Thermal Physics

1. Thermal Physics Energy in Thermal Processes

2. Energy Transfer • When two objects of different temperatures are placed in thermal contact, the temperature of the warmer decreases and the temperature of the cooler increases • The energy exchange ceases when the objects reach thermal equilibrium • The concept of energy was broadened from just mechanical to include internal • Made Conservation of Energy a universal law of nature

3. Heat Compared to Internal Energy • Important to distinguish between them • They are not interchangeable • They mean very different things when used in physics

4. Internal Energy • Internal Energy, U, is the energy associated with the microscopic components of the system • Includes kinetic and potential energy associated with the random translational, rotational and vibrational motion of the atoms or molecules • Also includes any potential energy bonding the particles together

5. Thermal Energy • Thermal Energy is the portion of the Internal Energy, U, that is associated with the motion of the microscopic components of the system.

6. Heat • Heat is the transfer of energy between a system and its environment because of a temperature difference between them • The symbol Q is used to represent the amount of energy transferred by heat between a system and its environment

7. Units of Heat • Calorie • An historical unit, before the connection between thermodynamics and mechanics was recognized • A 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 Unit • A BTU is the amount of energy necessary to raise the temperature of 1 lb of water from 63° F to 64° F • 1 cal = 4.186 J • This 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 Energy • Relationship between heat and other forms of energy transfer

14. Specific Heat • Every substance requires a unique amount of energy per unit mass to change the temperature of that substance by 1° C • The specific heat, c, of a substance is a measure of this amount

15. Units of Specific Heat • SI units • J / kg °C • Historical units • cal / g °C

16. Heat and Specific Heat • Q = m c ΔT • ΔT is always the final temperature minus the initial temperature • When the temperature increases, ΔT and ΔQ are considered to be positive and energy flows into the system • When 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 land • On a hot day, the air above the land warms faster • The warmer air flows upward and cooler air moves toward the beach

18. Calorimeter • One technique for determining the specific heat of a substance • A 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 system • The energy that leaves the warmer substance equals the energy that enters the water • Qcold = -Qhot • Negative 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 heat • You can start with SQ = 0 • Each Q = m c DT • Use Tf – Ti • You don’t have to determine before using the equation which materials will gain or lose heat

21. Phase Changes • A phase change occurs when the physical characteristics of the substance change from one form to another • Common phases changes are • Solid to liquid – melting • Liquid to gas – boiling • Phases 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 L • L is the latent heat of the substance • Latent means hidden • L depends on the substance and the nature of the phase change • Choose 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 freezing • Latent heat of vaporization, Lv, is used for boiling or condensing • Table 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. Sublimation • Some substances will go directly from solid to gaseous phase • Without passing through the liquid phase • This process is called sublimation • There will be a latent heat of sublimation associated with this phase change

27. Graph of Ice to Steam

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 C • Use Q = m c ΔT • Will 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 added • Use Q = m Lf • Needs 333 J of energy

30. Warming Water • Between 0º C and 100º C, the material is liquid and no phase changes take place • Energy added increases the temperature • Use Q = m c ΔT • 419 J of energy are added

31. Boiling Water • At 100º C, a phase change occurs (boiling) • Temperature does not change • Use Q = m Lv • 2 260 J of energy are needed

32. Heating Steam • After all the water is converted to steam, the steam will heat up • No phase change occurs • The added energy goes to increasing the temperature • Use Q = m c ΔT • To raise the temperature of the steam to 120°, 40.2 J of energy are needed

33. Problem Solving Strategies • Make a table • A column for each quantity • A row for each phase and/or phase change • Use a final column for the combination of quantities • Use consistent units

34. Problem Solving Strategies, cont • Apply Conservation of Energy • Transfers in energy are given as Q=mcΔT for processes with no phase changes • Use Q = m Lf or Q = m Lv if there is a phase change • In Qcold = - Qhot be careful of sign • ΔT is Tf – Ti • Solve for the unknown

35. Your Turn • You start with 250. g of ice at -10 C. How much heat is needed to raise the temperature to 0 C? • 10.5 kJ • How much more heat would be needed to melt it? • 83.3 kJ

36. Your Turn • You 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 Problem • 100. 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 water • 199.2 kJ will be given up by skillet. • Final temperature will be 100. C • 182 kJ of heat from the skillet will be available to vaporize water • 81 grams of water will vaporize.