Energy and Changes in Matter

Presentation on theme: "Energy and Changes in Matter"— Presentation transcript:

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
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 marshmallow. Heat always flows from high temperatures to low temperatures.

Units of Measurement for Heat
1 cal = J 1000 cal = 1 Cal How many calories are in 54.0J?

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 Vaporization PE PE
Solid Liquid Gas Vaporization Melting KE PE Boiling Point Melting Point PE Potential Energy Changing with Phase Changes KE Kinetic Energy Changing with Temperature Changes

Energy and Changes in Matter
Today’s Objective Perform calculations related to heat and changes in temperature or phase.

DT = Change in Temperature = Tfinal - Tinitial
Calculating Heat To calculate the amount of heat energy when temperature changes, Q = m Cp DT Q = heat m = 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

Examples Un-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

Q > 0 = heat is being absorbed.
Example Problem #1 How much energy is required to raise a 34.0g sample of copper metal from 20.0oC to 45.0oC? The specific heat of copper is Q = m Cp DT Q = (34.0g)(0.385 ) (45.0oC -20.0oC) = (34.0g)(0.385 ) (25.0oC ) = 327 J J g oC “to” – “from” J g oC J g oC Q > 0 = heat is being absorbed.

Q < 0 = heat is being released.
Example Problem #2 How much heat is released when a 3.20g sample of water is cooled from 83.0oC to 54.0oC? The specific heat of water is Q = m Cp DT Q = (3.20g)(4.18 ) (54.0oC -83.0oC) = (3.20g)(4.18 ) (-29.0oC ) = -388 J J g oC “to” – “from” J g oC J g oC Q < 0 = heat is being released.

Heat of Vaporization (Hv)
Calculating Heat To calculate the amount of heat energy when phase changes, Q = mH Q = heat m = mass Phase Change H Value for H Melting Heat of Fusion (Hf) +Hf Freezing -Hf Vaporization Heat of Vaporization (Hv) +Hv Condensation -Hv

Q > 0 = endothermic phase changes
Example Problem #3 Use heat of fusion! 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

Example Problem #4 How much heat is released when 120g of steam condenses? Q = mHV Q = (120g) (-2260 ) = -270,000 J Use a negative heat of vaporization! J g Q < 0 = exothermic phase changes

Important Reminders When the Temperature Changes Q = mCpDT
When the Phase Changes Q = mH DT = Temperature You Are Going To – Temperature You are Coming From Phase Change H Value for H Melting Heat of Fusion (Hf) +Hf Freezing -Hf Vaporization Heat of Vaporization (Hv) +Hv Condensation -Hv

Energy and Changes in Matter
Today’s Objective Determine experimentally the heat of fusion of water.

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 mDT (53.8g)(22.3oC-18.0oC) 89.1J mDT Q (53.8g)(4.3oC) 89.1J J g oC Copper

Example #2 Determine 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

Step 1: Ice at -15oC  Ice at 0oC  Q=mCpDT
Example #2 - Continued Step 1: Ice at -15oC  Ice at 0oC  Q=mCpDT Q = m Cp DT Q = (4.5g)(2.05 )(0oC oC) Q = (4.5g)(2.02 )(15oC) Q = 138 J J g oC J g oC

Step 2: Ice at 0oC  Water at 0oC  Q=mHf
Example #2 - Continued Step 2: Ice at 0oC  Water at 0oC  Q=mHf Q = m Hf Q = (4.5g)(334 ) Q = J J g

Step 3: Water at 0oC  Water at 56oC  Q=mCpDT
Example #2 - Continued Step 3: Water at 0oC  Water at 56oC  Q=mCpDT Q = m Cp DT Q = (4.5g)(4.18 )(56oC - 0oC) Q = (4.5g)(4.18 )(56oC) Q = 1050 J J g oC J g 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 J Step 2  Q = J Step 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 100oC Step 2: Steam Condenses to Water Step 3: Water at 100oC to 90oC

Example #3 - Continued Step 1: Steam at 110oC to 100oC Q = mCpDT Q = (4.00g)(2.02 )(100oC - 110oC) Q = J Step 2: Steam Condenses to Water Q = mHv Q = (4.00g)(-2260 ) Q = J J g oC J g

Example #3 - Continued Step 3: Water at 100oC to 90oC Q = mCpDT Q = (4.00g)(4.18 )(90oC - 100oC) Q = -167 J Total Energy Released Q = J J J Q = J J g 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, The original chemical bonds are broken. This requires energy. 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 Exothermic Reactions Absorb Energy Feel cold to the touch Heat is treated as a reactant. Release Energy Feel warm to the touch Heat is treated as a product.

Potential Energy Diagram for an Endothermic Reaction
(kJ) PE of Products Heat of Reaction (DH) PE of Reactants PEReactants < PEProducts DH = Products – Reactants = 350kJ – 200kJ = 150 kJ

Potential Energy Diagram for an Exothermic Reaction
(kJ) PE of Reactants Heat of Reaction (DH) PE of Products PEReactants > PEProducts DH = Products – Reactants = 15kJ – 40kJ = -25kJ

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

Calorimetry Heat 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.