# THERMOCHEMISTRY: HEAT and CHANGE. When a material is heated (or cooled), it can undergo one of these changes: Its temperature changes OR Its physical.

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THERMOCHEMISTRY: HEAT and CHANGE

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

Type I : Changes In Temperature Heating and Cooling

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)

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

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

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)

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?

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

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)

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

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

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

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

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

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

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

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?

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 )

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 )

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

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.

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

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

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

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

#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

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.

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

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

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

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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