1. THERMODYNAMICS
Intro & Calorimetry

2. Terms Thermodynamics: Energy: Heat:
the study of heat changes that occur during chemical changes
Energy:
the capacity to do work
only detected by the effects it causes
ex. potential energy, kinetic energy, heat
Heat:
type of energy & transferred from one object to another
can only measure its effects, for example change in temp

3. System: Surroundings: the part which we focus our attention
ex. chemicals
Surroundings:
everything else
ex. water bath, flask, atmosphere
Heat can flow from the system to the surroundings OR from the surroundings to the system.

4. Three types of systems: 1. Open System:
Lets energy and matter in an out
Ex. Open flask
2. Closed System:
Allows energy in and out but not matter
Ex. Flask with stopper
3. Isolated System:
Allows neither energy nor matter in or out
Ex. Calorimeter

5. Laws of Thermodynamics:
No energy is created or destroyed (Law of Conservation of Energy)
Heat will flow from a warmer body to a cooler body, until both bodies have the same temp

6. Exothermic Process: Endothermic Process:
a process that releases heat to the surroundings
heat “exits” the system
“Q” is negative
Ex. burning candles, condensation
Endothermic Process:
a process that absorbs heat from the surroundings
heat “enters” the system
“Q” is positive
Ex. liquid water to water vapor, cooking

7. Calorimetry Calorimeter: Bomb Calorimeter:
a device used to measure the absorption or release of heat in processes
Bomb Calorimeter:
a device used to measure the heat released by burning a compound

8. We know that the more heat an object absorbs, the higher its temp can be.
The mass of an object and its ability to absorb heat (heat capacity) also affects heat absorption. So,
Tf - Ti in oC
Q = m c ΔT
Heat (J)
Specific Heat Capacity (J / goC)
Mass (g)

9. Example:
If 125g of copper forms a layer on the bottom of a frying pan, how much heat is needed to raise the temp of the copper from 25oC to 300oC. The specific heat capacity for Cu is 0.39J/goC.
Q = m c ΔT
m = 125g
Ti = 25oC
Tf = 300oC
c = 0.39J/goC
QCu = (125g)(0.39J/goC)( 300oC - 25oC)
QCu = J or 1.34 x 104J
Heat is absorbed by the frying pan.

10. Recall H2O density of 1g/mL
Example:
HCl was added to of NaOH in a calorimeter, producing 50mL of water. At the start, the solutions were at 25oC. During the reaction, the highest temp observed was 32oC. What is the heat of the reaction?
m = 50g
Ti = 25oC
Tf = 32oC
c = 4.19 J/goC
Recall H2O density of 1g/mL
So, 50mL x 1g/mL = 50g

11. Q = m c ΔT
QH2O = (50g)(4.19J/goC)( 32oC - 25oC)
QH2O = J
Heat is absorbed by the water.

12. Heat is given off by the iron nail
Example:
Calculate the heat involved when a 5.5g iron nail is cooled from 37oC to 25oC. The heat capacity of iron is 0.45J/goC.
Q = m c ΔT
QFe = (5.5g)(0.45J/goC)( 25oC - 37oC)
QFe = J
m = 5.5g
Ti = 37oC
Tf = 25oC
c = 0.45 J/goC
Heat is given off by the iron nail