Chapter 12 Thermal Energy

1. Chapter 12 Thermal Energy Thermodynamics - The study of heat Kinetic Theory - All matter consists of minute particles which are in constant motion Thermal Energy - The overall energy of motion of the particles that make up an object. Temperature Scale

2. Chapter 12 Thermal Energy Temperature - The average kinetic energy of the particles. Heat - The transfer of thermal energy because of a difference in temperature. • Thermal Energy Transfer • Conduction - The transfer of energy when particles collide. • Convection - The transfer of heat by means of motion in a fluid. • Radiation - The transfer of energy by electromagnetic waves.

3. Chapter 12Thermal Energy • Temperature Scales • Kelvin • Fahrenheit • Celsius Absolute Zero -273.15º C or 0 K Tk = Tc + 273 Absolute Zero - The temperature at which all the thermal energy has been removed from the gas.

4. Chapter 12 Thermal Energy Temperatures Fahrenheit ºF Celsius ºC Kelvin H20 Boils 212 100 373 H20 Freezes 32 0 273 CO2 Freezes -189 -123 150 Nitrogen Boils -320 -196 77 Absolute Zero -459 -273 0

5. Chapter 12 Thermal Energy 1st Law of Thermodynamics – When energy is converted to heat, all energy is conserved. • Rumford’s Experiment • Disproved Caloric Theory • Showed a relationship between heat and work • Joules’ Experiment • Related heat to energy • N(mgh) converts to heat

6. Chapter 12 Thermal Energy 2nd Law of Thermodynamics – Heat flows from hot to cold. Law of Entropy The universe is continuously going from a state of order to disorder.

7. Chapter 12 Thermal Energy ΔQ The transfer of thermal energy, measured in joules 1 calorie = 4.18 joules • Specific Heat ( C ) The amount of energy needed to to raise the temperature of a unit mass one temperature unit. • Specific Heat is measured in J/kg·K or J/kg·°C • Table Pg 279 Heat Transfer Q = mCΔT = mC(Tfinal-Tinitial)

8. Chapter 12 Thermal Energy How much heat is needed to raise 20 grams of water from 40ºC to 70 ºC? m = 20 g C = 4.18 J/g C º Ti = 40 ºC Tf = 70 ºC ΔQ = mCΔT ΔQ = (20g)(4.18 J/g C)(70 ºC - 40 ºC ) ΔQ = 2500 J

9. Chapter 12 Thermal Energy Find the specific heat of tungsten if it takes 100 joules of energy to raise 20 grams of the material from 20 ºC to 57 ºC. m = 20 g C = Tf = 57 ºC Ti = 20 ºC Q = 100 J ΔQ = mCΔT C = Q/(mΔT) C = 100 J/(20 g*37 ºC) C = .135 J/g ºC

10. Chapter 12 Thermal Energy 149,400 J of heat are added to a 5 kg mass of a substance that raises the temperature from -25ºC to 20º C. What is the material? m = 5000 g C = Tf = 20 ºC Ti = -25 ºC Q = 149,400 J ΔQ = mCΔT C = Q/(mΔT) C = 149,400 J/(5000 g*45 ºC) C = .644 J/g ºC The material is glass

11. Chapter 12 Thermal Energy Method of mixtures Heat gained plus heat loss in a closed system is zero. ΔQgained+ΔQlost = 0

12. Chapter 12 Thermal Energy What is the final temperature of a mixture where 100 grams of iron at 80ºC is added to 53.5 g of water at 20ºC? ΔQgained+ΔQlost = 0 (53.5 g)(4.18 J/gºC)(Tf - 20ºC ) + (100g)(.45 J/g ºC)(Tf - 80ºC) = 0 (223.63 J/ºC)(Tf – 20ºC )+(45 J/ ºC)(Tf – 80 ºC) = 0 Tf = 30ºC

13. Chapter 12 Thermal Energy What is the final temperature of a mixture where 400 grams of alcohol at 16ºC is added to 400 g of water at 85ºC? ΔQgained+ΔQlost = 0 (400 g)(2.45 J/g ºC)(Tf - 16 ºC ) + (400g)(4.18 J/g º C)(Tf - 85 ºC) = 0 (980 J/ ºC)(Tf – 16 ºC )+(1672 J/ ºC)(Tf – 85 ºC) = 0 980 Tf – 15680 + 1672 Tf -142120 = 0 2652 Tf = 157800 Tf= 59.5 ºC

14. Chapter 12 Thermal Energy What is the specific heat of a substance that absorbs 2.5 x 103 joules of heat when a sample of 1.0 x 104 g of the substance increases in temperature from 10.0º C to 70.0° C?

15. Chapter 12 Thermal Energy A 1.0 kg sample of metal with a specific heat of 0.50 J/g°C is heated to 100.0º C and then placed in a50.0 g sample of water at 20.0°C.What is the final temperature of the metal and the water?

16. Chapter 12 Thermal Energy A 1.0 kg sample of metal with a specific heat of 0.50 J/g°C is heated to 100.0º C and then placed in a50.0 g sample of water at 20.0°C.What is the final temperature of the metal and the water?

17. Chapter 12Change of State Sublimation Melting Vaporization Solid Liquid Gas Freezing Condensation Supercooled

18. Chapter 12Thermal Energy Heat of Fusion (Hf) The amount of energy needed to melt a unit mass of a substance. Melting Point Heat of Vaporization (Hv) The amount of heat needed to vaporize a unit mass of a liquid. Boiling Point There is no temperature change in changing between states. Q = m Hf Q = m Hv Chart Pg 287

19. Chapter 12Thermal Energy How much energy is needed to melt 20 grams of ice at 0º? Q = mHf = (20g)(334 J/g) = 6680 J How much energy is needed to change 30 grams of water at 100 ºC to steam? Q = mHv = (30g)(2260 J/g) = 67800 J

21. Chapter 12Thermal Energy How much energy is needed to melt 40 grams of ice at -60ºC to steam at 150ºC? • Warm the ice Q = mCΔT = (40 g)(2.06 J/gºC)(60ºC) = 4944 J • Melt the ice Q = mHf = (40 g)(334 J/ g) = 13360 J • Warm the water Q = mCΔT = (40g)(4.18 J/g ºC)(100 ºC) = 16720 J • Vaporize the water Q = mHv = (40g)(2260 J/g) = 90400 J • Warm the steam Q = mCΔT = (40g)(2.02 J/ gºC)(50ºC) = 4040 J Qtotal = 4944 J + 13360 J + 16720 J + 90400 J + 4040 J = 129464 J

22. Chapter 12Thermal Energy How much energy is needed to melt 20 grams of ice at -10ºC to steam at 130ºC? • Warm the ice Q = mCΔT = (20 g)(2.06 J/gºC)(10ºC) = 412 J • Melt the ice Q = mHf = (20 g)(334 J/ g) = 6680 J • Warm the water Q = mCΔT = (20g)(4.18 J/g ºC)(100 ºC) = 8360 J • Vaporize the water Q = mHv = (20g)(2260 J/g) = 45200 J • Warm the steam Q = mCΔT = (20g)(2.02 J/ gºC)(30ºC) = 1212 J Qtotal = 412 J + 6680 J + 8360 J + 45200 J + 1212 J = 61864 J