THERMOCHEMISTRY: HEAT and CHANGE

When a material is heated (or cooled), it can undergo one of these changes:

Its temperature changes OROR 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 TWhere 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 Specific Heat Capacity is measured in joules / g deg joules / g deg CelsiusCelsius

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 assuming the two materials have the same mass and same temperature change)mass and same temperature change)

Which will need more heat energy to warm?A pound of water from

20˚C to 30˚COr 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 materialDifference in temperature is represented by

the symbol ΔT. This is calculated by :

ΔT = Tf - Ti

where Tf represents the final temperature

and Ti represents the initial temperature

Calculate the amount of heat Calculate the amount of heat needed to raise the temperature needed to raise the temperature of 1.2g of water from 10of 1.2g of water from 1000C to C to 202000C. Specific heat of water is 4.2 C. Specific heat of water is 4.2

J/g ºC. J/g ºC.

Q = mCQ = mCTT = = 1. 2 g ( 4.2 J/g1. 2 g ( 4.2 J/g ooC ) ( 20 C ) ( 20 ooC – 10 C – 10

ooC)C)

= = 1.2 g ( 4.2 J/g1.2 g ( 4.2 J/g ooC) (10 C) (10 ooC)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: MeltingMelting Vaporization (Fusion) (Fusion) (Evaporation)(Evaporation)

SOLIDSOLID ↔↔ LIQUID LIQUID ↔↔ GAS GAS Freezing Freezing CondensationCondensation (Solidification)(Solidification)

Heat of Fusion Heat of Vaporization

MeltingMelting Vaporization (Fusion) (Fusion) (Evaporation)(Evaporation)

SOLIDSOLID ↔↔ LIQUID LIQUID ↔↔ GAS GAS Freezing Freezing CondensationCondensation ( Solidification) ( Solidification) Heat of CondensationHeat of Condensation

Heat of SolidificationHeat 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 (_______ The processes going to the left (_______ and condensation) are exothermic. They and condensation) are exothermic. They ________ energy. ( Q has a ______ sign )________ 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 (The processes going to the left (freezing freezing and condensation) are exothermic. They and condensation) are exothermic. They releaserelease energy. ( Q has a energy. ( Q has a negativenegative sign )sign )

Formulas for Heat Problems involving Changes of State

#1 Melting (or fusion) : solid liquid Freezing (or solidification) : liquid solid

Formula : Q = n ΔHf

Q = amount of heat absorbed or released n = number of moles

ΔHf = molar heat of fusion

Molar Heat of Fusion ΔHf

Molar Heat of Fusion- amount of energy needed to change 1 mole of solid to liquid at its melting temperature

Different materials have different ΔHf

Ex: Molar Heat of Fusion of H2O = 6.01 kJ/mol Molar Heat of Fusion of Lead = 4.77 kJ/mol Use ΔHf for melting (fusion) and freezing for melting (fusion) and freezing

problems . Remember problems . Remember ΔΔHf is is positivepositive for melting for melting

but but negativenegative for freezing. for freezing.

MELTINGSample Problem 1: What is the amount of heat needed to melt 4 moles of ice at its melting point? (ΔHf of water(ice) = 6.01 kJ/mol)

Q = n ΔHf

= (4 moles) (6.01 kJ/mole)

= 24.04 moles

MELTINGSample problem 2: What is the amount of energy needed to melt 5 grams of ice at its melting point? (ΔHf of water = 6.01 kJ/mol)

Q = n ΔHf

Number of moles:

5 grams 1 mole

1 18.01 grams or 0.28 mol

Q = nΔHf = (0.28 mol)(6.01 kJ/mol) = 1.68 kJ

FREEZING

The reverse process of melting (freezing) RELEASES energy

Use the same ΔHf 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? (ΔHf of water = -6.01 kJ/mol)

Q = n ΔHf = (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 = nΔHf

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

Hv = heat of vaporization

Q= n ΔHv

Formulas for Heat Problems involving Changes of State

Molar Heat of Vaporization

ΔHv 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 ΔHv

Molar Heat of vaporization of water= 40.6kJ/mol Molar Heat of vaporization of ethanol = 38.6kJ/mol Use ΔHv for vaporization (evaporation) and condensationfor vaporization (evaporation) and condensation

problems . Rememberproblems . Remember ΔΔHv isis positivepositive for vaporizationfor vaporization (evaporation) but(evaporation) but negativenegative for condensation.for condensation.

VAPORIZATIONSample 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? (ΔHv of water = 40.6 kJ/mol)

Q = n ΔHv

Q = (1.5 mol)(40.6 kJ/mol) Q = 60.9 kJ

VAPORIZATIONSample Problem 2: How much heat in kJ must be absorbed by 5 grams of water to evaporate completely at 100 °Celsius? (ΔHv of water = 40.6 kJ/mol)

Q = nΔHv

Number of moles:

5 grams 1 mole

1 18.01 grams or .28 mol Q = nΔHv = (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 Example 1: Calculate the amount of heat Calculate the amount of heat

released when 2 moles of water vapor change released when 2 moles of water vapor change completely to liquid water at 100°C ?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

Example 2: Calculate the amount of Calculate the amount of heat released when 50g of water heat released when 50g of water vapor changes completely to liquid vapor changes completely to liquid water at its boiling point? (water at its boiling point? (ΔΔ H Hv v of of water = 40.6 kJ/mol)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