Specific Heat Capacity. deals with heat changes that occur during chemical reactions deals with heat changes that occur during chemical reactions Heat.

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Presentation transcript:

Specific Heat Capacity

deals with heat changes that occur during chemical reactions deals with heat changes that occur during chemical reactions Heat – transfer of energy from one object to another Heat – transfer of energy from one object to another Thermochemistry

Endothermic – heat flows into the system Endothermic – heat flows into the system the test tube gets cold – heat is absorbed the test tube gets cold – heat is absorbed Exothermic – heat flows out of the system Exothermic – heat flows out of the system the test tube gets hot – heat is given off the test tube gets hot – heat is given off

“Burning” calories 1 calorie = quantity of heat needed to raise the temperature of 1 gram of water 1 o C 1 calorie = quantity of heat needed to raise the temperature of 1 gram of water 1 o C Calorie = dietary calorie = 1,000 calories = 1 kilocalorie Calorie = dietary calorie = 1,000 calories = 1 kilocalorie 1 Joule (metric unit) = calories 1 Joule (metric unit) = calories Joules = 1 calorie Joules = 1 calorie 10 g of sugar = 10 g of sugar = 41 Calories of heat or 41 Calories of heat or 41 kilocalories of heat 41 kilocalories of heat

Heat Capacity the amount of heat needed to increase the temperature of an object by 1 o C the amount of heat needed to increase the temperature of an object by 1 o C depends on the mass and chemical composition of the matter depends on the mass and chemical composition of the matter 250 gram steel cylinder 50 gram steel cylinder greater heat capacity wooden block iron block greater heat capacity

Specific Heat Capacity The amount of heat it takes to raise the temp. of 1 gram of a substance 1 o C The amount of heat it takes to raise the temp. of 1 gram of a substance 1 o C calorimetry = the measurement of heat changes calorimetry = the measurement of heat changes

Some specific heat capacities Substance J/(gX o C) water 4.18 ice 2.1 aluminum 0.90 glass 0.50 iron 0.46 silver 0.24

Calculation of the specific heat of a substance S = specific heat S = specific heat q = heat q = heat m = mass m = mass ∆T = change in temperature = T 2 – T 1 ∆T = change in temperature = T 2 – T 1 q = m S ∆T q = m S ∆T or q = m C ∆T or q = m C ∆T

Some problems! The temperature of a piece of copper with a mass of 95.4 g increases from 25.0 o C to 48.0 o C when the metal absorbs 849 J of heat. What is the specific heat of copper? The temperature of a piece of copper with a mass of 95.4 g increases from 25.0 o C to 48.0 o C when the metal absorbs 849 J of heat. What is the specific heat of copper? m = 95.4 g ∆T = (48.0 o C – 25.0 o C) = 23.0 o C q = 849 J S = ___q___ m ∆T S = ______849J_______ 95.4 g X 23.0 o C = J/(g o C)

When 435 J of heat is added to 3.4 g of olive oil at 21 o C, the temperature increases to 85 o C. What is the specific heat of olive oil? When 435 J of heat is added to 3.4 g of olive oil at 21 o C, the temperature increases to 85 o C. What is the specific heat of olive oil? S = ____435 J_____ 3.4 g X 64 o C S = 2.0 J/(gX o C)