1. Section 7.2—Calorimetry & Heat Capacity
Why do some things get hot more quickly than others?

4. v

6. Endothermic Exothermic
system
system

8. Temperature
Temperature – proportional to the average kinetic energy of the molecules
Energy due to motion
(Related to how fast the molecules are moving)
As temperature increases
Molecules move faster

9. Heat & Enthalpy
Heat (q)– The flow of energy from higher temperature particles to lower temperature particles
Heat and Enthalpy are the same as long as you have constant pressure conditions. In this class q=H=Heat
Enthalpy (H)– heat changes under constant pressure conditions.

10. Energy Units
The most common energy units are Joules (J) and calories (cal)
Energy Equivalents =
4.18 J
1.00 cal
1000 J =
1 kJ =
1000 cal
1 Cal (food calorie)
1 kcal
These equivalents can be used in dimensional analysis to convert units

11. We can track the energy of a reaction on a graph…

15. Endothermic Exothermic
system
system

16. Energy Units
The most common energy units are Joules (J) and calories (cal)
Energy Equivalents =
4.18 J
1.00 cal
1000 J =
1 kJ =
1000 cal
1 Cal (food calorie)
1 kcal
These equivalents can be used in dimensional analysis to convert units

17. HOW TO ID AN ENDOTHERMIC RXN from an EXOTHERMIC RXN:
There are many ways chemists show the heat of a reaction. Please be familiar with all 3 of these! The first way is GRAPHICAL!

18. Other Ways to ID Endo Vs. Exo
Another clue for the endo vs. exo
When heat is on the LEFT or REACTANT side it is ENDO:
6.01kJ + H2O(s)  H2O(l)
When heat is on the RIGHT or PRODUCT side it is EXO:
H2O(l)  H2O(s) kJ

19. Other Ways to ID Endo Vs. Exo
Yet another way to Endo and Exo!
H2O(s)  H2O(l) ΔH = kJ ENDO
H2O(l)  H2O(s) ΔH = kJ EXO
The + is used to indicate that energy is ADDED to the system for endo!
The – is used to indicate that energy is RELEASED or leaving the system for exo!

20. Releasing the joules/calories in a gummy bear

21. Heat Capacity
Heat Capacity – The amount of energy required to get a sample to move up by 1°C. (This changes for one substance depending on how much)
Specific Heat Capacity (Cp) – The amount of energy required to get 1 g of a substance to increase by 1°C (This stays the same for one substance because it is for a set amount.)
Cp for liquid water = 1.00 cal/g°C or 4.18 J/g°C

24. Ans: LOW – The Desert can go from 32 °F at night and 113 °F at day.
Heat Capacity
High Heat Capacity
Low Heat Capacity
Takes a large amount of energy to noticeably change temp
Small amount of energy can noticeably change temperature
Heats up slowly
Heats up quickly
Cools down slowly
Cools down quickly
Maintains temp better with small condition changes
Quickly readjusts to new conditions
Do you think Air has a high or low heat capacity? Why or Why Not?
Ans: LOW – The Desert can go from 32 °F at night and 113 °F at day.
Do you think the ocean has a high or low heat capacity? Why or Why Not?
Ans: HIGH - The heat capacity of ocean water is about four times that of air.
Why can you grab Aluminum foil out of the oven but not an aluminum cookie sheet?
Ans: Both have the same specific heat (about 0.9J/g°C) but the cookie pan is a much greater mass.

25. How can we calculate the amount of energy we put into a sample or take out?
This shows that how much energy something has absorbed or released is dependent upon three things
The Specific Heat Capacity of the substance (Cp)
The Mass of the substance
The temperature change of the substance (∆T = T2 - T1)
Specific heat capacity of substance
Energy added or removed
Change in temperature
Mass of sample

26. Let’s Practice #1 Example:
How many joules must be removed from 25.0 g of water at 75.0°C to drop the temperature to 30.0°? Cp water = 4.18 J/g°C

27. DH = - 4702.5 J BUT SIG FIGS WOULD BE
Let’s Practice #1
Example:
How many joules must be removed from 25.0 g of water at 75.0°C to drop the temperature to 30.0°? Cp water = 4.18 J/g°C
DH = change in energy
m = mass
Cp = heat capacity
DT = change in temperature (T2 - T1)
DH = J BUT SIG FIGS WOULD BE
DH = x 103 J

28. See page 259 for sample heat capacity calculations

29. Calorimetry

30. Calorimetry
Calorimetry – the science of heat flow between the system and the surroundings
Because of the Law of Conservation of Energy, the energy lost/gained by the surroundings is equal to but opposite of the energy lost/gained by the system.
DHsurroundings = - DHsystem
(m×Cp×DT)surroundings = - (m×Cp×DT)system
Don’t forget the “-” sign on one side
Make sure to keep all information about surroundings together and all
information about system together—you can’t mix and match!

31. Conservation of Energy
1st Law of Thermodynamics – Energy cannot be created nor destroyed in physical or chemical changes
This is also referred to as the Law of Conservation of Energy
If energy cannot be created nor destroyed, then energy lost by the system must be gained by the surroundings and vice versa
Qsystem = - Qsurrounding

33. Here’s a little tip to remember with calorimeters!
Thermal Equilibrium – Two objects at different temperatures placed together will come to the same temperature
Why is this important?
So you know that T2 for the system is the same as T2 for the surroundings!

35. Let’s Practice Example:
A 23.8 g piece of unknown metal is heated to 100.0°C and is placed in 50.0 g of water at 24°C water. If the final temperature of the water is 32.5°,what is the heat capacity of the metal?

36. An example of Calorimetry
A 23.8 g piece of unknown metal is heated to 100.0°C and is placed in 50.0 g of water at 24.0°C water. If the final temperature of the water is 32.5°,what is the heat capacity of the metal?
Metal:
Water:
M = 23.8 g
M = 50.0 g
T1 = °C
T1 = 24.0 °C
T2 = 32.5 °C
Cp = ?
Cp = 4.18 J/g°C
Cp = 1.10 J/g°C

37. Another example of Calorimetry
A 10.0 g of aluminum (specific heat capacity is J/g°C) at 95.0°C is placed in a container of g of water (specific heat capacity is 4.18 J/g°C) at 25.0°. What’s the final temperature?

38. Example:
A 10.0 g of aluminum (specific heat capacity is J/g°C) at 95.0°C is placed in a container of g of water (specific heat capacity is 4.18 J/g°C) at 25.0°C. What’s the final temperature?
Metal:
m = 10.0 g
T1 = 95.0°C
T2 = ?
Cp = J/g°C
Water:
m = g
T1 = 25.0°C
Cp = 4.18 J/g°C
T2 = 26.5 °C