Transcript PowerPoint Presentation - Chapter 19 Chemical Thermodynamics

Chemical
Thermodynamics
Chemical
Thermodynamics
First Law of Thermodynamics
• The total energy of the universe
is a constant.
• Energy cannot be created and
destroyed, however, it can be
converted from one form to
another.
Chemical
Thermodynamics
• Thermodynamics is concerned with the flow
of heat from the system to it’s surroundings,
and vice-versa.
• In studying thermodynamics, try to identify
these two parts:
the system - the part on which you focus
your attention, where the change is
occuring.
the surroundings - includes everything else
in the universe.
Together, the system and it’s surroundings
Chemical
represent everything in the universe. Thermodynamics
Endo or Exo?
• Heat flowing into a system from it’s
surroundings:
Object is gaining energy
q has a positive value
called endothermic
• system gains heat (gets warmer)
as the surroundings cool down
Chemical
Thermodynamics
Endo or Exo?
• Heat flowing out of a system into it’s
surroundings:
Object is losing energy
q has a negative value
called exothermic
• system loses heat (gets cooler)
as the surroundings heat up
Chemical
Thermodynamics
Humor
A small piece of ice lived in a test tube,
and fell in love with a Bunsen burner.
One day it decided to profess it’s love…
“Bunsen! Oh Bunsen my flame! I melt
whenever I see you” said the ice.
The Bunsen burner replied” , “Don’t worry,
it’s just a phase you’re going through”.
Chemical
Thermodynamics
• Heat transfers
between the
system and the
surrounding from
hot objects to
cooler objects.
• Eventually they
will reach a point
of equilibrium;
where there is a
balance betweenChemical
the two parts. Thermodynamics
Calorimetry
• Calorimetry - the measurement of the
transfer of heat into or out of a system
for chemical and physical processes.
Based on the fact that heat released by
an system is equal to the heat absorbed
by the surrounding!
• Two different parts are involved, but
the energy(Q) will be the same for both
parts.(1st Law of Thermodynamics)
Chemical
Thermodynamics
Calorimeter
• The device used • The material inside the
to measure the
cup will be the “System”
absorption or
and the water will be the
release of heat in
“Surroundings”
chemical or
physical
processes is
called a
“Calorimeter”
Chemical
Thermodynamics
• A 40.0 gram block is heated to 100.0 OC. It is
then inserted into an insulated calorimeter
containing 25.0 grams of water at 26.6 OC.
• After stirring, the water equalizes at a
temperature of 30.0 OC. What is the specific
heat of the metal?
•
•
•
•
What are we solving for?
Who is losing energy?
Who is gaining energy?
What do we know about
those values????
Surroundings
SYSTEM
H2O
m = 25.0 g
Ti = 26.6oC
??
Chemical
m = 40.0 g
Thermodynamics
Ti = 100.0oC
-QMetal(s) = QH2O(L)
• -QMetal = (M)(CMetal(s))(ΔT)
• -QMetal = (40.0 g) (CMetal(s))(30.0 OC -100 OC)
• * I have two unknowns! Lets look at the water.
• QH2O = (M)(CH20(L))(ΔT)
• QH2O = (25.0 g)(4.186 J/g OC)(30.0OC – 26.6OC)
• QH2O = 355.81 J
• Energy Gained is equal to the Energy Lost so…
QH2O = 355.81 J = -QMetal(s) = - 355.81 J
Chemical
Thermodynamics
Plug that back into the first equation
• -355.81 J = (40.0 g) (CMetal(s))(30.0OC -100OC)
• Why is the joules a negative value???
• CMetal(s)= 0.127075 =
• Sig Figs = 0.127 J/g OC
Chemical
Thermodynamics
Calorimetry Problem #2
• Problem: A 30.0 gram sample of Silver is
heated to 100.0 OC. It is placed in 75.0
grams of water at 25.0 OC. Determine what
the final temperature will be.
• Who is losing energy?
• Who is gaining energy?
• What do we know about
the amount of energy for
both of them?
Surroundings
SYSTEM
H2O
m = 75 g
T = 25oC
Ag
Chemical
m = 30 g
Thermodynamics
T = 100oC
A point of clarification
• You’re not solving for ΔT, instead we need the
final temperature, Tf.
• The warmer silver is going down from 100.0 to
X. So its ΔT = (X – 100)
• The cooler water is going up starting at 25 and
going X. So its ΔT = (x – 25.0)
25.0 OC
X
100.0 OC
To ensure we get an answer between these
values, you must make sure the Q’s have the
Chemical
correct sign. (Silver loses, Water gains) Thermodynamics
-QAg(s) = QH2O(L)
• -QAg = (M)(CAg(s))(ΔT)
• -QAg =(30.0 g)(0.236 J/g OC)(X–100.0OC)
• * I have two unknowns!
• QH2O = (M)(CH20(L))(ΔT)
• QH2O = (75.0 g)(4.18 J/g OC)(X – 25.0OC)
• Still have two unknowns, but what do we
know about the two Q’s?
Chemical
Thermodynamics
Calorimeter
- (30.0 g) (0.236 J/g OC) (x -100OC) =
(75.0 g) (4.186 J/g OC) (x – 25.0OC)
- 7.08X + 708 = 313.5x – 7848.75
(Combine Terms)
8556 = 320.58x
X = 26.7 OC – Final Temperature
Chemical
Thermodynamics
240.0 g of water (initially at 20.0oC) are mixed with an unknown mass of
iron (initially at 500.0oC). When thermal equilibrium is reached, the
system has a temperature of 42.0oC. Find the mass of the iron.
C Fe(s) = 0.4495 J/g OC
C H2O(l) = 4.18 J/g OC
Fe
T = 500oC
mass = ? grams
T = 20oC
mass = 240 g
- LOSE heat = GAIN heat
- [(Cp,Fe) (mass) (DT)] = (Cp,H O) (mass) (DT)
2
- [(0.4495 J/goC) (X g) (42oC - 500oC)] = (4.184 J/goC) (240 g) (42oC - 20oC)]
Drop Units:
- [(0.4495) (X) (-458)] = (4.184) (240 g) (22)
205.9 X = 22091
X = 107.3 g Fe
Chemical
Thermodynamics
Calorimetry Problems 2
question #5
A sample of ice at –12oC is placed into 68 g of water at 85oC.
If the final temperature of the system is 24oC, what was the
mass of the ice?
H2O
T = -12oC
mass = ? g
Ti = 85oC
mass = 68 g
ice
GAIN heat = - LOSE heat
qA = [(Cp,H O) (mass) (DT)]
2
qA = [(2.077 J/goC) (mass) (12oC)]
qB = (Cf,H O) (mass)
qB = (333 J/g) (mass)
Tf = 24oC
24.9 m
mass x [ qA + qB + qC ] = - [(Cp,H O) (mass) (DT)]
2
mass x [ qA + qB + qC ] = - [(4.184 J/goC) (68 g) (-61oC)]
2
333 m
458.2 mass =
458.2
17339
458.2
qC = [(Cp,H O) (mass) (DT)]
2
qC = [(4.184 J/goC) (mass) (24oC)]
100.3 m
qTotal = qA + qB + qC
458.2 m
m = 37.8 g
Chemical
Thermodynamics
Calorimetry Problems 2
question #13