Chapter 3
Specific Heat
Part 1 and Part 2
Specific Heat
• Different substances require different
amounts of heat to change their temperature.
• In general the specific heat of a substance
indicates how hard something is to heat up
or cool down.
• Scientifically speaking the specific heat is the
amount of heat required to change the
temperature of 1 gram of a substance by 1ºC.
•Which substance is the easiest to
heat up or cool down?
→
•Which substance is the easiest to
heat up or cool down?
•Which substance is the hardest to
heat up or cool down?
→
•Which substance is the hardest to heat up or cool down?
→
•Which substance is the hardest to heat up or cool down?
•How might we observe the high specific heat of water?
“Flying Freddie Spooner” 2014
Tri-Valley Cannonball Champion
→
•Scientific interpretation of the specific heat of water.
The specific heat of water
• The specific heat of
water is 4.184 J/g•ºC.
• To simplify our
discussion we will
round this to 4 J/g•ºC.
The specific heat of water
• If we have one gram
of water at 25ºC and
we add 4J of energy
to it what will the
temperature be?
• Answer: 26ºC
The specific heat of water
• If we have two
grams of water at
30ºC and we remove
16J of energy from it
what will the
temperature be?
• Answer 28ºC
The specific heat of water
• The high specific
heat of water has
many implications
to our everyday
lives.
Importance of Water’s High
Specific Heat
• Maintenance of Body
Temperature
Importance of Water’s High
Specific Heat
• Heat Storage
on a Larger
Scale.
Importance of Water’s High
Specific Heat
• Temperature
fluctuations
that permit
life.
Homework
 Worksheet Chapter 3 – 2 (due tomorrow)
 Study Guide Chapter 3 (due Friday)
How Much Heat?
• I want to heat some
water.
• What do I need to
know in order to
determine the amount
of heat required?
• Mass of water.
• Temperature Change
• Specific Heat of Water
H = m ∆T Cp
Heat = (mass) x (change in temperature) x (specific heat)
Kilojoules
• kilo (k) = 1000
• kJ = 1000 J
• 4.08 kJ = 4080 J
•How much heat is necessary to heat 258g of
water from 25.0ºC to 100.0ºC?
(258g)(75.0ºC)(4.184J/gºC) = 80,960.4J = 81,000J
= 81.0 kJ
•How much heat must be removed to cool
1058g of copper from 325.0ºC to 18.0ºC?
(1058g)(307.0ºC)(0.385J/gºC) = 125,050J = 125,000J
= 125 kJ
How “much” water do I have?
52.8 ml =
52.8 cm3
or 52.8 g
• The mass of an unknown metal is 14.9 g. It is heated
to 100.0C and dropped into 75.0 mL of water at
20.0C. The final temperature of the system is
28.5C. What is the specific heat of the metal?
• The mass of unknown metal is 17.19 g. It is heated
to 100.00C and dropped into 25.00 mL of water at
24.50C. The final temperature of the system is
30.05C. What is the specific heat of the metal?
Heat gained by the water = Heat lost by the metal
m ∆T Cp = m ∆T Cp
(25.00g)(5.55°C)(4.184 J/g C) = (17.19g)(69.95°C)(x)
C p of metal =
q
m x DT
=
580.53J
o
(17.19g )(69.95 C)
Cp = 0.4828 J/g · °C
=
0.48279
J
g oC
Homework
 Worksheet Chapter 3 – 3 (due tomorrow)
 Study Guide Chapter 3 (due Thursday)