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

2. 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

3. Heat & Enthalpy
Heat (q)– The flow of energy from higher temperature particles to lower temperature particles
Enthalpy (H)– Takes into account the internal energy of the sample along with pressure and volume
Under constant pressure (lab-top conditions), heat and enthalpy are the same…we’ll use the term “enthalpy”

4. 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)
These equivalents can be used in dimensional analysis to convert units

5. Heat Capacity
Specific Heat Capacity (Cp) – The amount of energy that can be absorbed before 1 g of a substance’s temperature has increased by 1°C
Cp for liquid water = 1.00 cal/g°C or 4.18 J/g°C

6. 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
A pool takes a long time to warm up and remains fairly warm over night.
The air warms quickly on a sunny day, but cools quickly at night
A cast-iron pan stays hot for a long time after removing from oven.
Aluminum foil can be grabbed by your hand from a hot oven because it cools so quickly

7. What things affect temperature change?
Heat Capacity of substance
The higher the heat capacity, the slower the temperature change
Mass of sample
The larger the mass, the more molecules there are to absorb energy, so the slower the temperature change
Specific heat capacity of substance
Energy added or removed
Change in temperature
Mass of sample

8. Positive & Negative DT
Change in temperature (DT) is always T2 – T1 (final temperature – initial temperature)
If temperature increases, DT will be positive
A substance goes from 15°C to 25°C.
25°C - 15°C = 10°C
This is an increase of 10°C
If temperature decreases, DT will be negative
A substance goes from 50°C to 35°C
35°C – 50°C = -15°C
This is a decrease of 15°C

9. Positive & Negative DH
Energy must be put in for temperature to increase
A “+” DT will have a “+” DH
Energy must be removed for temperature to decrease
A “-” DT will have a “-” DH

10. Example
Example:
If 285 J is added to 45 g of water at 25°C, what is the final temperature? Cp water = 4.18 J/g°C

11. Example
Example:
If 285 J is added to 45 g of water at 25°C, what is the final temperature? Cp water = 4.18 J/g°C
DH = change in energy
m = mass
Cp = heat capacity
DT = change in temperature (T2 - T1)
T2 = 27°C

12. Let’s Practice #1 Example:
How many joules must be removed from 25 g of water at 75°C to drop the temperature to 30°? Cp water = 4.18 J/g°C

13. 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

14. Let’s Practice #2 Example:
If the specific heat capacity of aluminum is J/g°C, what is the final temperature if 437 J is added to a 30.0 g sample at 15°C

15. Let’s Practice #2 Example:
If the specific heat capacity of aluminum is J/g°C, what is the final temperature if 437 J is added to a 30.0 g sample at 15.0°C
DH = change in energy
m = mass
Cp = heat capacity
DT = change in temperature (T2 - T1)
T2 = 31.2°C

16. Calorimetry

17. 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

18. Calorimetry
Calorimetry – Uses the energy change measured in the surroundings to find energy change of the system
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!

19. Two objects at different temperatures
Thermal Equilibrium – Two objects at different temperatures placed together will come to the same temperature
So you know that T2 for the system is the same as T2 for the surroundings!

20. 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°C water. If the final temperature of the water is 32.5°,what is the heat capacity of the metal?

21. 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°C water. If the final temperature of the water is 32.5°,what is the heat capacity of the metal?
Metal:
m = 23.8 g
T1 = 100.0°C
T2 = 32.5°C
Cp = ?
Water:
m = 50.0 g
T1 = 24°C
Cp = 4.18 J/g°C
Cp = 1.04 J/g°C

22. Let’s Practice #3 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°. What’s the final temperature?

23. Let’s Practice #3 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