Download - ENERGY
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ENERGY
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Energy Review
• Temperature – measurement of the random motion of the components of a substance
• Heat – flow of energy due to temperature differences
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In general, the universe is made up of two parts for thermodynamic purposes.
• System – part of the universe in which you are interested
• Surroundings – everything outside of that system
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EXOTHERMIC REACTIONS• Energy is released. (Negative value)
Examples: Combustion:
2C26H54 + 79 O2 52 CO2 + 54 H2O + Heat
Precipitating: Na+
(aq) + CH3COO-(aq) NaCH3COO(s) + Heat
Phase change:
H2O(l) H2O(s) + Heat
• Energy of reactants is greater than products. (See diagram on next slide.)
• Energy flows out of the system into the surroundings.
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Exo vs. Endo
Exothermic reactions get
HOT
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Endothermic Reactions
• Energy is absorbed. (Positive value)Examples:
Phase changes:Heat + H2O(s) H2O(l)
Dissolving:Heat + NH4Cl(s) NH4
+(aq) + Cl-
(aq)
• Energy of products is greater than energy of reactants. (See next slide.)
• Energy flows INTO the system from the surroundings.
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Endothermic reactions get
COLD
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Measuring Energy Changes
• Units – Calorie and Joules
– 1 calorie = 4.184 Joules
– Example – Convert 60.1 calories of energy to joules
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SPECIFIC HEAT CAPACITY
• Amount of heat needed to raise 1 gram of a substance 1 Celsius.
• Measures the ability of a substance to store heat energy.
• When the temp of something, is changed heat is required.
• The amount of heat depends on the amount (mass) and nature of the substance.
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Heat equation
q=mCDT q = heat (J, Joule)m = mass (g, grams)C = Specific heat (J/g°C)DT = change in temperature (°C)
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Rearranging the heat equation
Solve q = mCDT for each of the other variables:
m =
C =
DT = qmC
qCDT
qmDT
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Practice problems
1) How much heat is released when a 100g piece of iron (CFe =0.45 J/g°C) goes from 80°C to 25°C?
q = mCDTq = m = C = DT =
q = (100g)(0.45 J/g°C)(-55°C)q = - 2475 J
?100g0.45 J/g°CTf – Ti = 25°C - 80°C = -55°C
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Practice problems
2) How much heat is required to heat a 75g piece of iron (CFe = 0.45 J/g°C) from 20°C to 105°C?
q = mCDTq = m = C = DT =
q = (75g)(0.45 J/g°C)(85°C)q = 2868.8 J
?75g0.45 J/g°CTf – Ti = 105°C - 20°C = 85°C
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Thermodynamics
• Study of matter and energy interactions
H: enthalpy – heat content of a substanceS: entropy – disorder of a substanceG: Gibb’s free energy – chemical potential
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Types of Thermodynamic Reactions
• Exothermic – heat is given off– ∆H<0 - number
• Endothermic– Heat is absorbed– ∆H>0 + number
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• ∆H˚rxn = ∑∆Hf˚products - ∑∆Hf˚reactants
• ∆H˚rxn = enthalpy change for a rxn
• ∆Hf˚ = heat of formation, how much E it takes to put substance together
• ˚ = standard conditions (25˚C, 101.3kPa, 1.0M)• ∑ = “sum of”
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• Use tables to look up ∑∆Hf values.• Unit – KJ
mol
• All lone elements in a rxn: ∆Hf = 0
• ∆Hf Al = 0 ∆Hf O2 = 0
• Need Balanced equations
• Must account for moles
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Practice problemsCalculate the ∆Hrxn for the following rxn:
Cl2 (g) + HBr (g) → 2HCl (g) + Br2 (g)
∆Hrxn = ∑Products - ∑Reactants
Cl2= HBr = HCl = Br2 =
∆Hrxn = (2mol(-92.30KJ/mol) - (2mol(-36.23KJ/mol)
∆Hrxn = - 112.14KJ
0-36.23 KJ/mol-92.30 KJ/mol0
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Heat of Vaporization – energy change from Liquid → gas
Calculate the heat of vaporization for water:H2O(l) → H2O(g)
∆Hrxn = ∑Products - ∑Reactants
H2O(g) =
H2O(l) =
∆Hrxn = -241.8 KJ/mol – (-285.85KJ/mol)
∆Hrxn = - 44.05 KJ/mol
Endothermic
-241.8 KJ/mol -285.85 KJ/mol
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Homework – Due 5/3
Food assignment:For one entire day keep track of what and how much you eat in a table. Use the food labels or the USDA website to determine how many calories each item contains. Due Tuesday.
Time/meal Food item Amount Calories
Breakfast Honey bunches of oats
1 cup 350
1% milk ½ cup 90