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# Thermodynamics

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1. Thermodynamics Brief walk-through of temperature, heat, and energy transfer

2. Temperature • Temperature: A measure of how hot (or cold) something is • Specifically, a measure of the average kinetic energy of the particles in an object.

3. Thermometers • Thermometer: an instrument that measures and indicates temperature a. b. c. Bimetallic strip

4. Temperature Scales • Kelvin • International System (SI) of measurement • Fahrenheit • Based on 30 being the freezing point of water and 100 as the normal body temperature (later revised to 32 and 98.6) • Celsius • Based on water freezing at 0°C, boiling at 100°

5. Kelvin and Absolute Zero • The Kelvin scale is based on absolute zero • Absolute Zero: the temp at which molecular movement stops • 0 K on the Kelvin scale = -273.16ºC • Tc + 273 = TK

6. What is heat? • Heat: ENERGY created by the motion of atoms and molecules. All matter has heat. • Put energy into a system and it heats up. Take energy away and it cools down. (applet) • The more kinetic energy the particles of a substance have, the greater the temperature of the object

7. Temperature and Thermal Energy Temperature Limits • The wide range of temperatures present in the universe is shown in the figure. • Temperatures do not appear to have an upper limit. The interior of the Sun is at least 1.5×107°C. Temperatures do, however, have a lower limit.

8. Energy Transfer • The transfer of heat is normally from a high temperature object to a lower temperature object.

9. 1) Conduction • Thermal Conduction: the transfer of heat within a substance, molecule by molecule.

10. 2) Convection • Convection: the movement of matter due to the differences in density that are caused by temp. variations applet

11. 3) Radiation • Radiation: the energy that is transferred as electromagnetic waves, Doesn’t need matter • Most radiation comes from the sun

12. Conductor vs. Insulator • Conductor: any material through which energy can be transferred as heat • Insulator: poor conductors

13. Energy conversion • Mechanical energy is converted into thermal energy whenever you bounce a ball. Each time the ball hits the ground, some of the energy of the ball's motion is converted into heating up the ball, causing it to slow down at each bounce

14. Heat Transfer • The specific heat of a material is the amount of energy that must be added to the material to raise the temperature of a unit mass by one temperature unit. • Heat Transfer = Q = mC(Tf – Ti) • Where: m = mass of object C = specific heat of object; T = temperature in Kelvins

15. Specific Heat • Liquid water has a high specific heat compared to the specific heat of other substances. • A mass of 1 kg of water requires 4180 J of energy to increase its temperature by 1 K. The same mass of copper requires only 385 J to increase its temperature by 1 K.

16. Example • A 5.10 kg cast iron skillet is heated on the stove from 295 K to 450 K. How much heat had to be transferred to the iron? The specific heat of iron is 450 J/kg*K • M = 5.10 kg; ∆T = 450 – 295 = 155 K Q = 5.10(450)(155) = 3.6 E 5 Joules

17. Conservation of Energy • A system is composed on two model blocks at different temps that are initially separated. When they are brought together, heat flows from the hotter block to the colder block. Total energy remains constant.

18. Conservation of Energy • Qbefore = Qafter • mACaTi+ mBCbTi = mACaTf + mBCbTf

19. Example – Your Turn • A calorimeter contains 0.50 kg of water at 15 degrees Celsius. A 0.040 kg block of zinc at 115 degrees Celsius is placed in the water. What is the final temperature of the system?

20. Section Temperature and Thermal Energy 12.1 Transferring Heat in a Calorimeter A calorimeter contains 0.50 kg of water at 15°C. A 0.040-kg block of zinc at 115°C is placed in the water. What is the final temperature of the system?

21. Section Temperature and Thermal Energy 12.1 Transferring Heat in a Calorimeter Step 1:Analyze and Sketch the Problem

22. Section Temperature and Thermal Energy 12.1 Transferring Heat in a Calorimeter Let zinc be sample A and water be sample B. Sketch the transfer of heat from the hotter zinc to the cooler water.

23. Section Temperature and Thermal Energy 12.1 Transferring Heat in a Calorimeter Identify the known and unknown variables. Known: mA = 0.040 kg CA = 388 J/kg·ºC TA = 115 ºC mB = 0.500 kg CB = 4180 J/kg·ºC TB= 15.0 ºC Unknown: Tf = ?

24. Section Temperature and Thermal Energy 12.1 Transferring Heat in a Calorimeter Step 2:Solve for the Unknown

25. Section Temperature and Thermal Energy 12.1 Transferring Heat in a Calorimeter Determine the final temperature using the following equation.

26. Section Temperature and Thermal Energy 12.1 Transferring Heat in a Calorimeter Substitute mA = 0.040 kg, CA = 388 J/kg·ºC, TA = 115 ºC, mB = 0.500 kg, CB = 4180 J/kg·ºC, TB= 15.0 ºC.

27. Section Temperature and Thermal Energy 12.1 Transferring Heat in a Calorimeter Step 3:Evaluate the Answer

28. Section Temperature and Thermal Energy 12.1 Transferring Heat in a Calorimeter • Are the units correct? Temperature is measured in Celsius. • Is the magnitude realistic? The answer is between the initial temperatures of the two samples, as is expected when using a calorimeter.

29. Section Temperature and Thermal Energy 12.1 Transferring Heat in a Calorimeter • Step 1:Analyze and Sketch the Problem • Let zinc be sample A and water be sample B. • Sketch the transfer of heat from the hotter zinc to the cooler water. The steps covered were:

30. Section Temperature and Thermal Energy 12.1 Transferring Heat in a Calorimeter • Step 2:Solve for the Unknown • Determine the final temperature using the following equation. • Step 3:Evaluate the Answer The steps covered were:

31. Section Temperature and Thermal Energy 12.1 Calorimetry: Measuring Specific Heat • Animals can be divided into two groups based on their body temperatures. • Most are cold-blooded animals whose body temperatures depend on the environment. • The others are warm-blooded animals whose body temperatures are controlled internally. • That is, a warm-blooded animal’s body temperature remains stable regardless of the temperature of the environment.

32. Changes of State • At normal atmospheric pressure, water boils at 373 K. The thermal energy needed to vaporize 1 kg of a liquid is called the heat of vaporization.

33. Section Changes of State and the Laws of Thermodynamics 12.2 Changes of State • The figure below diagrams the changes of state as thermal energy is added to 1.0 g of water starting at 243 K (ice) and continuing until it reaches 473 K (steam).

34. Section Changes of State and the Laws of Thermodynamics 12.2 Melting Point • This can be observed between points B and C in the figure, where the added thermal energy melts the ice at a constant 273 K. • Because the kinetic energy of the particles does not increase, the temperature does not increase between points B and C.

35. Section Changes of State and the Laws of Thermodynamics 12.2 Boiling Point • Once a solid is completely melted, there are no more forces holding the particles in the solid state. • Adding more thermal energy again increases the motion of the particles, and the temperature of the liquid rises. • In the diagram, this process occurs between points C and D.

36. Section Changes of State and the Laws of Thermodynamics 12.2 Heat of fusion • The amount of energy needed to melt 1 kg of a substance is called the heat of fusion of that substance. • The added energy causes a change in state but not in temperature. • The horizontal distance in the figure from point B to point C represents the heat of fusion.

37. Section Changes of State and the Laws of Thermodynamics 12.2 Heat of Vaporization • At normal atmospheric pressure, water boils at 373 K. • The thermal energy needed to vaporize 1 kg of a liquid is called the heat of vaporization. For water, the heat of vaporization is 2.26106 J/kg. • The distance from point D to point E in the figure represents the heat of vaporization. • Every material has a characteristic heat of vaporization.

38. Section Changes of State and the Laws of Thermodynamics 12.2 Heat of Vaporization • Between points A and B, there is a definite slope to the line as the temperature is raised. This slope represents the specific heat of the ice. • The slope between points C andD represents the specific heat of water, and the slope above point E represents the specific heat of steam. • Note that the slope for water is less than those of both ice and steam. This is because water has a greater specific heat than those of ice and steam.

39. Section Changes of State and the Laws of Thermodynamics 12.2 Heat of Vaporization • The values of some heats of fusion, Hf, and heats of vaporization, Hv, are shown in the table below.

40. Section Changes of State and the Laws of Thermodynamics 12.2 Heat Suppose that you are camping in the mountains. You need to melt 1.50 kg of snow at 0.0°C and heat it to 70.0°C to make hot cocoa. How much heat will be needed?

41. Section Changes of State and the Laws of Thermodynamics 12.2 Heat Step 1:Analyze and Sketch the Problem

42. Section Changes of State and the Laws of Thermodynamics 12.2 Heat Sketch the relationship between heat and water in its solid and liquid states.

43. Section Changes of State and the Laws of Thermodynamics 12.2 Heat Sketch the transfer of heat as the temperature of the water increases.

44. Section Changes of State and the Laws of Thermodynamics 12.2 Heat Identify the known and unknown variables. Known: mA = 1.50 kg Ti = 0.0 ºC C = 4180 J/kg· ºC Hf= 3.34×105 J/kg mB = 0.500 kg Tf= 70.0 ºC Unknown: Qmelt ice = ? Qheat liquid = ? Qtotal = ?

45. Section Changes of State and the Laws of Thermodynamics 12.2 Heat Step 2:Solve for the Unknown

46. Section Changes of State and the Laws of Thermodynamics 12.2 Heat Calculate the heat needed to melt ice. Qmelt ice = mHf

47. Section Changes of State and the Laws of Thermodynamics 12.2 Heat Substitute mA = 1.50 kg, Hf = 3.34×105 J/kg. Qmelt ice= (1.50 kg) (3.34×105 J/kg) = 5.01×105 J = 5.01×102 kJ

48. Section Changes of State and the Laws of Thermodynamics 12.2 Heat Calculate the temperature change. ΔT = Ti– Tf

49. Section Changes of State and the Laws of Thermodynamics 12.2 Heat Substitute Tf = 70.0 ºC, Ti = 0.0 ºC ΔT= 70.0 ºC –0.0 ºC = 70.0 ºC