1. Calorimetry
Chapter 5

2. Formal assessment – analyze student responses to the quiz and problems
Objectives
Today I will be able to:
Apply knowledge of the polyatomic ions to a quiz
Distinguish between the terms heat capacity, specific heat capacity and molar heat capacity.
Solve calorimetry problems.
Informal assessment – monitoring student questions as they complete the practice problems
Formal assessment – analyze student responses to the quiz and problems
Common Core Connection
Make sense of problems and persevere in solving them
Build strong content knowledge

3. Lesson Sequence Evaluate: Polyatomic Ions Quiz
Explain: Calorimetry Notes
Elaborate: Calorimetry Practice Problems
Evalaute: Exit Ticket

4. Warm Up Take everything off of your desk for your polyatomic ions quiz
Place your electronic device in the bin on Ms. Ose’s desk with a sticky note
Make sure your device is on silent

5. objectives Today I will be able to:
Apply knowledge of the polyatomic ions to a quiz
Distinguish between the terms heat capacity, specific heat capacity and molar heat capacity.
Solve calorimetry problems.

6. Homework Bring lab notebook to school Thursday Reading
Read the Calorimetry section (5.5) in the textbook before Monday (Chapter 5)
Read through the Handwarmer Lab before Monday
Extra Calorimetry Practice Problems
Solubility Rules Quiz Thursday, September 18

7. AGENDA Polyatomic Ions Quiz Calorimetry Notes
Calorimetry Practice Problems
Exit Ticket

8. Calorimetry
Chapter 5

9. Calorimetry
Calorimetry is the study of the heat released or absorbed during physical and chemical reactions.
For a certain object, the amount of heat energy lost or gained is proportional to the temperature change.
The initial temperature and the final temperature in the calorimeter are measured and the temperature difference is used to calculate the heat of reaction.

10. Heat Capacity Symbol = C
is the amount of energy required to raise the temperature of an object 1 Kelvin or 1 °C
Units = J/K or J/ °C

11. Heat Capacity
Heat capacity is an extensive property, meaning it depends on the amount present
a large amount of a substance would require more heat to raise the temperature 1 K than a small amount of the same substance.
Heat capacity depends upon the amount of the substance you have.

12. Molar Heat Capacity = C molar.
Symbol = C molar.
For pure substances, the heat capacity for one mole of the substance may be specified as the molar heat capacity.
Units = C/moles
q = molar heat x moles x DT

13. Specific Heat Capacity = c or s
Symbol = c or s
The specific heat capacity, c or s, is often used since it is the heat capacity per one gram of the substance
Units=J/gK or J/gC.
The specific heat capacity of each substance is an intensive property which relates the heat capacity to the mass of the substance.

14. Remember…
An extensive property is a property that changes when the size of the sample changes.
Examples are mass, volume, length, and total charge.
An intensive property doesn't change when you take away some of the sample.
Examples are temperature, color, hardness, melting point, boiling point, pressure, molecular weight, and density.

15. Specific Heat Capacity “c or s”
q = mcDT
This is the main equation for calorimetry calculations
mass will be in grams
Units for “c” are J/g K or J/g °C
The specific heat of water is 1 cal/g ºC

16. Is q positive or negative?
If a process results in the sample losing heat energy, the loss in heat is designated as - q.
The temperature of the surroundings will increase during this exothermic process
If the sample gains heat during the process, then q is positive.
The temperature of the surroundings will decrease during an endothermic process

17. Direct proportionality
The amount of heat that an object gains or loses is directly proportional to the change in
temperature.
remember q = mcΔT
q and T are directly proportional

18. Dt = tfinal – tinitial = 50C – 940C = -890C
How much heat is given off when an 869 g iron bar cools from 940C to 50C?
s of Fe = J/g • 0C
Dt = tfinal – tinitial = 50C – 940C = -890C
q = msDt
= 869 g x J/g • 0C x –890C
= -34,000 J
6.4

19. Try this Problem The specific heat of graphite is 0.71 J/gºC.
Calculate the energy needed to raise the temperature of 75 kg of graphite from 294 K to 348 K.

20. Answer to Problem
q = mcΔT 348 K K= q = (75 kg ) (1000g/kg) (0.71 J/gºC)(54 ºC) q = J or kJ

21. Extra Problem
A 46.2 g sample of copper is heated to 95.4ºC and then placed in a calorimeter containing g of water at 19.6ºC. The final temperature of both the water and the copper is 21.8ºC.
What is the specific heat of copper?

22. Answer to Problem
qcu = mcΔT qh2o = mcΔT mcΔT = qcu = qh2o = mcΔT (46.2 g) ccu (73.6 °C) = (75 g) (4.184 J/g°C) (2.2°C) ccu = 0.20 J/g°C

23. Summary of Definitions:
Heat capacity is the amount of energy required to raise the temperature of an object 1 kelvin or 1 °C.
Specific heat capacity is the heat capacity of 1 gram of a substance.
Molar heat capacity is the heat capacity of 1 mole of a substance.

24. Heat (q) absorbed or released:
The specific heat (s) of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius.
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius.
C = ms
Heat (q) absorbed or released:
q = msDt
q = CDt
Dt = tfinal - tinitial
6.4

25. Practice problems Try the calorimetry problems at the end of Ch 5
# 51,53, 55, 57

26. Wrap up
Calorimetry is an experimental way of determining heat values for substances.
Please make sure that you have a bound composition notebook. We will perform a lab on Monday.
You will need to wear close toed shoes.