 220 V).

15. 1 kwh = 3600 J/sec
The Kilowatt Hour (kwh) is not a unit of power but a unit of energy (because it has a time component).
Notice that a kilowatt is a unit of power and hour is a unit of time. Therefore, E = P x t.
A kilowatt-hour is equal to 1 kw (or watts) delivered continuously for one hour (3600 sec).
1 kwh = 1000 J/sec x 3600 sec = 3,600,000 J or 3600 kJ
So 1 kwh = 3600 J/sec

16. Other units of energy include:
1 calorie (cal) = J
1 BTU = 1.05 kJ
1 therm = 100,000 BTU

17. Problem 1
Dr. Smith’s power bill shows that his home used kwh over a 30 day period.
1 kwh = 3600 kJ
Find the energy used (in kJ) for the 30-day period.
Find the energy used in J/day.
At the rate of $.075/kwh, what is Dr. Smith’s power bill (without tax)?

18. 1a
(1355 kwh in 30 days)
a) Find the energy used (in kJ) for the 30 day period.
1355 kwh 3600 kJ x 106 kJ
1 kwh
Multiplication short-cut: 1355 x 36 (take off the two zeros) = 48780, then put the two zeros back = To put into scientific notation, move the decimal six places to the left.

19. 1b b) Find the energy used in J/day.
Start with previous calculation
4.88 x 106 kJ 1000 J 1 month x 108 J.day
Month 1 kJ 30 days
4.88 x 106 x 1000 is equal to 4.88 x 109
4.88 x 109
3 x 101 (this is the same as 30)
Calculate in parts: 4.88 / 3 = 1.63, subtract the exponents to get 1.63 x 108

20. 1c
c) At the rate of $.075/kwh, what is Dr. Smith’s power bill (without tax)?
1355 kwh $ $101.63
1 kwh

21. Problem 2
A current through a toaster (110 V) is 8 A. Remember, P = V x I 
What is the power (in watts) of the toaster?
How much energy (in J) will the toaster use in 5 minutes of operation?

22. 2a P = V x I where V = voltage and I = amps P = 110V x 8 A = 880 watts
A current through a toaster (110 V) is 8 A.
a) What is the power (in watts) of the toaster?
P = V x I
where V = voltage and I = amps
P = 110V x 8 A = 880 watts

23. (1 watt = 1 joule per second)2b
b) How much energy (in J) will the toaster use in 5 minutes of operation?
P = E/t so E = P x t
(1 watt = 1 joule per second)
E = 880 W (1 J/sec) 60 sec 5 min ,000 J
1 watt min
2.64 x 105 J

24. Problem 3
A 100 watt light bulb is 20% efficient. That means 20% of the energy used is converted to light, while 80% of the energy used is lost as heat.
How much energy does it use in 12 hours of operation?
How much energy does the bulb convert into light over the 12-hour period?
How much energy does the bulb convert into heat over the 12-hour period?
Convert the total energy use into kwh

25. 3a A 100 watt light bulb is 20% efficient.
a) How much energy does it use in 12 hours of operation?
E = P x t
E = 100 W (1 J/sec) 60 sec 60 min 12 hrs = 4,320,000 J
1 watt min 1 hr
4,320,000 J = 4,320 kJ

26. 3b E = 4.32 x 106 J (.20 efficiency) = 864,000 J = 8.64 x 105 J
b) How much energy does the bulb convert into light over the 12-hour period?
E = 4.32 x 106 J (.20 efficiency) = 864,000 J = 8.64 x 105 J
Math shortcut:
Multiply 4.32 x 0.2 to get 0.864
Move the decimal six places to the right (because 106) to get 864,000 J
(or move the decimal one place to the right and drop the exponent by one to have it in scientific notation)

27. 3c
3c) How much energy does the bulb convert into heat over the 12-hour period?
E = 4.32 x 106 J (.80 heat) = 3,456,000 J =
3.46 x 106 J

28. 3d
d) Convert the total energy use into kwh. (the energy calculated in 3a was 4,320 kJ) 1 kwh = 3600 kJ 4320 kJ x 1 kwh = 1.2 kwh 3600 kJ

29. Problem 4
An electric clothes dryer has a power rating of 4000W. Assume that a family does five loads of laundry each week for 4 weeks.
Further assume that each load takes one hour.
Remember, 1 W = 1 J/sec.
Find the energy used in both J and kwh
If the cost of electricity is $.075/kwh, find the cost of operating the dryer for a month (4 weeks).

30. 4a a) Find the energy used in both J and kwh
5 loads x 4 weeks x  1 hr = 20 hrs
week load
E = P x t
E = 4000 W x (1 J/sec) x 60 sec x 60 min x 20 hrs = 2.88 x 108 J watt  min hr
2.88 x 105 kJ x 1 kwh = 80 kwh
3600 kJ

31. 4b
b) If the cost of electricity is $.075/kwh, find the cost of operating the dryer for a month (4 weeks). 80 kwh x $.075 = $6.00 kwh

32. Problem 5
Dr. Smith’s natural gas bill states that his household used 110 therms of energy over a 30-day period.
Convert 110 therms to kwh.
His charge for the energy was $ Find the cost of this natural gas in $/kwh.
Using the information about electricity costs in the problems above, which form of energy (electricity or natural gas) is more expensive? How many times more expensive is it?

33. 5a 5a) Convert 110 therms to kwh
110 therms x 100,000 BTU x 1.05 kJ x 1 kwh = 3208 kwh
1 therm BTU kJ

34. 5b
5b) His charge for the energy was $ Find the cost of this natural gas in $/kwh.
$88.78 / 3208 kwh = $.028/kwh

35. 5c
5c) Which form of energy (electricity or natural gas) is more expensive? How many times more expensive is it? elec: $.0749 gas: $.028 Electricity is about 2.5 times more expensive