Transcript Slide 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
Calorimetry
CP Unit 9 Chapter 17
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
Calorimeter
Device used to measure the heat absorbed or released during a chemical or physical process Styrofoam cup
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?
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).
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!)
The numbers in these two boxes are
To do calorimetry problems…
(+/-). What heat one lost, the other gained.
Make a Chart: Water Object/Reaction Heat Mass Specific Heat Final Temp Initial Temp 4.184
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 4.184 J/g
C.
Water Pebble Heat Mass Specific Heat Final Temp Initial Temp
25.0 g
4.184
26.4 o C 25.0 o C
(#13) The numbers in these two boxes are always the same, but with different signs (+/-). What heat one lost, the other gained.
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)
Water Pebble Heat Mass Specific Heat Final Temp Initial Temp
25.0 g
4.184
26.4 o C 25.0 o C
q water = m water c water T water q water = (25.0g)(4.184J/g o C)(26.4
o C-25.0
o C) q water = 150 J If the water ABSORBED 150 J of heat, then the pebble _______ ______ J of heat.
q pebble = - 150 J
Example 2 (LAB type of CALC)
Suppose that 100.00 g of water at 22.4 °C is placed in a calorimeter. A 75.25 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
4.184 J/g
C.
MAKE YOUR CHART
Suppose that 100.00 g of water at 22.4 °C is placed in a calorimeter. A 75.25 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 4.184 J/g C.
Water Pebble Heat Mass Specific Heat Final Temp Initial Temp
100.00 g
4.184
32.9 o C 75.25 g 32.9 o C 22.4 o C 99.3 o C
1. Make chart 2. Calculate q for water 3. Q for water is the same (but with different sign) as q for metal.
4. Using q metal, calculate c metal
0.879 J/g °C
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 4.184 J/g
C.
502 J = 5.0 x 10
2