Chemical Thermodynamics

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Chemical Thermodynamics
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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.

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• Thermodynamics is concerned with the flow of heat
  from the system to its 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 its surroundings
  represent everything in the universe.

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Endo or Exo?

• Heat flowing into a system from its
  surroundings
• Object is gaining energy
• q has a positive value
• called endothermic
• system gains heat (gets warmer) as the
  surroundings cool down

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Endo or Exo?

• Heat flowing out of a system into its
  surroundings
• Object is losing energy
• q has a negative value
• called exothermic
• system loses heat (gets cooler) as the
  surroundings heat up

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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 its love
• Bunsen! Oh Bunsen my flame! I melt whenever I
  see you said the ice.
• The Bunsen burner replied , Dont worry, its
  just a phase youre going through.

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• 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 between the
  two parts.

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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)

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Calorimeter

• The device used to measure the absorption or
  release of heat in chemical or physical processes
  is called a Calorimeter

• The material inside the cup will be the System
  and the water will be the Surroundings

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• 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
m 40.0 g Ti 100.0oC
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-QMetal(s) QH2O(L)

• -QMetal (M)(CMetal(s))(?T)
• -QMetal (40.0 g) (CMetal(s))(30.0 OC -100 OC)
• I have two unknowns! Lets look at the water.

• QH2O (M)(CH20(L))(?T)
• 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
  soQH2O 355.81 J -QMetal(s) - 355.81 J

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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

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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 aboutthe amount of energy
  forboth of them?

Surroundings
SYSTEM
H2O
Ag
m 75 g T 25oC
m 30 g T 100oC
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A point of clarification

• Youre not solving for ?T, instead we need the
  final temperature, Tf.
• The warmer silver is going down from 100.0 to X.
  So its ?T (X 100)
• The cooler water is going up starting at 25 and
  going X. So its ?T (x 25.0)
• 25.0 OC X 100.0 OC
• To ensure we get an answer between these values,
  you must make sure the Qs have the correct sign.
  (Silver loses, Water gains)

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-QAg(s) QH2O(L)

• -QAg (M)(CAg(s))(?T)
• -QAg (30.0 g)(0.236 J/g OC)(X100.0OC)
• I have two unknowns!

• QH2O (M)(CH20(L))(?T)
• QH2O (75.0 g)(4.18 J/g OC)(X 25.0OC)
• Still have two unknowns, but what do we know
  about the two Qs?

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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

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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
mass ? grams
-
LOSE heat GAIN heat
- (Cp,Fe) (mass) (DT) (Cp,H2O) (mass) (DT)
- (0.4495 J/goC) (X g) (42oC - 500oC)
(4.184 J/goC) (240 g) (42oC - 20oC)
- (0.4495) (X) (-458) (4.184) (240 g) (22)
Drop Units
205.9 X 22091
X 107.3 g Fe
Calorimetry Problems 2 question 5
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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?
T -12oC
mass ? g
H2O
ice
Tf 24oC
GAIN heat - LOSE heat
qA (Cp,H2O) (mass) (DT)
mass x qA qB qC - (Cp,H2O)
(mass) (DT)
qA (2.077 J/goC) (mass) (12oC)
24.9 m
mass x qA qB qC - (4.184 J/goC)
(68 g) (-61oC)
qB (Cf,H2O) (mass)
333 m
qB (333 J/g) (mass)
458.2 mass 17339
qC (Cp,H2O) (mass) (DT)
m 37.8 g
qC (4.184 J/goC) (mass) (24oC)
100.3 m
qTotal qA qB qC
458.2 m
Calorimetry Problems 2 question 13