1. Chemical Thermodynamics

2. 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.

3. 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 represent everything in the universe.

4. 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

5. 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

6. 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”.

7. 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.

8. 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)

9. 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”

10. What do we know about those values????
A 40.0 gram block is heated to 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

11. -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 = J
Energy Gained is equal to the Energy Lost so… QH2O = J = -QMetal(s) = J

12. Plug that back into the first equation
J = (40.0 g) (CMetal(s))(30.0OC -100OC)
Why is the joules a negative value???
CMetal(s)= =
Sig Figs = J/g OC

13. Calorimetry Problem #2
Problem: A 30.0 gram sample of Silver is heated to 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 Ag
m = 75 g
T = 25oC
m = 30 g
T = 100oC

14. A point of clarification
You’re not solving for ΔT, instead we need the final temperature, Tf.
The warmer silver is going down from 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 OC
To ensure we get an answer between these values, you must make sure the Q’s have the correct sign. (Silver loses, Water gains)

15. -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?

16. Calorimeter - 7.08X + 708 = 313.5x – 7848.75 (Combine Terms)
- (30.0 g) (0.236 J/g OC) (x -100OC) =
(75.0 g) (4.186 J/g OC) (x – 25.0OC)
- 7.08X = 313.5x – (Combine Terms)
8556 = x
X = 26.7 OC – Final Temperature

17. 240. 0 g of water (initially at 20
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) = 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,H2O) (mass) (DT)
- [( 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 =
X = g Fe
Calorimetry Problems 2
question #5

18. A sample of ice at –12oC is placed into 68 g of water at 85oC
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
H2O
mass = ? g
Ti = 85oC
mass = 68 g
ice
Tf = 24oC
GAIN heat = - LOSE heat
qA = [(Cp,H2O) (mass) (DT)]
qA = [(2.077 J/goC) (mass) (12oC)]
24.9 m
mass x [ qA + qB + qC ] = - [(Cp,H2O) (mass) (DT)]
mass x [ qA + qB + qC ] = - [(4.184 J/goC) (68 g) (-61oC)]
qB = (Cf,H2O) (mass)
qB = (333 J/g) (mass)
333 m
458.2 mass =
458.2
qC = [(Cp,H2O) (mass) (DT)]
qC = [(4.184 J/goC) (mass) (24oC)]
100.3 m
m = g
qTotal = qA + qB + qC
458.2 m
Calorimetry Problems 2
question #13