1. Chapter 15 Temperature and Heat

2. Mechanics vs. Thermodynamics Mechanics: obeys Newton’s Laws key concepts: force kinetic energy static equilibrium Newton’s 2nd Law Thermodynamics: will find new ‘laws’ key concepts: temperature, heat internal energy thermal equilibrium 2nd Law of Thermodynamics

3. Temperature (T) Temperature = a macroscopic quantity (see later: T is related to KE of particles) many properties of matter vary with T: (length, volume, pressure of confined gas)

4. Temperature (T) Human senses can be deceiving On a cold day: iron railings feel colder than wooden fences, but both have the same T How can we define T ? Look for macroscopic changes in a system when heat is added to it

5. Two Thermometers Add heat to (a) and (b). (a) liquid thermometer liquid level rises T is measured by L (b) constant volume gas thermometer gas pressure p rises T is measured by p

6. Using Thermometers put the bulb of (a) in contact with a body wait until the value of L (i.e. T) settles out the thermometer and the body have reached thermal equilibrium (they have the same T)

7. Consider thermal interactions of systems in (a). red slab = thermal conductor (transmits interactions) blue slab = thermal insulator (blocks interactions) Demonstration

8. Let A and C reach thermal equilibrium (T A =T C ). Let B and C reach thermal equilibrium (T B =T C ). Then are A and B in thermal equilibrium (T A =T B )? Demonstration

9. In (a), are A and B in thermal equilibrium? Yes, but it’s not obvious! It must be proved by experiment! Demonstration

10. Experimentally, consider going from (a) to (b): Thermally couple A to B and thermally decouple C. Experiments reveal no macroscopic changes in A, B! Demonstration

11. This suggests the Zeroth Law of Thermodynamics: If C is in thermal equilibrium with both A and B, then A and B in thermal equilibrium with each other. Demonstration

12. This means: If two systems A and B are in thermal equilibrium, they must have the same temperature (T A =T B ), and vice versa Demonstration

13. Temperature Scales

14. Three scales: Fahrenheit, Celsius, Kelvin To define a temperature scale, we need one or more thermodynamic fixed points fixed point = a convenient, reproducible thermodynamic environment

15. Temperature Scales Both Fahrenheit and Celsius scales are defined using two fixed points: freezing point and boiling point of water Kelvin scale defined using one fixed point: ‘triple point’ of water (all three phases coexist: ice, liquid, vapor)

16. Temperature Scales: Summary Relations among temperature scales: Fahrenheit temperature Celsius temperature Kelvin temperature

17. Temperature Scales: Kelvin vs. Celsius triple point of water: we measure T C, triple = 0.01 o C we define T K, triple = 273.16 K (  T) K = (  T) C so the unit of  T is K or o C the scales differ only by an offset, so: T K = T C +273.15

18. Kelvin Temperature Scale Fixed point = triple point of water: T K, triple p = pressure of ‘ideal’ (i.e. low density) gas (on a constant volume gas thermometer) (has value p triple at T K, triple ) We define:

19. At low density, see same graph for all gases Extrapolate to p=0 (at T = absolute zero K) Demonstration

20. Thermal Expansion

21. Empirical law for solids, valid for small  T (simple case: all directions expand equally) For  > 0: If  T > 0:  L > 0, material expands If  T < 0:  L < 0, material compresses

22. Thermal Expansion  = coefficient of linear expansion > 0 (almost always) characterizes thermal properties of matter varies with material (and range of T) unit: 1/K, or 1/ o C since (  T) K = (  T) C

23. Thermal Expansion Example: two different materials have different  L They can be used to build a thermometer or a thermostat

24. Atomic explanation of thermal expansion! Recall ‘spring’ model for diatomic molecule: Van der Waals potential energy, U Demonstration

25. Thermal Expansion Similar for a solid made of many atoms Each pair of atoms has a potential energy U The asymmetry of U explains thermal linear expansion!

26. Thermal Volume Expansion: Solids and Liquids  = coefficient of volume expansion varies with material (and range of T) unit: 1/K, or 1/ o C since (  T) K = (  T) C

27. Thermal Volume Expansion: Solids Find a simple relationship between linear and volume expansion coefficients:  = 3 

28. Thermal Expansion of Water ‘unusual’ state:  < 0 if 0 o C < T < 4 o C (it’s why lakes freeze from the top down)

29. Thermal Stress Thermal stress= stress required to counteract (balance) thermal expansion Tensile thermal stress:

30. Announcements Midterms: will probably be returned Monday Homework 5: is returned at front Homework Extra Credit: is on record (but not yet listed on classweb if it brings a score over the maximum)

31. Temperature Scales: Kelvin vs. Celsius triple point of water: we measure T C, triple = 0.01 o C we define T K, triple = 273.16 K (  T) K = (  T) C so the unit of  T is K or o C the scales differ only by an offset, so: T K = T C +273.15

32. Heat and Heat Transfer

33. Quantity of Heat (Q) Heat = energy absorbed or lost by a body due to a temperature difference Heat = energy ‘in transit’ SI unit: J other units: 1 cal = 4.186 J 1 kcal = ‘calorie’ on food labels

34. Quantity of Heat (Q) Q > 0: heat is absorbed by a body Q < 0: heat leaves a body (we will see several expressions for Q)

35. Quantity of Heat (Q) Conservation of energy (‘calorimetry’): For an isolated system, the algebraic sum of all heat exchanges add to zero Q 1 + Q 2 + Q 3 +... = 0

36. Absorption of Heat Q = heat energy required to change the temperature of material (mass m) by  T c = ‘specific heat capacity’ of the material (treat as independent T) unit: J/(kg ·K)

37. Absorption of Heat If Q and  T positive: heat absorbed by m If Q and  T negative: heat leaves m Do Exercise 15-35

38. Phase Changes ‘phase’ = state of matter = solid, liquid, vapor energy is needed to change phase of matter under a phase transition of matter: only its phase changes, not its temperature!

39. Phase Changes in Water

40. Solid-Liquid Phase Change: Q = ± mL f ± mL f = heat needed for phase change L f = ‘(latent) heat of fusion’ of the material = (heat/unit mass) needed for transition unit: J/kg + for melting (solid to liquid) – for freezing (liquid to solid) Do Exercise 15-51

41. Liquid-Vapor Phase Change: Q = ± mL v ± mL v = heat needed for phase change L v = ‘(latent) heat of vaporization’ = (heat/unit mass) needed for transition unit: J/kg + for evaporating (liquid to vapor) – for condensing (vapor to liquid)

42. Heat Transfer

43. dQ/dt = rate of heat flow = ‘heat current’ Three mechanisms for achieving heat transfer: Conduction Convection Radiation

44. Heat Transfer Mechanisms Conduction: Collisions of molecules, no bulk motion Convection: Bulk motion from one region to another Radiation: Emission of electromagnetic waves

45. Conduction

46. k = thermal conductivity of material unit: W/(m·K) A = cross sectional area of material L = length of material

47. Conduction Do Exercises 15-57, 15-58 Notes on a composite conducting rod

48. Convection (usually complicated)

49. Radiation (e.g. emitted by the sun)

50. Radiation = Electromagnetic Waves

51. Emission of Radiation all bodies emit electromagnetic radiation A = surface area of body T = surface temperature of body e = emissivity of body (0 < e < 1) Do Exercise 15-67

52. Absorption of Radiation In general, bodies emit radiation and also absorb radiation from their surroundings T = surface temperature of body T S = surface temperature of surroundings Example of net radiation and Problem 15-89