Different forms of energy include electrical, heat, work, potential, and kinetic energy. Heat energy can be calculated using the formula Q=mcΔT, where Q is heat energy in Joules, m is mass in grams, c is specific heat capacity in J/g°C, and ΔT is the change in temperature in °C. Potential energy is calculated as PE=mgh, where m is mass in kg, g is acceleration due to gravity (9.81 m/s2), and h is height in meters. Kinetic energy is calculated as KE=1/2mv2, where m is mass in kg and v is velocity in m/s. Total mechanical energy is the sum of

1. Different Forms of Energy
Objective: Calculate energy in it
different forms:
Electrical
Heat
Work
Potential
Kinetic

2. Q = mC∆T
Heat
Energy
Joules (J)
mass
grams
(g)
Change in
Temperature
Degrees
o
C
Specific
Heat
Capacity
(J/g•o
C)

3. Specific Capacity (c)
Specific Heat Capacity (c) - the amount of
energy absorbed by a substance or
object.
It is a characteristic property.
Substances Joules/(g• 0
C)
Copper 0.383
Iron 0.452
Aluminum 0.896
Antifreeze 2.2
Methanol 2.547
Water 4.19

4. A pot with 500g of water is placed on a
stove. The temperature of the water
increased from 20o
C to 100o
C. How much
heat was absorbed by the water?
m = 500g
Ti = 20o
C
Tf = 100o
C
c = 4.19 J/go
C
Q = mc∆T
Q = (500)(4.19)(100-20)
Q = 167,600 J
Recall: The “c” of water
is 4.19 J/go
C

5. A 250g glass of cold ice water is placed
outside in the sun. The temperature of the
water rises from 5o
C to 25o
C. How much
heat was absorbed by the water?
m = 250g
Ti = 5o
C
Tf = 25o
C
c = 4.19 J/go
C
Q = mc∆T
Q = (250)(4.19)(25-5)
Q = 20,950 J

6.  Q = mc∆T
(J) (g)(J/g•o
C)
o Heat capacity of water is 4.19 J/g•o
C
o 1 g of water = 1 ml of water
o Q is always in Joule (j) not kJ!
Tf –Ti
(o
C)

7. E = P•t
Energy
Joules (J)
Power
Watts (W)
Time
Seconds (s)
Recall:
P = V•I
Power
Watts (W)
Potential
Difference
Volts (V)
Current
Amps (A)
If we combine the two previous formulas, we get….
E = • tPV•I

8. E =V•I•t
Recall: Q = mc∆T
Heat Energy
Joules (J)
Mass
(g)
Heat Capacity
(J/g•o
C)
Temperature
(o
C)
Heat energy (Q) is a form of energy.
So using the previous energy formula (E=VIt) &
the heat energy formula (Q=mc∆T) we get…
Q =E = QV•I•t mc∆Tmc∆T
=

9. Jason heats 300 grams of distilled
water in an electric calorimeter from 20o
c to
46o
c. He notes that the potential difference
across the terminals of the power supply of
the calorimeter is 15 Volts. If the current
flows for 25 minutes, what is the current
intensity?
GIVEN INFORMATION
m = 300 g
c = 4.19J/go
C
Ti = 20 o
C
Tf = 46 o
C
V = 15 V
I = ???
t = 25 min or 1500s
V•I•t = mc∆T
15•I•1500 = 300(4.19)(46-20)
(22500)I = 32682
I = 1.45 A
E = Q

10. How long will it take a heating
coil with a resistance of 14 ohms,
connected to a 2.3 V battery to bring
360 grams of distilled water from 12o
C
to 95o
C?
GIVEN INFORMATION
m = 360 g
c = 4.19J/go
C
Ti = 12 o
C
Tf = 95 o
C
V = 2.3 V
I = ???
t = ???
You need to solve for time .
You first need “I”… SO,
I
V
R =
I
3.2
14 = A0.164I =
Plug in information and solve for
time….
E = Q
V•I•t = mc∆T
2.3•0.164•t = 360(4.19)(95-12)
(0.3772)t = 125197.2
t = 331911.98 seconds or 5531.87 minutes

11. Key Points to Remember:
E = Q so…
V•I•t = mc∆T
 Remember:
 Potential difference: measured in Volts
 Current Intensity: measured in Amps
 Time: measured in Seconds
 Mass: measured in Grams
 Heat Capacity: measure in J/o
C•g
 Temperature: measure in o
C

12. Potential Energy
• If we lift up an object against gravity, it now
has the ability to move; it has the potential
to fall down and use up the energy we put
into it.
• Ep , PE = mgh, unit is Joules, J
• m is the mass in kg
• g is the acceleration due to gravity, 9.81N/kg
• h is the height above the Earth’s surface, m.

13. Activity
• What is the PE of a 10 kg weight, 8 m
above the ground?
• PE = mgh
• = 10 kg x 9.81N/kgx 8m
• = 784.8 J
• Calculate the PE of a 20 kg weight, 20 m
above the ground.

14. Kinetic Energy
• Kinetic Energy is the energy an object has
due to its motion.
• The KE depends on the mass and the
speed.
• Ek or KE= ½ mv2
,
• E is Energy in Joules, J
• m is mass in kg, v is velocity in m/s

15. Example
• What is the KE of a 6 kg curling stone
moving at 4 m/s?
• KE = ½ mv2
• = ½ x 6kg x (4 m/s)2
• = ½ x 6 x 16
• = 48 J
• Calculate the KE of a 5 kg ball moving at
20 m/s.

16. Total Mechanical Energy
• The energy of a system transfers between
Potential Energy and Kinetic Energy.
• Total Energy = PE + KE
• The PE of an object gets
transferred to KE as it
speeds up.
• As the PE decreases, the
KE increases.

17. Total Mechanical Energy
• What is the speed of a 500g rock that
drops from a height of 78.4 m, just before
it hits the ground?
• ET = KE + PE, at first, v = 0 m/s
• = ½ mv2
+ mgh, since v = 0, KE = 0
• = 0.5kgx9.81m/s2
x78.4m, ET = PE only
• = 384.6 J
• As the rock approaches the ground all its
PE is transferred to KE, so PE = 0. So…

18. Total Energy, Part Deux
• ET = PE + KE
• 384.6 J = KE
• 384.6 = ½ mv2
• 384.6 = 1/2x 0.5kg x v2
• 1538.4 J = v2
• v = 39.2 m/s
• So just before it hits the ground, the rock
has a speed of 39.2 m/s

19. Measuring Work
• Work is defined as the energy that comes
from applying a force in one direction over
a certain distance.
• W = F Δd = mad (horizontal)
• = magd (against gravity)
• Work is in Joules, J
• Force is in Newtons, N
• Distance is in metres, m

20. Activity
• E.g How much work is done by a boy
pushing a car with a force of 800 N over a
distance of 200m?
• W = F d
• = 800 N x 200 m
• = 160 000 J
• = 160 kJ

21. Exam Question
A 200 g brick falls from a wall 4.0 metres above the ground. It hits the ground with a
velocity of 8.5 m/s.
4.0 m
How much work did gravity do on the brick?
A) 8.0 J
B) 7.2 J
C) 3.4 J
D) 1.7 J

22. Effective Force
• Work Energy is calculated using the
Effective Force, the Force in the direction
of motion.

23. Question
• Calculate the work energy exerted by a
girl pulling a wagon with a force of 40 N at
an angle of 60° over a distance of 75 m.
• W = Fd
• = 40 cos 60 (75 m)
• = 40 x 0.5 x 75
• = 1500 J