Energy Efficiency Lecture Notes and Lab
A device is efficient when most of its input energy is converted into useful
output (the primary purpose) energy.
Some transformations are more efficient that others – remember that a light bulb
converts only 5% of electrical energy to light energy, while a gas furnace
converts 80% of chemical energy to thermal energy.
Review of formula:
efficiency = useful energy output X 100
energy input
LAB: Conversion of electrical energy into thermal energy
How much of the input energy is actually used to boil the water?
Question: How do we calculate the energy efficiency of a device used to heat
water?
Materials:
kettle, 250 ml of water, thermometer, stopwatch, insulated glass jar
Procedure: Record the temperature of the water
Place thermometer in the kettle
Heat water to 90o C
Count the seconds to reach 90oC and record
Remove water from the heat and stir for 1 minute
Record the highest temperature reached in the water
Calculate efficiency
Notes:
Energy is measured in joules
First, we observe the amount of input energy from the kettle entering the water
The formula for input energy is:
Energy (joules) = power (watts) X time (seconds)
Power is measured by the number of watts (look on the bottom of any electrical
device for this number); in this case, the kettle puts out _______ watts of
electrical power.

The original temperature of the water is ________ oC

The water reached 90oC in ______ seconds
Therefore, the power, or input energy, from the kettle during this experiment was:
______ watts x _____ sec. = ________ joules
Now, let’s calculate the efficiency of the kettle to convert input energy to output
energy:
The highest water temperature reached was ______oC.
The original temperature of the water was ______oC, so the temperature gain
was
_____ (highest) - _____ (original) = _____oC
How much thermal energy was absorbed by the water?
The formula to calculate thermal efficiency is:
Thermal energy = volume of water x temperature change x 4.2 joules/ml – o C*
Thermal energy = 250 ml x _____
x 4.2 = ______ joules
*The 4.2 figure is a constant in this equation, just like pi is a constant value in
calculating circumference of a circle
Now, you complete the efficiency equation:
efficiency = useful energy output X 100
energy input
Useful energy input =
_________ joules
Energy input =
_________ joules
(_________divided by _________) = . _______ X 100 = _______
Therefore, the energy efficiency rating of the kettle is ____, or the water is
absorbing only _____% of the input energy.
How does this compare with a light bulb or gas furnace?
________________________________________________________________
Was all of the electrical energy supplied by the kettle transformed into thermal
energy in the water? Where else do you think the energy went?
________________________________________________________________
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