ONACD provides an editable curriculum designed by teachers for teachers. The document discusses energy efficiency and calculations for determining power and energy usage. Examples include the efficiency of light bulbs, costs for running electronics, and determining energy lost when heating water on a hotplate. The hotplate example calculates that 61% of the energy was lost to the surroundings, and the efficiency could be improved by using better insulation or a lid to retain more heat.

1. ONACD - Editable curriculum designed for teachers by teachers
Efficiency
Where does the energy go?
ONACD –Editable Curriculum designed for teachers by teachers
LED bulbs are more efficient than any other light bulb

2. ONACD - Editable curriculum designed for teachers by teachers
Depending on the information given you can use either of
the power equations to determine power consumption.
or
How much energy does a 1500W hair dryer being used for
60 seconds use?
E = Pt
E = (1500)(60)
E = 90 000 Joules
Was all of the heat energy from the hair dryer transferred to
the hair it was drying?
No, most of it was lost to heating the air surrounding the hair.
P IV=
E
P
t
=
E
P
t
=

3. ONACD - Editable curriculum designed for teachers by teachers
Heat loss to the surroundings often causes devices to be
inefficient. Energy is transferred from an input source
and transformed to output energy at the load.
For example, in one second a 60W bulb uses an input of
60J of energy to output 3J of light energy.

4. ONACD - Editable curriculum designed for teachers by teachers
Efficiency
The efficiency of the 60W bulb in the previous example is:
Efficiency = 0.05 or 5%
Examples:
1. An old computer monitor operates at 1.3A. How many
kilowatts does it use? P = IV
P = (1.3)(110)
P = 143W
P = 0.143kW
2. At $0.12 per kWh how much does it cost to run for 4
hours?
Cost = kW x hr x rate
Cost = (0.143)(4)(0.12); Cost = $0.069 or 7 cents
3
60
J
efficiency
J
=

5. ONACD - Editable curriculum designed for teachers by teachers
3. Consider a 300 W hotplate heats 400 mL of water for 6
minutes (360 seconds). The temperature raised 25o
C.
a) How many joules of energy were used?
b) It requires 1675J to heat 400 mL of water 1o
C. How
much energy should have been used to heat the water
25o
C?
25 times 1675 = 41875J
c) How much energy was “lost”?
108000 - 41875 = 66125J
108000E Joules=300
360
E
=
E
P
t
=

6. ONACD - Editable curriculum designed for teachers by teachers
d) Where did this energy go?
Heat was lost to the surroundings around the edges of the
element and off the surface of the water and container.
e) What is the efficiency of the hotplate set-up?
f) What could be done to improve the efficiency?
Use a container that covers the entire element, use a lid, use a
better insulating container.
output
efficiency
input
=
41875
108000
efficiency = 39%efficiency =

7. ONACD - Editable curriculum designed for teachers by teachers
d) Where did this energy go?
Heat was lost to the surroundings around the edges of the
element and off the surface of the water and container.
e) What is the efficiency of the hotplate set-up?
f) What could be done to improve the efficiency?
Use a container that covers the entire element, use a lid, use a
better insulating container.
output
efficiency
input
=
41875
108000
efficiency = 39%efficiency =