1. Warm-up… EXPLAIN HOW THE ICE CREAM LAB WORKS
What does the salt do (collig. Pro and temp gap)
How does the ice melt– explain the heat movement

2. Calorimetry
CP Unit 9
Chapter 17

3. CALORIMETRY
The enthalpy change associated with a chemical reaction or process can be determined experimentally.
Measure the heat gained or lost during a reaction at CONSTANT pressure

4. Calorimeter
Device used to measure the heat absorbed or released during a chemical or physical process
Styrofoam cup

5. Example
If you leave your keys and your chemistry book sitting in the sun on a hot summer day, which one is hotter?
Why is there a difference in temperature between the two objects?

6. Because…
Different substances have different specific heats (amount of energy needed to raise the temperature of 1 g of a substance by 1 degree Celsius).

7. Need to know how to calc. heat (review)
Heat (q) = mCT
If the specific heat of Al is 0.90 J/gC, how much heat is required to raise the temperature of 10,000 g of Al from 25.0 C to 30.0 C?
HIDDEN

8. Calculator says 45,000 J BUT 1 sig fig…
HIDDEN

9. For example (review): HIDDEN
If 418 J is required to increase the temperature of 50.0 g of water by 2.0 C, what is the specific heat of water?
HIDDEN

10. What happens in a calorimeter
One object will LOSE heat, and the other will ABSORB the heat
System loses heat to surroundings = EXO = -q
System absorbs heat from surroundings = ENDO = +q
When a hot chunk of metal is dropped in a cool glass of water, the metal cools off. Where did the heat from the metal go?
Did the metal lose more heat then the water gained?
HEAT GAINED = HEAT LOST (ALWAYS!)

11. To do calorimetry problems…
The numbers in these two boxes are always the same, but with different signs (+/-). What heat one lost, the other gained.
Make a Chart:
Water
Object/Reaction
Heat
Mass
Specific Heat
4.184
Final Temp
Initial Temp

12. EXAMPLE: A small pebble is heated and placed in a foam cup calorimeter containing 25.0 g of water at 25.0 C. The water reaches a maximum temperature of 26.4 C. How many joules of heat were released by the pebble? The specific heat of water is J/g C. (#13)
Water
Pebble
Heat
Mass
Specific Heat
4.184
Final Temp
Initial Temp
The numbers in these two boxes are always the same, but with different signs (+/-). What heat one lost, the other gained.
25.0 g
26.4 oC
25.0 oC

13. The pebble because the water heated up from 25.0 C to 26.4 C.
Pebble loses heat (-q, exothermic) while water gains heat (+q, endothermic)
Do you calculation based on water (since the problem gave all the water’s information) 

14. Water Pebble Heat Mass Specific Heat 4.184 Final Temp Initial Temp
25.0 g
26.4 oC
25.0 oC
qwater = mwatercwaterTwater
qwater = (25.0g)(4.184J/goC)(26.4oC-25.0oC)
qwater = 150 J
If the water ABSORBED 150 J of heat,
then the pebble _______ ______ J of heat.
qpebble = J

15. Example 2 (LAB type of CALC)
Suppose that g of water at 22.4 °C is placed in a calorimeter. A g sample of Al is removed from boiling water at a temperature of 99.3 °C and quickly placed in a calorimeter. The substances reach a final temperature of 32.9 °C . Determine the SPECIFIC HEAT of the metal.The specific heat of water is J/g C.
 MAKE YOUR CHART

16. Suppose that 100. 00 g of water at 22. 4 °C is placed in a calorimeter
Suppose that g of water at 22.4 °C is placed in a calorimeter. A g sample of Al is removed from boiling water at a temperature of 99.3 °C and quickly placed in a calorimeter. The substances reach a final temperature of 32.9 °C . Determine the SPECIFIC HEAT of the metal.The specific heat of water is J/g C.
Make chart
Calculate q for water
Q for water is the same (but with different sign) as q for metal.
Using q metal, calculate c metal
Water
Pebble
Heat
Mass
Specific Heat
4.184
Final Temp
Initial Temp g
75.25 g
32.9 oC
32.9 oC
22.4 oC
99.3 oC

17. 0.879 J/g °C

18. Example 3
A lead mass is heated and placed in a foam cup calorimeter containing 40.0 g of water at 17C. The water reaches a temperature of 20.0 C.
How many joules of heat were released by the lead?
The specific heat of water is J/g C.

19. 502 J = 5.0 x 102 J for 2 sig figs