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Chapter 16 Calorimetry

Calorimetry

The study of heat flow and heat measurement.

Heat Capacity

The amount of heat needed to raise the temperature of an object by 1 Celsius degree.

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

1 Calorie = 1000 calories = 1 kilocalorie

1 Food Calorie = 1000 calories

Calorimetry Experiments

Determine the heats of reaction (ENTHALPY CHANGES) by making accurate measurements of temperature changes using a calorimeter.

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

qsur = m x Cp x (Tf –Ti)

Mass of WaterSpecific heat

of Water

Temperature change

HEAT

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.

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

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.

Convert grams KCl to moles.

12.8 g KCl

74 g KCl

1 mol KCl= 0.173 mol KCl

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!

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