1. Heat Energy: Heat (thermal) energy is simply a type of energy
What is heat
energy?
Heat Energy:
Heat (thermal) energy is simply a type of energy
Ex. heat from a stove
Ex. heat from a light bulb
Heat energy can be transferred from one object to another
Ex. from a hot coffee cup to your cold hands

2. Heat Energy
It is possible to calculate the amount of heat energy absorbed or released by a given substance using the following formula:

3. Specific Heat Capacity
FORMULA
Specific Heat Capacity
(J/g•oC)
Q = mC∆T
Heat Energy
Joules (J)
Change in
Temperature
Degrees oC
mass
grams (g)

4. Specific Capacity (c)
specific heat capacity (c) - the amount of energy required to heat up or cool down a substance.
It is a characteristic property.
Every substance has their own specific heat capacity that does not change. However, water is most commonly used in exam questions.
Substances
Joules/(g• 0C)
Copper
0.383
Iron
0.452
Aluminum
0.896
Antifreeze
2.2
Methanol
2.547
Water
4.19

5. Example #1 Write Given Info. m = 500g Ti = 20oC Tf = 100oC
A pot with 500g of water is placed a stove. The temperature of the water increased from 20oC to 100oC. How much heat was absorbed by the water?
Write Given Info.
m = 500g
Ti = 20oC
Tf = 100oC
c = 4.19 J/goC
Q = mc∆T
Q = (500)(4.19)(100-20)
Q = 167,600 J
Recall: The “c” of water is 4.19 J/goC

6. Example #2 Write Given Info. m = 250g Ti = 5oC Tf = 25oC
A 250g glass of cold ice water is placed outside in the sun. The temperature of the water rises from 5oC to 25oC. How much heat was absorbed by the water?
Write Given Info.
m = 250g
Ti = 5oC
Tf = 25oC
c = 4.19 J/goC
Q = mc∆T
Q = (250)(4.19)(25-5)
Q = 20,950 J

7. Example #3 Write Given Info. m = 20g Ti = 21oC Tf = 85oC
A 20g iron frying pan is heated from 21oC to 85oC. If the specific heat capacity of iron is 0.45 J/goC, how much heat was absorbed by the firing pan?
Write Given Info.
m = 20g
Ti = 21oC
Tf = 85oC
c = 0.45 J/goC
Q = mc∆T
Q = (20)(0.45)(85-21)
Q = 576 J

8. Example #4
Note for water: 1mL = 1g
A 250mL of water at 30oC is heated until it boils at 100oC. How much heat was absorbed by the water?
m = 250 g
Ti = 30oC
Tf = 100oC
c = 4.19 J/goC
To go from volume to mass: times volume of water by the density of water. So,
1g/mL x 250 mL =
250 g
Q = (250)(4.19)(100-30)
Q = mc∆T
Q = 73,325 J

9. Example #5
Note for water: 1mL = 1g
A 1500mL of water at 5oC is heated until it reaches 37oC. How much heat was absorbed by the water?
m = 1500 g
Ti = 5oC
Tf = 37oC
c = 4.19 J/goC
Q = mc∆T
Q = (1500)(4.19)(37-5)
Q = 201,120J

10. Example #6
20,112J of energy is used to heat water from 2oC to 50oC. How many grams of water was heated?
Q = mc∆T
m = ? g
Ti = 2oC
Tf = 50oC
c = 4.19 J/goC
Q = 20, 112J
20,112 = (?)(4.19)(50-2)
20,112 = (?)(201.12)
20,112 = (?)(201.12)
m = 100 g

11. Example #7
360J of energy is used to heat a 20g piece of iron from 20oC to 60oC. What is the heat capacity of iron?
Q = mc∆T
m = 20 g
Ti = 20oC
Tf = 60oC
Q = 360J
c = ? J/goC
360 = (20)(?)(60-20)
360 = (800)(?)
360 = (800)(?)
c = 0.45 J/goC

12. Tricky
Question
Example # 8
A calorimeter was used to heat 300ml of water to 70oC. If the thermal (heat) energy was equal to J, what was the temperature of the water before being heated?
Q = mc∆T
= x 4.19(70 - Ti)
= (70 - Ti)
= – 1257Ti
= Ti
m = 300 g (remember 1 g = 1ml)
c = 4.19 Joules/(g• 0C)
Ti= ?
TF= 70 0C
Q = J
20 0C= Ti

13. Tricky
Question
Example # 9
A calorimeter was used to heat 400ml of water to 40oC. If the thermal (heat) energy was equal to 5.4 kJ, what was the initial temperature of the water before being heated?
Q = mc∆T
5400 = x 4.19(40 - Ti)
5400 = (40 - Ti)
5400 = – 1676Ti
= Ti
m = 400 g (remember 1 g = 1ml)
c = 4.19 Joules/(g• 0C)
Ti= ?
TF= 40 0C
Q = 5400 (5.4kJ = 5400 J)
36.8= Ti

14. KEY POINTS Q = mc∆T (J) (g)(J/g•oC)
Heat capacity of water is 4.19 J/g•oC
1 g of water = 1 ml of water
Q is always in Joule (j) not kJ!
Tf –Ti
(oC)