Thermal Energy A. Temperature & Heat 1. Temperature is related to the average kinetic energy of the particles in a substance. 1

Temperature and Thermal Energy Why does water burn your skin so much quicker than air? Why is falling into a 32º F lake more dangerous than standing outside naked on a 32º F? Temperature - measure of the average kinetic energy of the particles in a substance - particles in box on right have higher temperature - higher velocity = more KE = higher temperature Both boxes have same temperature - particles have same average velocity/KE - box on right has more thermal energy - energy contained in a substance - more particles 2

2. SI unit for temp. is the Kelvin a. K = C (10C = 283K) b. C = K – 273 (10K = -263C) 3. Thermal Energy – the total of all the kinetic and potential energy of all the particles in a substance. 3

4. Thermal energy relationships a. As temperature increases, so does thermal energy (because the kinetic energy of the particles increased). b. Even if the temperature doesn’t change, the thermal energy in a more massive substance is higher (because it is a total measure of energy). 4

5. Heat a. The flow of thermal energy from one object to another. b. Heat always flows from warmer to cooler objects. Ice gets warmer while hand gets cooler Cup gets cooler while hand gets warmer 5

Heat A form of energy transfer between two objects External energy - total potential and kinetic energy of an every-day sized object Internal energy - total kinetic energy of the molecules in that object External can be transferred to internal, resulting in a temperature increase 6

Heat versus temperature Temperature A measure of hotness or coldness of an object Based on average molecular kinetic energy Heat Based on total internal energy of molecules Doubling amount at same temperature doubles heat 7

Resistance of water to temperature changes Ex.: lake – bird ¿Were is more internal energy? 8

6. Specific Heat a. Some things heat up or cool down faster than others. Land heats up and cools down faster than water 9

b. Specific heat is the amount of heat required to raise the temperature of 1 kg of a material by one degree (C or K). 1) C water = 4184 J / kg C 2) C sand = 664 J / kg C This is why land heats up quickly during the day and cools quickly at night and why water takes longer. 10

Why does water have such a high specific heat? Water molecules form strong bonds with each other; therefore it takes more heat energy to break them. Metals have weak bonds and do not need as much energy to break them. water metal 11

How to calculate changes in thermal energy Q = m Cp  T Q = change in thermal energy m = mass of substance  T = change in temperature (Tf – Ti) Cp = specific heat of substance 12

Equivalente mecánico del calor James Joule encontró la equivalencia entre el trabajo y el calor, para ello se ayudo de un dispositivo como el de la figura. 1 cal = J 13

c. A calorimeter is used to help measure the specific heat of a substance. First, mass and temperature of water are measured Then heated sample is put inside and heat flows into water  T is measured for water to help get its heat gain This gives the heat lost by the substance Knowing its Q value, its mass, and its  T, its Cp can be calculated 14

Calorimetry Calorimetry means “measuring heat” –In practice, it is a technique used to measure specific heat Technique involves: –Raising temperature of object(s) to some value –Place object(s) in vessel containing cold water of known mass and temperature –Measure temperature of object(s) + water after equilibrium is reached A calorimeter is a vessel providing good insulation that allows a thermal equilibrium to be achieved between substances without any energy loss to the environment (styrofoam cup or thermos with lid) Conservation of energy requires that: ( Q > 0 ( < 0 ) when energy is gained (lost)) 15

EX1: A 28.5 g sample of metal with a temperature of  C is placed in a calorimeter holding g of water at 0.00  C. The temperature of the water rises to 2.00  C. a)What is the specific heat of the metal? b)What metal is it? (use the table of text) (For these problems, the final temperature of the water and the object are the same.) (Remember that the specific heat of water is always J/gºC) 16

EX2: A 12.5 g sample of metal with a temperature of  C is placed in a calorimeter holding 75.0 g of water at 22.0  C. The temperature of the water rises to 24.0  C. a)What is the specific heat of the metal? b)What metal is it? 17

Phase Transitions A phase transition occurs when the physical characteristics of the substance change from one form to another Common phase transitions are –Solid  liquid (melting) –Liquid  gas (boiling) Phase transitions involve a change in the internal energy, but no change in temperature –Kinetic energy of molecules (which is related to temperature) is not changing, but their potential energy changes as work is done to change their positions Energy required to change the phase of a given mass m of a pure substance is:  L = latent heat – depends on substance and nature of phase transition  + (–) sign used if energy is added (removed) 18

Phase Transitions All phase changes can go in either direction –Heat flowing into a substance can cause melting (solid to liquid) or boiling (liquid to gas) –Heat flowing out of a substance can cause freezing (liquid to solid) or condensation (gas to liquid) Latent heat of fusion L f is used for melting or freezing Latent heat of vaporization L v is used for boiling or condensing (somewhat larger for lower pressures) Large L f of water is partly why spraying fruit trees with water can protect the buds from freezing –In process of freezing, water gives up a large amount of energy and keeps bud temperature from going below 0°C 19

Heat vs. Temperature Graph (for water) Examples of Phase Change Question: So why do you feel so “cold” when you step out of the shower soaking wet? 20

T vs. Q for Transition from Ice to Steam Part A: Temperature of ice changes from –30°C to 0°C – Q = mc ice DT = (1.00  10 –3 kg)(2090 J/kg  °C)(30.0°C) = 62.7 J Part B: Ice melts to water at 0°C – Q = mL f = (1.00  10 –3 kg)(3.33  10 5 J/kg) = 333 J Part C: Temperature of water changes from 0°C to 100°C – Q = mc water DT = (1.00  10 –3 kg)(4.19  10 3 J/kg  °C)(100°C) = 419 J Part D: Water changes to steam at 100°C – Q = mL v = (1.00  10 –3 kg)(2.26  10 6 J/kg) = 2.26  10 3 J Part E: Temperature of steam changes from 100°C to 120°C – Q = mc steam DT = (1.00  10 –3 kg)(2.01  10 3 J/kg  °C)(20°C) = 40.2 J Initial state: 1 g of ice at –30°C Final state: 1 g of steam at 120°C Q tot = 3.11  10 3 J 21