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THERMOCHEMISTRY: HEAT and CHANGE

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When a material is heated (or cooled), it can undergo one of these changes: Its temperature changes OR Its physical state changes

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Type I : Changes In Temperature Heating and Cooling

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Heating or Cooling Only Involves only an increase or decrease in temperature No change in state involved Formula: Q = m C T Where q = heat, in cal, J, or kcal,kJ m = mass, in g or kg c = specific heat capacity, (value depends on the substance) T = temperature change (final temperature - initial temperature)

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The amount of heat energy that must be supplied so as to warm a material depends on three things: Mass m Specific heat capacity C Temperature change T

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Which will need more heat in order to boil? Or a bucket of water at room temperature? A cup of water at room temperature?

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You will need more heat energy to warm an object with a bigger mass. (assuming you have the same material and the same temperature change)

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Which needs more heat to warm up to 75 degrees Celsius? 1 lb. of water at room temp? Or 1 lb. of iron at room temp?

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Different materials have different Specific Heat capacities (C) Iron ( C= 0.449) To warm 1 gram of iron from 10 to 11 deg Celsius, you must supply 0. 449 joules Water (C = 4.2) To warm 1 gram of water from 10 to 11 deg Celsius, you must supply 4.2 joules Specific Heat Capacity is measured in joules / g deg Celsius Specific Heat Capacity is measured in joules / g deg Celsius

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Some materials just warm up faster than others. You will need more heat to warm a material which has a high specific heat capacity assuming the two materials have the same mass and same temperature change) (assuming the two materials have the same mass and same temperature change)

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Which will need more heat energy to warm? A pound of water from 20˚C to 30˚C Or a pound of water from 20˚C to 130˚C

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The bigger the difference in temperature, the more the heat energy needed to warm the material Difference in temperature is represented by the symbol ΔT. This is calculated by : ΔT = T f - T i where T f represents the final temperature and T i represents the initial temperature

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Calculate the amount of heat needed to raise the temperature of 1.2g of water from 10 0 C to 20 0 C. Specific heat of water is 4.2 J/g ºC. Q = mC T = 1. 2 g ( 4.2 J/g o C ) ( 20 o C – 10 o C) = 1. 2 g ( 4.2 J/g o C ) ( 20 o C – 10 o C) = 1.2 g ( 4.2 J/g o C) (10 o C) = 1.2 g ( 4.2 J/g o C) (10 o C) = 50.4 J = 50.4 J

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Type 2 : Changes In State * Freezing/Melting * Vaporization/Condensation

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Changes in State (Phase) Most substances can exist in three states— solid, liquid, and gas—depending on the temperature and pressure.

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When energy is added to or taken away from a system, one phase can change into another. Some phase changes NEED energy ( + Q) * melting (fusion) * evaporation (vaporization) Some phase changes RELEASE energy (-Q) * freezing (solidification) * condensation

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To simplify this graphic: Melting Melting Vaporization (Fusion) (Evaporation) (Fusion) (Evaporation) SOLID ↔ LIQUID ↔ GAS SOLID ↔ LIQUID ↔ GAS Freezing Condensation Freezing Condensation (Solidification) (Solidification)

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Heat of Fusion Heat of Vaporization Melting Melting Vaporization (Fusion) (Evaporation) (Fusion) (Evaporation) SOLID ↔ LIQUID ↔ GAS SOLID ↔ LIQUID ↔ GAS Freezing Condensation Freezing Condensation ( Solidification) Heat of Condensation ( Solidification) Heat of Condensation Heat of Solidification Heat of Solidification What are the terms for each of the heat change (ΔH) associated with each process?

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The processes going to the right (melting and _________) are endothermic. They _______ energy. ( Q has a _____ sign ) The processes going to the left (_______ and condensation) are exothermic. They ________ energy. ( Q has a ______ sign ) The processes going to the left (_______ and condensation) are exothermic. They ________ energy. ( Q has a ______ sign )

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The processes going to the right (melting and vaporization ) are endothermic. They need energy. ( Q has a positive sign ) The processes going to the left ( freezing and condensation) are exothermic. They release energy. ( Q has a negative sign ) The processes going to the left ( freezing and condensation) are exothermic. They release energy. ( Q has a negative sign )

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Formulas for Heat Problems involving Changes of State #1 Melting (or fusion) : solid liquid Freezing (or solidification) : liquid solid Formula : Q = n ΔH f Q = amount of heat absorbed or released n = number of moles ΔH f = molar heat of fusion

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Molar Heat of Fusion ΔH f Molar Heat of Fusion- amount of energy needed to change 1 mole of solid to liquid at its melting temperature Different materials have different ΔH f Ex: Molar Heat of Fusion of H 2 O = 6.01 kJ/mol Molar Heat of Fusion of Lead = 4.77 kJ/mol for melting (fusion) and freezing Use ΔH f for melting (fusion) and freezing problems. Remember Δis positive for melting problems. Remember ΔH f is positive for melting but negative for freezing. but negative for freezing.

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MELTING Sample Problem 1: What is the amount of heat needed to melt 4 moles of ice at its melting point? (ΔH f of water(ice) = 6.01 kJ/mol) Q = n ΔH f = (4 moles) (6.01 kJ/mole) = 24.04 moles

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MELTING Sample problem 2: What is the amount of energy needed to melt 5 grams of ice at its melting point? (ΔH f of water = 6.01 kJ/mol) Q = n ΔH f Number of moles: 5 grams 1 mole 1 18.01 grams or 0.28 mol Q = nΔH f = (0.28 mol)(6.01 kJ/mol) = 1.68 kJ

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FREEZING The reverse process of melting (freezing) RELEASES energy Use the same ΔH f as in melting but make the sign negative Sample 1: How much energy is released when 5 moles of water freezes at 0 deg Celsius? (ΔH f of water = -6.01 kJ/mol) Q = n ΔH f = (5 moles) ( -6.01 kJ/mol) = -30.05 kJ

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Sample problem 2: How much energy is released when 50 grams of water freezes at 0 degrees Celsius? Q = nΔH f n : 50 g 1 mole 1 18.01 g Q = (2.8 mol)( - 6.01 kJ/mol) Q = - 16.8 kJ

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#2 Vaporization : liquid gas Condensation: gas liquid Formula: Where n = no. of moles H v = heat of vaporization Q= n ΔH v Formulas for Heat Problems involving Changes of State

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Molar Heat of Vaporization ΔH v Molar Heat of Vaporization- Molar Heat of Vaporization- amount of energy needed to change 1 mole of liquid to gas at its boiling temperature Different Materials have different ΔH v Molar Heat of vaporization of water= 40.6kJ/mol Molar Heat of vaporization of ethanol = 38.6kJ/mol for vaporization (evaporation) and condensation problems. Remember Δis positive for vaporization (evaporation) but negative for condensation. Use ΔH v for vaporization (evaporation) and condensation problems. Remember ΔH v is positive for vaporization (evaporation) but negative for condensation.

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VAPORIZATION Sample problem 1: How much heat in kJ must be absorbed by 1.5 moles of water in order to evaporate completely at its boiling point? (ΔH v of water = 40.6 kJ/mol) Q = n ΔH v Q = (1.5 mol)(40.6 kJ/mol) Q = 60.9 kJ

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VAPORIZATION Sample Problem 2: How much heat in kJ must be absorbed by 5 grams of water to evaporate completely at 100 °Celsius? (ΔH v of water = 40.6 kJ/mol) Q = nΔH v Number of moles: 5 grams 1 mole 1 18.01 grams or.28 mol Q = nΔH v = (0.28 mol)(40.6 kJ/mol) = 11.37 kJ

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CONDENSATION The reverse process of vaporization (condensation) RELEASES energy Use the same ΔHv as in vaporization but make the sign negative Calculate the amount of heat released when 2 moles of water vapor change completely to liquid water at 100°C ? Example 1: Calculate the amount of heat released when 2 moles of water vapor change completely to liquid water at 100°C ? (ΔHv of water = 40.6 kJ/mol) Q = nΔHv Q = (2 mol)( - 40.6 kJ/mol) Q = - 81.2 kJ

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Calculate the amount of heat released when 50g of water vapor changes completely to liquid water at its boiling point? (Δ H v of water = 40.6 kJ/mol) Example 2: Calculate the amount of heat released when 50g of water vapor changes completely to liquid water at its boiling point? (Δ H v of water = 40.6 kJ/mol) Q = nΔHv n: 50 g 1 mole 1 18.01 g Q = (2.8 mols) ( -40.6 kJ/mol) = -113.68 kJ

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