Energy Efficiency
Energy Efficiency
 the percentage of “wasted” energy to productive
(output) energy in your isolated system
(Inputs)
(Outputs)
(“Wasted” Energy)
Energy Efficiency Equation
 How much energy is required to change the temperature
 q=m(ΔT)c
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q=amount of energy (measured in Joules)
m=mass (measured in grams)
ΔT= change in temperature (measured in degrees Celsius)
c= specific heat (measured in cal/g°C)
Δ means “change in”
Specific Heat
 A certain number that tells us how much energy is required to
raise 1 gram of a substance’s temperature 1 degree.
 These numbers will always be the same for that substance.
 Example: cwater= 1 cal/g°C OR cwater=4.184 J/g°C
 calories is the unit specific for heat energy (note: it is a
lowercase c… this is important!)
 We can still use Joules but c is a different number
Back to our equation…

q=m(ΔT)c
 q=amount of energy (measured in Joules)
 m=mass (measured in grams)
 ΔT= change in temperature (measured in degrees Celsius)
 c= specific heat (measured in cal/g°C)
Example: How many calories of heat will need to be added to a
200g sample of water to raise its temperature 30°C?
Step 1: Write down your givens:
q= ?
m= 200g
ΔT= 30°C
c=1 cal/g°C
Step 2: Rewrite your equation:
q=m(ΔT)c
Just like 1 Kilogram equals 1000 grams!
Step 3: Substitute your givens for all variables possible:
q=(200)(30)(1)
Step 4: Solve the algebra problem:
q=6000 calories OR q=6 Calories (the large “C” is equal to 1000 “c”’s)
Example 2:
Randy has a 500 g of water at 20°C. If he wants the final
temperature of the water to be 75°C, how many Joules of heat will
he need to add?
Step 1: Write down your givens:
q= ?
m= 500g
ΔT= 75°C-20°C = 55°C
c=4.184 J/g°C
Step 2: Rewrite your equation:
q=m(ΔT)c
To find the CHANGE in the temperature we
subtract final temperature minus the initial
temperature.
We use 4.184 here because the problem
asked for how many JOULES of heat we
needed, not calories
Step 3: Substitute your givens for all variables possible:
q=(500)(55)(4.184)
Just like 1 Kilogram equals 1000 grams!
Step 4: Solve the algebra problem:
q= 115,060 Joules OR
115.06 Kilojoules
Example 3:
Harry has a 50g lump of unknown material. He experiments with the material and
determines that the temperature of the substance increases by 6°C when 20 cal of
heat are added. What is the substance’s specific heat?
Step 1: Write down your givens:
q= 20 cal
m= 50g
ΔT= 6°C
c=?
Step 2: Rewrite your equation:
q=m(ΔT)c
Step 3: Substitute your givens for all variables possible:
20=(50)(6)(c)
Step 4: Solve the algebra problem:
20 = 300c
300
300
0.067 cal/g°C = c