Presentation on theme: "Energy and Changes in Matter Todays Objectives Differentiate between heat and temperature. Interpret the heating and cooling curves of various substances."— Presentation transcript:
Energy and Changes in Matter Todays Objectives Differentiate between heat and temperature. Interpret the heating and cooling curves of various substances.
Heat versus Temperature HeatTemperature Definition Energy Transfer between Substances, Related to the Total Kinetic Energy Average A Measure of the Average Kinetic Energy of a Substance Units of Measurement Joules (J) Calories (cal) Celsius ( o C) Kelvin (K) How to Measure? IndirectlyDirectly
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.
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 Melting Point Solid Liquid Gas Vaporization Melting PE Potential Energy Changing with Phase Changes KE Kinetic Energy Changing with Temperature Changes KE PE
Energy and Changes in Matter Todays 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 C p T Q = heat m = mass C p = specific heat T = Change in Temperature = T final - T initial T = Change in Temperature = T final - T initial
What is Specific Heat (C p )? Specific heat is the amount of energy required to change the temperature of a substance by 1 o C. high 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
Example Problem #1 How much energy is required to raise a 34.0g sample of copper metal from 20.0 o C to 45.0 o C? The specific heat of copper is Q = m C p T Q = (34.0g)(0.385 ) (45.0 o C o C) = (34.0g)(0.385 ) (25.0 o C ) = 327 J J g o C J to – from J g o C Q > 0 = heat is being absorbed.
Example Problem #2 How much heat is released when a 3.20g sample of water is cooled from 83.0 o C to 54.0 o C? The specific heat of water is Q = m C p T Q = (3.20g)(4.18 ) (54.0 o C o C) = (3.20g)(4.18 ) (-29.0 o C ) = -388 J J g o C J to – from J g o C Q < 0 = heat is being released.
Calculating Heat To calculate the amount of heat energy when phase changes, Q = mH Q = heatm = mass Phase ChangeHValue for H Melting Heat of Fusion (H f ) +H f Freezing-H f VaporizationHeat of Vaporization (H v ) +H v Condensation-H v
Example Problem #3 How much heat is needed to melt 56.2g of ice at 0 o C? Q = mH f 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 = mH V Q = (120g) (-2260 ) = -270,000 J J g Q < 0 = exothermic phase changes Use a negative heat of vaporization!
Important Reminders When the Temperature ChangesQ = mC p T When the Phase ChangesQ = mH Phase ChangeHValue for H Melting Heat of Fusion (H f ) +H f Freezing-H f VaporizationHeat of Vaporization (H v ) +H v Condensation-H v T = Temperature You Are Going To – Temperature You are Coming From
Energy and Changes in Matter Todays 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 -4 o C and 4 o C. 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 Todays 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 18.0 o C to 22.3 o C. Q = m C p T C p = = C p = = m T (53.8g)(22.3 o C-18.0 o C) 89.1J m T Q (53.8g)(4.3 o C) 89.1J J g o C Copper
Example #2 Determine the amount of heat energy required to change 4.50g of ice from o C to water at 56.0 o C. This problem is different because we are changing phase AND temperature.
Heating Curve of Water Step 1: Ice at -15 o C Ice at 0 o C Q=mC p T Step 2: Ice at 0 o C Water at 0 o C Q=mH f Step 3: Water at 0 o C Water at 56 o C Q=mC p T
Example #2 - Continued Q = m C p T Q = (4.5g)(2.05 )(0 o C o C) Q = (4.5g)(2.02 )(15 o C) Q = 138 J Step 1: Ice at -15 o C Ice at 0 o C Q=mC p T J g o C J
Example #2 - Continued Q = m H f Q = (4.5g)(334 ) Q = J Step 2: Ice at 0 o C Water at 0 o C Q=mH f J g
Example #2 - Continued Q = m C p T Q = (4.5g)(4.18 )(56 o C - 0 o C) Q = (4.5g)(4.18 )(56 o C) Q = 1050 J Step 3: Water at 0 o C Water at 56 o C Q=mC p T J g o C J
Example #2 - Continued Determine the amount of heat energy required to change 4.50g of ice from o C to water at 56.0 o C. 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 o C is cooled and condensed to form water at 90.0 o C? Step 1: Steam at 110 o C to 100 o C Step 2: Steam Condenses to Water Step 3: Water at 100 o C to 90 o C
Example #3 - Continued Step 1: Steam at 110 o C to 100 o C Q = mC p T Q = (4.00g)(2.02 )(100 o C o C) Q = J Step 2: Steam Condenses to Water Q = mH v Q = (4.00g)(-2260 ) Q = J J g o C J g
Example #3 - Continued Step 3: Water at 100 o C to 90 o C Q = mC p T Q = (4.00g)(4.18 )(90 o C o C) Q = -167 J Total Energy Released Q = J J J Q = J J g o C
Thermochemistry Thermochemistry the study of the heat changes during chemical reactions, and the effects on chemical and physical processes. Todays 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 of Reactants PE of Products Heat of Reaction ( H) H = Products – Reactants = 350kJ – 200kJ = 150 kJ (kJ) PE Reactants < PE Products
Potential Energy Diagram for an Exothermic Reaction PE of Reactants PE of Products Heat of Reaction ( H) H = Products – Reactants = 15kJ – 40kJ = -25kJ (kJ) PE Reactants > PE Products
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.