Presentation on theme: "Energy in Thermal Processes"— Presentation transcript:
1 Energy in Thermal Processes Thermal PhysicsEnergy in Thermal Processes
2 Energy TransferWhen two objects of different temperatures are placed in thermal contact, the temperature of the warmer decreases and the temperature of the cooler increasesThe energy exchange ceases when the objects reach thermal equilibriumThe concept of energy was broadened from just mechanical to include internalMade Conservation of Energy a universal law of nature
3 Heat Compared to Internal Energy Important to distinguish between themThey are not interchangeableThey mean very different things when used in physics
4 Internal EnergyInternal Energy, U, is the energy associated with the microscopic components of the systemIncludes kinetic and potential energy associated with the random translational, rotational and vibrational motion of the atoms or moleculesAlso includes any potential energy bonding the particles together
5 Thermal EnergyThermal Energy is the portion of the Internal Energy, U, that is associated with the motion of the microscopic components of the system.
6 HeatHeat is the transfer of energy between a system and its environment because of a temperature difference between themThe symbol Q is used to represent the amount of energy transferred by heat between a system and its environment
7 Units of HeatCalorieAn historical unit, before the connection between thermodynamics and mechanics was recognizedA calorie is the amount of energy necessary to raise the temperature of 1 g of water from 14.5° C to 15.5° C .A Calorie (food calorie) is 1000 cal
8 Units of Heat, cont. US Customary Unit – BTU BTU stands for British Thermal UnitA BTU is the amount of energy necessary to raise the temperature of 1 lb of water from 63° F to 64° F1 cal = JThis is called the Mechanical Equivalent of Heat
9 Problem: Working Off Breakfast A student eats breakfast consisting of two bowls of cereal and milk, containing a total of 3.20 x 102 Calories of energy. He wishes to do an equivalent amount of work in the gymnasium by doing curls with a 25 kg barbell. How many times must he raise the weight to expend that much energy? Assume that he raises it through a vertical displacement of 0.4 m each time, the distance from his lap to his upper chest.h
10 Problem: Working Off Breakfast Convert his breakfast Calories, E, to joules:
11 Problem: Working Off Breakfast Use the work-energy theorem to find the work necessary to lift the barbell up to its maximum height.The student must expend the same amount of energy lowering the barbell, making 2mgh per repetition. Multiply this amount by n repetitions and set it equal to the food energy E:
12 Problem: Working Off Breakfast Solve for n, substituting the food energy for E:
13 James Prescott Joule 1818 – 1889 British physicist Conservation of EnergyRelationship between heat and other forms of energy transfer
14 Specific HeatEvery substance requires a unique amount of energy per unit mass to change the temperature of that substance by 1° CThe specific heat, c, of a substance is a measure of this amount
15 Units of Specific HeatSI unitsJ / kg °CHistorical unitscal / g °C
16 Heat and Specific Heat Q = m c ΔT ΔT is always the final temperature minus the initial temperatureWhen the temperature increases, ΔT and ΔQ are considered to be positive and energy flows into the systemWhen the temperature decreases, ΔT and ΔQ are considered to be negative and energy flows out of the system
17 A Consequence of Different Specific Heats Water has a high specific heat compared to landOn a hot day, the air above the land warms fasterThe warmer air flows upward and cooler air moves toward the beach
18 CalorimeterOne technique for determining the specific heat of a substanceA calorimeter is a vessel that is a good insulator which allows a thermal equilibrium to be achieved between substances without any energy loss to the environment
19 Calorimetry Analysis performed using a calorimeter Conservation of energy applies to the isolated systemThe energy that leaves the warmer substance equals the energy that enters the waterQcold = -QhotNegative sign keeps consistency in the sign convention of ΔT
20 Calorimetry with More Than Two Materials In some cases it may be difficult to determine which materials gain heat and which materials lose heatYou can start with SQ = 0Each Q = m c DTUse Tf – TiYou don’t have to determine before using the equation which materials will gain or lose heat
21 Phase ChangesA phase change occurs when the physical characteristics of the substance change from one form to anotherCommon phases changes areSolid to liquid – meltingLiquid to gas – boilingPhases changes involve a change in the internal energy, but no change in temperature
22 Latent Heat During a phase change, the amount of heat is given as Q = ±m LL is the latent heat of the substanceLatent means hiddenL depends on the substance and the nature of the phase changeChoose a positive sign if you are adding energy to the system and a negative sign if energy is being removed from the system
23 Latent Heat, cont. SI units of latent heat are J / kg Latent heat of fusion, Lf, is used for melting or freezingLatent heat of vaporization, Lv, is used for boiling or condensingTable 11.2 gives the latent heats for various substances
24 Problem: Boiling Liquid Helium Liquid helium has a very low boiling point, 4.2 K, as well as low latent heat of vaporization, 2.09 x 104 J/kg. If energy is transferred to a container of liquid helium at the boiling point from an immersed electric heater at a rate of 10 W, how long does it take to boil away 2 kg of the liquid?
25 Problem: Boiling Liquid Helium Find the energy needed to vaporize 2 kg of liquid helium at its boiling point:Divide this result by the power to find the time:
26 SublimationSome substances will go directly from solid to gaseous phaseWithout passing through the liquid phaseThis process is called sublimationThere will be a latent heat of sublimation associated with this phase change
28 Warming Ice Start with one gram of ice at –30.0º C During A, the temperature of the ice changes from –30.0º C to 0º CUse Q = m c ΔTWill add 62.7 J of energy
29 Melting Ice Once at 0º C, the phase change (melting) starts The temperature stays the same although energy is still being addedUse Q = m LfNeeds 333 J of energy
30 Warming WaterBetween 0º C and 100º C, the material is liquid and no phase changes take placeEnergy added increases the temperatureUse Q = m c ΔT419 J of energy are added
31 Boiling Water At 100º C, a phase change occurs (boiling) Temperature does not changeUse Q = m Lv2 260 J of energy are needed
32 Heating SteamAfter all the water is converted to steam, the steam will heat upNo phase change occursThe added energy goes to increasing the temperatureUse Q = m c ΔTTo raise the temperature of the steam to 120°, 40.2 J of energy are needed
33 Problem Solving Strategies Make a tableA column for each quantityA row for each phase and/or phase changeUse a final column for the combination of quantitiesUse consistent units
34 Problem Solving Strategies, cont Apply Conservation of EnergyTransfers in energy are given as Q=mcΔT for processes with no phase changesUse Q = m Lf or Q = m Lv if there is a phase changeIn Qcold = - Qhot be careful of signΔT is Tf – TiSolve for the unknown
35 Your TurnYou start with 250. g of ice at -10 C. How much heat is needed to raise the temperature to 0 C?10.5 kJHow much more heat would be needed to melt it?83.3 kJ
36 Your TurnYou start with 250. g of ice at -10 C. What will happen if we add 50. kJ of heat?10.5 kJ will be used to warm it up to the MP, and the rest will start melting the ice.0.119 kg will be melted
37 Problem: Partial melting A 5 kg block of ice at 0o C is added to an insulated container partially filled with 10 kg of water at 15 o C.(a) Find the temperature, neglecting the heat capacity of the container.(b) Find the mass of the ice that was melted.
38 Problem: Partial melting (a) Find the equilibrium temperature.First, Compute the amount of energy necessary to completely melt the ice:
39 Problem: Partial melting Next, calculate the maximum energy that can be lost by the initial mass of liquid water without freezing it:This is less than half the energy necessary to melt all the ice, so the final state of the system is a mixture of water and ice at the freezing point:
40 Problem: Partial melting (b) Compute the mass of the ice melted.Set the total available energy equal to the heat of fusion of m grams of ice, mLf:
41 Final Problem100. grams of hot water ( 60. C) is added to a 1.0 kg iron skillet at 500 C. What is the final temperature and state of the mixture?
42 Final Problem 16.7 kJ needed to warm water to BP. 226 kJ needed to vaporize water199.2 kJ will be given up by skillet.Final temperature will be 100. C182 kJ of heat from the skillet will be available to vaporize water81 grams of water will vaporize.
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