Name: _____________
Energy Calculation Problems
BACKGROUND INFORMATION
Energy:
The basic unit of energy is the Joule (J). Other units are calorie, kilojoule, British Thermal Unit (BTU), and therm.
1,000 J = 1 kJ
1,000 cal = 1 kcal
1 cal = 4.184 J
1 BTU = 1.05 kJ
1 therm = 100,000 BTU
1 quad = 1×1015 BTU
Power:
Power is the rate at which energy is used. The basic unit of power is the Watt (1 Watt = 1 Joule per second).
(P = E ÷ t)
If the wattage is not given, then some information about the current can usually be found. To find the power (in Watts) of any
electrical appliance in your home use the equation W = V x A. V is the voltage (Volts) and A is the current in amps.
American household voltage is 110 V AC (lights, TVs, computers, etc.) Electric stoves, clothes driers and air conditioners are
220 V.
Efficiency Example:
A 100 Watt light bulb uses 100 Joules of electrical energy every second. If it is 20% efficient, then the bulb converts 20% of
the electrical energy into light and 80% into heat (waste energy). To calculate output, multiply the wattage by the efficiency
(in decimal form):
100 W × 0.2 = 20 W of light energy (and 80 W of heat energy)
Energy:
A kilowatt-hour is a unit of energy, not power. Notice that kilowatt is a unit of power and hour is a unit of time (E = P × t). A
kilowatt-hour is equal to 1 kW of power delivered continuously for 1 hour.
PRACTICE PROBLEMS
Perform the energy calculations below. Please show your work – no points will be given for guesswork.
1.
How much energy (in Wh) does a 75 Watt light bulb use when it is turned on for 25 minutes? How many kWh is this?
25 min × 1 h/60 min = 0.417 h
75 W × 0.417 h = 31.275 Wh
31.275 Wh = 0.031275 kWh
2.
A current of 11 Amps at 240 Volts flows through an electric range. If it is used an average of 1 hour/day:
a. Calculate the watts of the electric range.
11 A × 240 V = 2,640 W
3.
b.
Calculate the kWh used per month. (30 days)
2,640 W = 2.64 kW
1 h/day × 30 days = 30 h
2.64 kW × 30 h = 79.2 kWh
c.
What is the cost to run the range for one month at $0.08/ kWh?
79.2 kWh × $0.08/ kWh = $6.34
d.
What is the cost to run the range for one year at $0.08/ kWh?
365 days × 1 h/day = 365 h
2.64 kW × 365 h = 963.6 kWh
963.6 kWh × $0.08/ kWh = $77.09
A 100 Watt light bulb is 20% efficient.
a. How much energy (in Wh) does it use in 12 hours of operation?
100 W × 12 h = 1,200 Wh
4.
b.
How much energy (in Wh) does the bulb convert to light during 12 hours?
1,200 Wh × 0.2 = 240 Wh
c.
How many kWh does it use in 12 hours of operation?
1,200 Wh = 1.2 kWh
An electric clothes dryer has a power rating of 4000 W. Assume a family does 5 loads of laundry each week for 4
weeks. Each dryer load takes 1 hour to complete.
a. Find the total energy used (in Joules and kWh).
1 h/load × 5 loads/week × 4 weeks = 20 h
20 h × 60 min/h × 60 s/min = 72,000 s
4,000 W = 4,000 J/s
4,000 J/s × 72,000 s = 288,000,000 J
4,000 W × 20 h = 80,000 Wh
80,000 Wh = 80 kWh
b.
5.
6.
Find the operating cost of the dryer for 4 weeks of use. Assume the cost of electricity is $0.0758/kWh.
80 kWh × $.0758/kWh = $ 6.06
Your 7.25 kW air conditioner is used for a total of 137 days (24 hours per day). Assume the cost per kWh is $0.0825
and 1 kWh = 3,400 BTUs.
a.
Calculate the total number of kWh used per year.
137 days/ yr × 24 h/day × 7.25 kWh/h = 23,838 kWh/yr
b.
Determine the cost of air conditioning for one year.
23,838 kWh/yr × $0.0825/ kWh = $1,966.64/yr
c.
How many kcal are used per year?
23,838 kWh/yr × 860 kcal/kWh = 20,500,680 kcal/yr
d.
How many BTUs are used in one year?
23,838 kWh/yr × 3,400 BTU/1 kWh = 81,049,200 BTU/yr
Thorpeville is a rural community with a population of 8,000 homes. It gets its electricity from a small, municipal
coal-burning power plant just outside of town. The power plant’s capacity is rated at 20 megawatts, with the average
home consuming 10,000 kilowatt hours (kWh) of electricity per year. Residents of Thorpeville pay $0.12 per kWh.
a.
The existing power plant runs 8,000 hours per year. How many kWh of electricity is the current plan
capable of producing in a year?
20 MW × 1,000 kW/MW = 20,000 kW
20,000 kW × 8,000 hours = 160,000,000 kWh
b.
How many kWh of electricity do the residents of Thorpeville consume in one year?
8,000 homes × 10,000 kWh/home = 80,000,000 kWh
c.
Compare answers (a) and (b). What conclusions can you make concerning the differences between the
energy consumed and the energy needed by the town?
Power plants produces 1.6 × 108 kWh per year. The residents, however, only use 8 × 107 kWh per year.
This leaves a surplus of 8 × 107 kWh in one year, which can be sold to other towns.
d.
If it takes 1 ton of coal to produce 2,460 kWh of electricity, how many tons of coal are burned by the power
plant ever year?
160,000,000 kWh × 1 ton/2,460 kWh = 65,040 tons