Post on 26-Jun-2015
Module 2 – Heat Energy
Objectives5.2 Calculate heat energy, using a
mathematical relationship.
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
What is heat energy?
Q = mC∆T
Heat Energy
Joules (J)
mass
grams (g)
Change in
Temperature
Degrees
oC
Specific Heat
Capacity
(J/g•oC)
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• 0C)
Copper 0.383
Iron 0.452
Aluminum 0.896
Antifreeze 2.2
Methanol 2.547
Water 4.19
Every substance has their own specific heat capacity that does not change. However, water is most commonly used in exam
questions.
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?
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
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?
m = 250g
Ti = 5oC
Tf = 25oC
c = 4.19 J/goC
Q = mc∆T
Q = (250)(4.19)(25-5)
Q = 20,950 J
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?
m = 20g
Ti = 21oC
Tf = 85oC
c = 0.45 J/goC
Q = mc∆T
Q = (20)(0.45)(85-21)
Q = 576 J
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
Q = mc∆T
Q = (250)(4.19)(100-30)
Q = 73,325 J
To go from volume to mass: times volume of water by the
density of water. So,
1g/mL x 250 mL = 250 g
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
Note for water: 1mL = 1g
20,112J of energy is used to heat water from 2oC to 50oC. How many grams of water was heated?
m = ? g
Ti = 2oC
Tf = 50oC
c = 4.19 J/goC
Q = 20, 112J
Q = mc∆T
20,112 = (?)(4.19)(50-2)
m = 100 g
20,112 = (?)(201.12)
20,112 = (?)(201.12) 201.12 201.12
360J of energy is used to heat a 20g piece of iron from 20oC to 60oC. What is the heat capacity of iron?
m = 20 g
Ti = 20oC
Tf = 60oC
Q = 360J
c = ? J/goC
Q = mc∆T
360 = (20)(?)(60-20)
c = 0.45 J/goC
360 = (800)(?)
360 = (800)(?) 800 800
A calorimeter was used to heat 300ml of water to 70oC. If the thermal (heat) energy was equal to 62850 J, what was the final temperature of the water before being heated?
m = 300 g (remember 1 g = 1ml)
c = 4.19 Joules/(g• 0C)
Ti= ?
TF= 70 0C
Q = 62850 J
Q = mc∆T62850 = 300 x 4.19(70 - Ti)62850 = 1257(70 - Ti)62850 = 87990 – 1257Ti
-25140 = - 1257Ti
20 0C= Ti
TrickyQuestion
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?
m = 400 g (remember 1 g = 1ml)
c = 4.19 Joules/(g• 0C)
Ti= ?
TF= 40 0C
Q = 5400 (5.4kJ = 5400 J)
Q = mc∆T5400 = 400 x 4.19(40 - Ti)5400 = 1676(40 - Ti)5400 = 67040 – 1676Ti
-61640 = - 1676Ti
36.8= Ti
TrickyQuestion
Q = mc∆T (J) (g)(J/g•oC)
o Heat capacity of water is 4.19 J/g•oC
o 1 g of water = 1 ml of water
o Q is always in Joule (j) not kJ!
Tf –Ti
(oC)
Activities
Worksheet # 17
First Law of Thermodynamics
This law states that “in an isolated system energy remains constant”.
This means that energy can be transformed (neither created nor destroyed).
Example: When you eat a candy bar the bar is transformed into energy within your body.
Different Types of Energy
Thermal
Light
Electrical
Different Types of Energy
Magnetic
Nuclear
Kinetic
Potential
Chemical
Sound
Aeolian
Different Types of Energy
Tidal
Geothermal
Hydro
Solar
Different Types of Energy