Energy and Changes in Matter

Today’s Objectives

• Differentiate between heat and temperature.

• Interpret the heating and cooling curves of various substances.

Heat versus Temperature

Heat Temperature

Definition

Energy Transfer between

Substances, Related to the Total Kinetic

Energy

A Measure of the Average

Kinetic Energy of a Substance

Units of Measurement

Joules (J)Calories (cal)

Celsius (oC)Kelvin (K)

How to Measure?

Indirectly Directly

Match versus Ice Sculpture

Which Has a Higher Temperature?A Match

Which Has More Heat Energy?The Ice Sculpture

Heat

Heat always flows from high temperatures to low temperatures.

Heat flows from the fire

to the marshmallo

w.

Units of Measurement for Heat

1 cal = 4. 18 J1000 cal = 1 Cal

How many calories are in 54.0J?

calJ

calJ9.12

18.4

10.54

Heat and Physical Changes

Add or Remove Heat

Change in Kinetic Energy of the

Substance

Change in Potential Energy of the Substance

Change in Temperature

Change in Phase

Heating Curve

Boiling Point

MeltingPoint

Solid

Liqu

id

Gas

Vaporization

Melting

PE Potential Energy Changing

with Phase Changes

KEKinetic Energy Changing

with Temperature Changes

KE

KE

KE

PE

PE

Energy and Changes in Matter

Today’s Objective

Perform calculations related to heat and changes in temperature or phase.

Calculating Heat

To calculate the amount of heat energy when temperature changes,

Q = m Cp DT

Q = heatm = mass

Cp = specific heat

DT = Change in Temperature = Tfinal - Tinitial

What is Specific Heat (Cp)?

Specific heat is the amount of energy required to change the temperature of a substance by 1oC.

A substance with a high specific heat requires more energy to change its temperature than a substance with a low specific heat.

Low Specific Heat = Good Conductors

High Specific Heat = Good Insulators

ExamplesUn-Popable Balloon• Which balloon will pop first?– One with Only Air – One with Air and Water

Which substance is the best conductor?– Copper– Gold– Air

Example Problem #1How much energy is required to raise a 34.0g sample of copper metal from 20.0oC to 45.0oC? The specific heat of copper is0.385 .

Q = m Cp DT

Q = (34.0g)(0.385 ) (45.0oC -20.0oC)

= (34.0g)(0.385 ) (25.0oC )

= 327 J

Jg oC

Jg oC

“to” – “from”

Jg oC

Q > 0 = heat is being absorbed.

Example Problem #2How much heat is released when a 3.20g sample of water is cooled from 83.0oC to 54.0oC? The specific heat of water is4.18 .

Q = m Cp DT

Q = (3.20g)(4.18 ) (54.0oC -83.0oC)

= (3.20g)(4.18 ) (-29.0oC )

= -388 J

Jg oC

Jg oC

“to” – “from”

Jg oC

Q < 0 = heat is being released.

Calculating Heat

To calculate the amount of heat energy when phase changes,

Q = mHQ = heat m = mass

Phase Change H Value for H

MeltingHeat of Fusion (Hf)

+Hf

Freezing -Hf

Vaporization Heat of Vaporization (Hv)

+Hv

Condensation -Hv

Example Problem #3

How much heat is needed to melt 56.2g of ice at 0oC?

Q = mHf

Q = (56.2g)(334 )

= 18,800 J

J g

Q > 0 = endothermic phase changes

Use heat of fusion!

Example Problem #4

How much heat is released when 120g of steam condenses?

Q = mHV

Q = (120g) (-2260 )

= -270,000 J

J g

Q < 0 = exothermic phase changes

Use a negative heat of vaporization!

Important Reminders

When the Temperature Changes

Q = mCpDT

When the Phase Changes Q = mH

Phase Change H Value for H

MeltingHeat of Fusion (Hf)

+Hf

Freezing -Hf

Vaporization Heat of Vaporization (Hv)

+Hv

Condensation -Hv

DT = Temperature You Are Going To – Temperature You are Coming From

Energy and Changes in Matter

Today’s Objective

Determine experimentally the heat of fusion of water.

Important Reminders about Your Experiment

• Make sure that you always have ice in your calorimeter at ALL times!

• Your constant temperature should be between -4oC and 4oC.

• If you spill the water, you will have to re-do the lab.– This is especially important at the end when you

are measuring the volume. Your volume will exceed 100mL!

• For your calculations, only use the accepted value for the heat of fusion when you are calculating percent error.

Energy and Changes in Matter

Today’s Objectives

• Solve problems related to heat and physical changes.

Example #1

Determine the identity of a substance that requires 89.1J to raise 53.8g sample from 18.0oC to 22.3oC.

Q = m Cp DT

Cp = =

Cp = = 0.385

mDTmDT

(53.8g)(22.3oC-18.0oC)

89.1J

mDT

Q

(53.8g)(4.3oC)

89.1J Jg oC

Copper

Example #2Determine the amount of heat energy required to change 4.50g of ice from -15.0oC to water at 56.0oC.

This problem is different because we are changing phase AND

temperature.

Heating Curve of Water

Step 1: Ice at -15oC Ice at 0oC Q=mCpDT

Step 2: Ice at 0oC Water at 0oC Q=mHf

Step 3: Water at 0oC Water at 56oC Q=mCpDT

Example #2 - Continued

Q = m Cp DT

Q = (4.5g)(2.05 )(0oC - - 15oC)

Q = (4.5g)(2.02 )(15oC)

Q = 138 J

Step 1: Ice at -15oC Ice at 0oC Q=mCpDT

Jg oC

Jg oC

Example #2 - Continued

Q = m Hf

Q = (4.5g)(334 )

Q = 101700 J

Step 2: Ice at 0oC Water at 0oC Q=mHf

J g

Example #2 - Continued

Q = m Cp DT

Q = (4.5g)(4.18 )(56oC - 0oC)

Q = (4.5g)(4.18 )(56oC)

Q = 1050 J

Step 3: Water at 0oC Water at 56oC Q=mCpDT

Jg oC

Jg oC

Example #2 - Continued

Determine the amount of heat energy required to change 4.50g of ice from -15.0oC to water at 56.0oC.Calculate the total amount of heat.

Step 1 Q = 138 JStep 2 Q = 101700 JStep 3 Q = 1050J

103, 000J

Example #3

How much energy is released when 4.00g of steam at 110.0oC is cooled and condensed to form water at 90.0oC?

Step 1: Steam at 110oC to 100oCStep 2: Steam Condenses to Water

Step 3: Water at 100oC to 90oC

Example #3 - ContinuedStep 1: Steam at 110oC to 100oC

Q = mCpDT

Q = (4.00g)(2.02 )(100oC - 110oC)

Q = -80.8 J

Step 2: Steam Condenses to WaterQ = mHv

Q = (4.00g)(-2260 )

Q = -9040 J

J

g oC

J

g

Example #3 - Continued

Step 3: Water at 100oC to 90oCQ = mCpDT

Q = (4.00g)(4.18 )(90oC - 100oC)

Q = -167 J

Total Energy ReleasedQ = -80.8 J + - 9040 J + -167 J

Q = -9290 J

Jg oC

Thermochemistry

the study of the heat changes during chemical reactions, and the effects on

chemical and physical processes.

Today’s Objectives

• Differentiate between exothermic and endothermic reactions

• Interpret the potential energy diagrams for chemical reactions.

Heat and Chemical Reactions

In chemical reactions two things occur,

1) The original chemical bonds are broken. This requires energy.

2) New chemical bonds are formed. This releases energy.

Because the energy it takes to break the bonds and the energy that is released when the new bonds are formed are not always equal, heat is either absorbed or released in chemical reactions.

Heat and Chemical Reactions

Endothermic Reactions

• Absorb Energy

• Feel cold to the touch

• Heat is treated as a reactant.

Exothermic Reactions

• Release Energy

• Feel warm to the touch

• Heat is treated as a product.

Potential Energy Diagram for an Endothermic Reaction

PE ofReactants

PE of Products

Heat of Reaction

(DH)

DH = Products – Reactants = 350kJ – 200kJ = 150 kJ

(kJ)

PEReactants < PEProducts

Potential Energy Diagram for an Exothermic Reaction

PE ofReactants

PE of Product

s

Heat of Reaction

(DH)

DH = Products – Reactants = 15kJ – 40kJ = -25kJ

(kJ)

PEReactants > PEProducts

Law of Conservation of Energy

Any chemical or physical process does not create or destroy

energy.

Energy Released or Absorbed by the System

Energy Absorbed or Released by

the Surroundings

=

System = What You are Studying

Surroundings = Everything Else/Generally Water

CalorimetryHeat must be calculated indirectly. In other words, scientists use experimental means to calculate the heat energy associated

with a process. These experimental means are generally referred to as calorimetry.

Tips for Solving Calorimetry Problems Successfully

• Calculate how much heat is gained or lost by the water.

• Assume that the amount of heat gained or lost by the water is equal to the amount of heat lost or gained by the other substance or reaction.

• Solve for the missing value for the other substance or reaction.