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I II III
Chapter 16 Calorimetry
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Calorimetry
The study of heat flow and heat measurement.
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Heat Capacity
The amount of heat needed to raise the temperature of an object by 1 Celsius degree.
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Specific Heat
The heat capacity of 1 gram of a substance
Specific Heat of liquid water is 4.184 J/gºC
4.184 J = 1 calorie
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1 Calorie = 1000 calories = 1 kilocalorie
1 Food Calorie = 1000 calories
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Calorimetry Experiments
Determine the heats of reaction (ENTHALPY CHANGES) by making accurate measurements of temperature changes using a calorimeter.
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Scientists use q to denote measurements made in a calorimeter.
Heat transferred in a reaction is EQUAL, but OPPOSITE in sign to heat absorbed by the surroundings.
qrxn = - qsur
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qsur = m x Cp x (Tf –Ti)
Mass of WaterSpecific heat
of Water
Temperature change
HEAT
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Practice Problem: #1When a 12.8g sample of KCl
dissolves in 75.0g of water in a calorimeter, the temperature drops from 31.0ºC to 21.6ºC. Calculate H for the process.
KCl(s) K+ (aq) + Cl- (aq)
First, calculate qsur and then calculate H.
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Practice Problem: #1When a 12.8g sample of KCl
dissolves in 75.0g of water in a calorimeter, the temperature drops from 31.0ºC to 21.6ºC. Calculate H for the process.
KCl (s) K+ (aq) + Cl-(aq)
qsur = m x Cp x (Tf –Ti)
qsur = 75.0g x 4.184 J/gºC x (21.6 ºC –31.0 ºC )
qsur = -2949.72 J
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qsur is negative (as expected)
based on the temperature drop of the water.
KCl (s) K+ (aq) + Cl- (aq)
qrxn = - qsur = +2949.72 J
qrxn represents the heat absorbed
due to the reaction of 12.8g KCl.
Now you must convert the KCl to moles.
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Convert grams KCl to moles.
12.8 g KCl
74 g KCl
1 mol KCl= 0.173 mol KCl
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Calculate H for the reaction
=0.173 mol KCl
KCl (s) K+ (aq) + Cl- (aq)
H+2949.72 J
x 1 mol KCl
Coefficient from balanced
equationH = +17050 J
H = +17.1 kJ H Must be positive because it was an endothermic reaction!
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Practice Problem: #2What is the specific heat of
aluminum if the temperature of a 28.4 g sample of aluminum is
increased by 8.1ºC when 207 J of heat is added?
qsur = m x Cp x (Tf –Ti)
207J = 28.4g x Cp x 8.1oC
Cp = 28.4 g x 8.1 oC 207J
= 0.90 J/g oC