Presentation on theme: "Applications of Heat and Energy"— Presentation transcript:
1Applications of Heat and Energy Thermochemistry Branch of chemistry dealing with the relationship between chemical action and heat.Applications of Heat and Energy
2Energy All energy can be classified as either potential or kinetic. Potential energy is any type of energy that is stored.Examples: batteries (stored chemicalconverts to electrical)Top of a roller coaster hill (gravitational)Candy Bar (stored energy in its chemical bonds)Kinetic energy – any energy from the movement of matter.
3Which has more heat - Lake Erie in December of a drop of boiling oil? Frozen Lake ErieDrop of boiling oil
4Which has a higher temperature- Lake Erie in December of a drop of boiling oil? Frozen Lake ErieDrop of boiling oil
6Heat vs. TemperatureHeat is the amount of energy that is transferred from one substance to another due to the difference in temperature.Temperature is the average kinetic energy of an object.
7Heat TransferHeat always flows from an object with more heat (hotter) to an object with less heat.Note: Something that is cold just lacks heat. There is no unit for “cold”.When you feel cold, it is because you are losing heat (not gaining “cold”)
8True or False: When you place ice cream in the freezer, heat is transferred from the ice cream to the freezer.TrueFalse
9True or False: When we open the window, heat is transferred from your body to the air outside.
10Units of Heat Heat is measure in Joules (J). (The joule is the SI unit for all types of energy.)Example: when you heat a cup of tea, you use about 75,000J (or 75 kilojoules) of heat.Joule is pronounced jewel
11Other units of heat Heat is also measured in calories (cal). 1000 calories equals a kilocalorie (Cal).A calorie is the amount of heat needed to raise the temperature of 1 gram of water by 1 degree Celsius.Also, 1 calorie = JoulesAnd … 1 Cal = kJ
12What is the boiling point of water? 273K32oF212oC100oC
13What is the freezing point of water? 32K100oF273K32oC
16Specific HeatDifferent substances absorb (and lose) heat at different rates.Specific heat (SH) is the amount of heat (q) needed to raise the temperature of 1 gram of a substance by 1 degree Celsius.
17SPECIFIC HEAT q = s x m x DT q = thermal energy (J) SPECIFIC HEAT: The quantity of heat required to raise the temperature of one gram of a substance by one degree Celsius (or one Kelvin).q = s x m x DTq = thermal energy (J)sh = specific heat (J/goC)m = mass (g)DT = change in temperature (Tfinal – Tinitial) (oC)
18SPECIFIC HEATDetermine the energy (in kJ) required to raise the temperature of g of water from 20.0 oC to 85.0 oC?m = g DT = Tf -Ti = oC = 65.0 oCq = m x s x DT s (H2O) = J/ g - oCq = (100.0 g) x (4.184 J/g-oC) x (65.0oC)q = J (1 kJ / 1000J) = 27.2 kJDetermine the specific heat of an unknown metal that required 2.56 kcal of heat to raise the temperature of g from 15.0 oC to oC?S = cal /g -oC
19LAW OF CONSERVATION OF ENERGY The law of conservation of energy (the first law of thermodynamics), when related to heat transfer between two objects, can be stated as:The heat lost by the hot object = the heat gained by the cold object-qhot = qcold-mh x sh x DTh = mc x sc x DTcwhere DT = Tfinal - Tinitial
20Measuring Heat Changes A calorimeter is an instrumentused to measure heat changes.By placing an object in an insulatedcontainer, the heat loss can bemeasured by the temperature gainof the water.Heat lost by object = Heat gained by water-qobject = qwater
21PRACTICE PROBLEM #71. Iron metal has a specific heat of J/goC. How much heat is transferred to a 5.00 g piece of iron, initially at 20.0 oC, when it is placed in a beaker of boiling water at 1 atm?2. How many calories of energy are given off to lower the temperature of g of iron from oC to 35.0 oC?3. If 3.47 kJ were absorbed by 75.0 g H2O at 20.0 oC, what would be the final temperature of the water?4. A 100. g sample of water at 25.3 oC was placed in a calorimeter g of lead shots (at 100 oC) was added to the calorimeter and the final temperature of the mixture was 34.4 oC. What is the specific heat of lead?5. A 17.9 g sample of unknown metal was heated to oC. It was then added to g of water in an insulted cup. The water temperature rose from oC to 23.98oC. What is the specific heat of the metal in J/goC?180. J1.23 x 103 cal31.1 oC1.28 J/g oC0.792 J/goC
22LAW OF CONSERVATION OF ENERGY Assuming no heat is lost, what mass of cold water at 0.00oC is needed to cool g of water at 97.6oC to 12.0 oC?-mh x sh x DTh = mc x sc x DTc- (100.0g) (1 cal/goC) ( oC) = m (1 cal/goC) ( oC)8560 cal = m (12.0 cal/g)m = cal / (12.0 cal/g)m = 713 gCalculate the specific heat of an unknown metal if a g piece at 100.0oC is dropped into mL of water at 17.8 oC. The final temperature of the mixture was 39.4oC.s (metal) = cal/g oC
23GROUP STUDY PROBLEM #7_____1. A g metal bar requires kJ to change its temperature from 22.0oC to 100.0oC. What is the specific heat of the metal in J/goC?_____2. How many joules are required to lower the temperature of g of iron from 75.0 oC to 25.0 oC?_____ 3. If 40.0 kJ were absorbed by g H2O at 10.0 oC, what would be the final temperature of the water?_____ 4. A 250 g of water at oC is mixed with mL of water at 5.0 oC. Calculate the final temperature of the mixture._____5. A 400 g piece of gold at 500.0oC is dropped into 15.0 L of water at 22.0oC. The specific heat of gold is J/goC or cal/goC. Calculate the final temperature of the mixture assuming no heat is lost to the surroundings.
24A solid becoming a liquid is called: FreezingMeltingEvaporationSublimationCondensationDeposition
25A gas becoming a liquid is called: FreezingMeltingEvaporationSublimationCondensationDeposition
26A solid becoming a gas is called: FreezingMeltingEvaporationSublimationCondensationDeposition
27A liquid becoming a gas is called: FreezingMeltingEvaporationSublimationCondensationDeposition
28Attractive Forces In gases, these attractive forces are minimal. In solids and liquids, the forces are strong enough to keep the materials from scattering everywhere.These attractive forces also determine the melting point and boiling point of different compounds. (ex. NaCl melts at 801o C)
29Changes in StateMaterials experience a change in state when enough heat energy is applied to break apart (or form) the attractions between molecules.When intermolecular bonds are broken, heat is absorbed from the surroundings; when the bonds are formed, heat is taken from the system and released to the surroundings.
30Changes in StateDifferent states of matter (solids, liquids, and gases) have very different properties due to attractive forces that exist between atoms.To change from a solid to a liquid, for example, these attractive forces in solids must be broken so that the liquid molecules have more freedom to move. In gases, the molecules have even more mobility.
31When a solid melts, heat is removed from the surrounding environment to break those intermolecular forces.TrueFalse
32When a gas condenses, heat is removed from the surrounding environment? TrueFalse
40Melting Point / Freezing Point The melting point (same temperature as freezing point) is the temperature at which a solid turns to a liquid.Latent Heat of Fusion – the amount of heat needed to freeze one gram of a substance (or the heat released when one gram of a substance melts.)Note: units in J/g
41Heat = mass x Heat of Fusion Latent Heat of FusionQ = m x HfusHeat = mass x Heat of Fusion****For ice to water: Hfus = 334 J/g(every substance has a difference heat of fusion)
42Boiling Point / Condensation Point The boiling point (same temperature as condensation point) is the temperature at which a liquid turns to a gas.Latent Heat of Vaporization – the amount of heat needed to vaporize one gram of a substance (or the heat release when one gram of a substance condenses.)Note: units in J/g
43Latent Heat of Vaporization Q = m x HvapHeat = mass x Heat of Vaporization****For water to steam: Hvap = 2260 J/g(every substance has a difference heat of vaporization)
45Heat Graph calculations When the graph is flat, use latent heat equations because of change of state.For melting: Q = m x HfusFor evaporating: Q = m x Hvap
46Specific HeatDifferent substances absorb (and lose) heat at different rates.Specific heat (SH) is the amount of heat (q) needed to raise the temperature of 1 gram of a substance by 1 degree Celsius.
47Heat Graph calculations When the graph is sloped, use specific heat equations because of change of temperatureQ = SH x m x Temp. Change or
48Measuring Heat Changes A calorimeter is an instrumentused to measure heat changes.By placing an object in an insulatedcontainer, the heat loss can bemeasured by the temperature gainof the water.Heat lost by object = Heat gained by water-qobject = qwater
49Why do chemical reactions occur between some substances and not in others?
50Chemical reactions occur so that the atoms in the elements involved attain a more stable state of being.
51Collision TheoryCollision theory – molecules must collide with the proper orientation and sufficient energy to react.
55Condition Necessary for Reactions to Occur Collision: Reactants must collide.2) Orientation: The reactants must align properly to react.3) Energy: The activation energy must be attained to react.
56Energy in Chemical Reactions Many chemical reactions also produce energy changes.Definitions:System – the reactants and products in the reactionSurroundings – everything else around the reaction (eg air in the room, reaction flask)
57Heat of ReactionHeat of Reaction (ΔH) – the amount of heat lost or gained in a reactionHeat of Reaction: ΔH = Hproducts – Hreactants
58Exothermic ReactionsExothermic Reactions – energy is produced by a reaction; energy flows from the system to the surroundingsΔH is negative because the reaction loses heat.
62Classify the reaction: 2H2O + 572kJ -> 2H2+ O2 ExothermicEndothermic
63ΔH of the reaction = -560kJExothermicEndothermic
64After the reaction, your hand gets burnt from the heat After the reaction, your hand gets burnt from the heat. The reaction must be:ExothermicEndothermic
65Heat Values in Chemical Reactions Heat of Reaction is a stoichiometric value and is proportional to the coefficients of the reactants and products.2H2O + 572kJ -> 2H2+ O2Therefore, for every 2 moles of water that react, 572kJ of energy are required.
66Presence of a Catalyst – a substance that increases the rate without being permanently changed - lowers activation energy
67Also used:inhibitors – “tie up” a reaction so that it does not occur (opposite of a catalyst)- preservatives- anti-rust agents