Temperature and Thermometers. Temperature is a measure of how hot or cold something is. Most materials expand when heated.

Transcription

1 Heat Energy

2 Temperature and Thermometers Temperature is a measure of how hot or cold something is. Most materials expand when heated.

3 Thermometers are instruments designed to measure temperature. In order to do this, they take advantage of some property of matter that changes with temperature. Early thermometers:

4 Common thermometers used today include the liquid-in-glass type and the bimetallic strip.

5 Temperature is generally measured using either the Fahrenheit or the Celsius scale. The Kelvin scale is widely used in science. It has the same increments as celsius. Absolute zero is the coldest theoretical temperature that matter can approach. 0 K= C

6 The freezing point of water is 0 C, or 32 F; the boiling point of water is 100 C, or 212 F. The boiling and freezing points of water are good points for calibration of instruments. Knowing the temperatures in Celsius and Fahrenheit at these points can enable one to obtain a relationship relating the two scales. Both scales are linear. Let y= temp in Fahrenheit Let x= temp in Celsius General equation of a line: y= mx +b

7 We have two points on the line relating the two variables: (0,32) and (100,212) The slope m=(y 2 -y 1 )/(x 2 -x 1 )=9/5 The y intercept b is then found to be 32. The relationship can now be stated: y = 9 5 x + 32 There are easier genral rules of thumb for converting between the two temperature scales.

8 Heat We often speak of heat as though it were a material that flows from one object to another; it is not. Rather, it is a form of energy. Unit of heat: calorie (cal) or the Joule (J) 1 cal is the amount of heat necessary to raise the temperature of 1 g of water by 1 Celsius degree. Don t be fooled the calories on our food labels are really kilocalories (kcal or Calories), the heat necessary to raise 1 kg of water by 1 Celsius degree.

9 If heat is a form of energy, it ought to be possible to equate it to other forms. The experiment below found the mechanical equivalent of heat by using the falling weight to heat the water:

10 Definition of heat: Heat is energy transferred from one object to another because of a difference in temperature. Remember that the temperature of a gas is a measure of the kinetic energy of its molecules.

11 The sum total of all the energy of all the molecules in a substance is its internal (or thermal) energy. Temperature: measures molecules average kinetic energy Internal energy: total energy of all molecules Heat: transfer of energy due to difference in temperature

12 Methods of Heat transfer There are 3 methods of heat energy transfer: Conduction: require contact between matter and the energy moves through the matter. Convection:mass moves from one region to another(e.g. air thermals--hot air rises cold air sinks) Radiation: This refers to electromagnetic radiation and it can be transmitted through a vacuum.

13 Specific Heat The amount of heat required to change the temperature of a material is proportional to the mass and to the temperature change: The specific heat, c, is characteristic of the material. Some values are listed at left.

14 Specific heats of gases are more complicated, and are generally measured at constant pressure (c P ) or constant volume (c V ). Some sample values:

15 Calorimetry Closed system: no mass enters or leaves, but energy may be exchanged Open system: mass may transfer as well Isolated system: closed system where no energy in any form is transferred For an isolated system, Energy out of one part = energy into another part Or: heat lost = heat gained

16 Thermal Equilibrium (Zeroith Law) Lets take two objects (A and B) that are in physical contact, and are isolated from the environment. Let the initial temperatures of objects A and B be T Ai and T Bi respectively. Also let T Ai > T Bi.

17 Heat energy will flow from object A to B and in the process T A will decrease while T B increases. Eventually they will reach a final equilibrium temperature Teq. The only heat transfer occurred between the two objects but not with the environment. The heat lost by object A equals the heat gained by object B. If Q A is the heat object A, and Q B is the heat of object B, one can state: Q A +Q B =0 m A c A (T eq -T Ai )+m B c B (T eq -T Bi )=0

18 Example: 1) How much heat is required to raise the temperature of a 20 kg vat made of iron contaning 20 kg of water, from 10 0 C to 90 0 C? (C iron = 450 J/kg/ o C) a) Q=m 1 c 1 ΔT+m 2 c 2 ΔT =(20 kg)(450 J/kg/ o C)(80 o C) +(20 kg) (4186 J/kg/ o C)(80 o C) =720 kj+ 6700kJ= 7400 kj

19 Example 2) 100 ml of water is heated from 25.0oC to 50.0oC. What is the thermal energy added to the water m = 100 g = kg c = 4.19 kj/kg.oc. Δt = ( )oC = 25.0oC. Q =? Q = mcδt = (0.100 kg) (4.19 kj/kg.oc) (25oC) = 10.5 kj 10.5 kj is used to heat 100 ml of water from 25.0oC to 50.0oC Physics 11 - Unit 4- Work and Energy 19

20 Example 3) If 200 cm 3 of tea at 95 0 C is poured into a 150 g glass cup initially at 25 0 C, what will be the final temperature T of the mixture when equilibrium is reached, assuming no heat is lost to the surroundings? (C glass = 840 J/kg/ o C)

21 Q 1 = heat of tea Q 2 = heat of cup Q 1 +Q 2 =0 m tea c tea (T-95 0 )+m glass c glass (T-25 0 )=0 Other option: T= 86 0 C heat lost by tea =heat gained by glass m tea c tea (95 0 -T)= m glass c glass (T-25 0 )

22 Example [efficiency]: A 1500 W electric kettle took 6.00 minutes to raise the temperature of 1.50 kg of water from 20 0 C to 95 0 C. What is the efficiency of the kettle? Solution : e = useful work input energy x100 = mcδt Pt = 1.5(4180)(75) (1500)(360) x 100 = 87%

23 Latent Heat Energy is required for a material to change phase, even though its temperature is not changing.

24 Heat of fusion, L F : heat required to change 1.0 kg of material from solid to liquid Heat of vaporization, L V : heat required to change 1.0 kg of material from liquid to vapor

25 The total heat required for a phase change depends on the total mass(m) and the latent (L) heat:

26 Calorimetry-with Latent heat Ex) 1.20 kg of iron at C is added to 3.40 kg of water at 80 0 C. Determine the final equilibrium temperature, if it is at C state how much has been converted to steam.(use c water =4180J/kg/ 0 C,c steam =2010 J/kg/ 0 C,,c iron =450 J/kg/ 0 C L v = 22.6x10 5 J/ kg,l f = 3.33x10 5 J/kg)

27 Solution: First one must determine of the equilibrium temperature T is greater than, equal, or less than C. Let Q 1 = heat to raise water to C Q 2 = heat to convert all the water to steam Q 3 = heat lost from iron cooling from C to C

28 Q 1 = (3.4)(4180)(20)= 2.84x10 5 J Q 2 = (3.4)(22.6x10 5 )= 7.684x10 6 J Q 3 = (1.2)(450)(741)= 4.00x10 5 J In this case Q 3 >Q 1, and Q 3 <Q 1 +Q 2 Therefore the final equilibrium temperature is at C. There is enough energy given up from the cooling iron to heat the water and convert some of the water to steam. The equation to solve can now be written: (3.4)(4180)(100-80)+m(22.6x10 5 )+(1.2)(450) ( )=0

29 m= 5.13x10-2 kg Other points: 1) If Q 3 = Q 1 +Q 2, all the water would be converted to steam, and T final =100 0 C. 2) If Q 3 >Q 1 +Q 2, then all the water would be converted to steam and the temperature raised above C. 3) If Q 3 <Q 1, there would only be water at a temperature below C.

30 Example 3) How much energy does a refrigerator have to remove from 1.50 kg of water at 20 0 C to make ice at C? (C ice = 2100J/kg/ o C, L F = 3.33x10 5 J/kg) Solution: Q=mc w (20 0 C)+mL F + mc ice (12 0 C) = 6.6x10 5 J

31 Problem Solving: Calorimetry 1. Is the system isolated? Are all significant sources of energy transfer known or calculable? 2. Apply conservation of energy. 3. If no phase changes occur, the heat transferred will depend on the mass, specific heat, and temperature change. 4. If there are, or may be, phase changes, terms that depend on the mass and the latent heat may also be present. Determine or estimate what phase the final system will be in.

32 5. Make sure that each term is in the right place. 6. There is only one final temperature when the system reaches equilibrium. 7. Solve.

33 Heat Transfer: Conduction Heat conduction can be visualized as occurring through molecular collisions. The heat flow per unit time is given by: H=

34 The constant k is called the thermal conductivity. Materials with large k are called conductors; those with small k are called insulators.

35 Building materials are measured using R values rather than thermal conductivity: Here, l is the thickness of the material.

36 Example problem: Calculate the rate of heat flow through a glass window 2.0m x 1.5m in area and 3.2 mm thick, if the temperatures at the inner and outer surfaces are 15 0 C and 14 0 C respectively.(k glass =0.84 J/s m C 0 )

37 Solution: ΔQ Δt = ka T T 1 2 l = (0.84J /s m C 0 )(3.0m 2 )( C C) (3.2x10 3 m) = 790 J/s

38 Effective Thermal Conductivity Lets assume one had two materials of given thickness (L 1 and L 2 ) and thermal conductivities(k 1 and k 2 ), and one wanted to replace both materials with on material with an equivalent thermal conductivity(k eff ) with a thickness (L 1 +L 2 ).The temperatures at the boundaries are T 1, T 2, and T 3.

39 It turns out that: L k eff = L 1 k 1 + L 2 k 2

40 Thermal Expansion Linear expansion occurs when an object is heated. ΔL = L 0 αδt or L = L 0 (1+ αδt) Here, α is the coefficient of linear expansion.

41 Volume expansion is similar, except that it is relevant for liquids and gases as well as solids: ΔV = V 0 βδt V = V 0 (1+ βδt) Here, β is the coefficient of volume expansion. For uniform solids,

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43 Thermal Expansion of Water Water behaves differently from most other solids its minimum volume occurs when its temperature is 4 C. As it cools further, it expands, as anyone who has left a bottle in the freezer to cool and then forgets about it can testify.

44 Thought Experiment: If you were to heat up a plate of metal with a hole cut in it, what would happen to the size of the hole? Answer: The hole would get larger

45 β = 3α Question: Show that if one ignores insignificant terms. Start with a cube with initial dimensions L 0 and final dimensions L.

46 The Gas Laws The relationship between the volume, pressure, temperature, and mass of a gas is called an equation of state. We will deal here with gases that are not too dense. Boyle s Law: the volume of a given amount of gas is inversely proportional to the pressure as long as the temperature is constant.

47 The volume is linearly proportional to the temperature, as long as the temperature is somewhat above the condensation point and the pressure is constant: Extrapolating, the volume becomes zero at C; this temperature is called absolute zero.

48 The concept of absolute zero allows us to define a third temperature scale the absolute, or Kelvin, scale. This scale starts with 0 K at absolute zero, but otherwise is the same as the Celsius scale. Therefore, the freezing point of water is K, and the boiling point is K. Finally, when the volume is constant, the pressure is directly proportional to the temperature:

49 We can combine the three relations just derived into a single relation: What about the amount of gas present? If the temperature and pressure are constant, the volume is proportional to the amount of gas:

50 A mole (mol) is defined as the number of grams of a substance that is numerically equal to the molecular mass of the substance: 1 mol H 2 has a mass of 2 g 1 mol Ne has a mass of 20 g 1 mol CO 2 has a mass of 44 g The number of moles in a certain mass of material:

51 We can now write the ideal gas law: where n is the number of moles and R is the universal gas constant. Note on Units: One should generally use m 3, Pa, and K. It is often common to use kpa with litres(l).

52 Problem Solving with the Ideal Gas Law Useful facts and definitions: Standard temperature and pressure (STP) Volume of 1 mol of an ideal gas is 22.4 L If the amount of gas does not change: Always measure T in kelvins P must be the absolute pressure

53 Since the gas constant is universal, the number of molecules in one mole is the same for all gases. That number is called Avogadro s number: The number of molecules in a gas is the number of moles (n) times Avogadro s number:

54 Therefore we can write: where k is called Boltzmann s constant.

55 Kinetic Theory and the Molecular Interpretation of Temperature

56 Assumptions of kinetic theory: large number of molecules, moving in random directions with a variety of speeds molecules are far apart, on average molecules obey laws of classical mechanics and interact only when colliding collisions are perfectly elastic

57 It can be shown that, See proof on pages of Ginacoli (5th ed) Recall that: so The average translational kinetic energy of the molecules in an ideal gas is directly proportional to the temperature of the gas.

58 We can invert this to find the average speed (root mean square) of molecules in a gas as a function of temperature:

59 Example: v avg versus v rms. One has 5 molecules travelling at speeds: 60 m/s, 55 m/s, 70 m/s, 73 m/s, and 62 m/s. a) What is the average speed of the molecules? b) What is v rms of the molecules? a) υ = = 64m /s b) υ rms = = 64.3m /s

60 Example: What is the average translational kinetic energy of molecules in a gas at 37 0 C? Solution E k = 3 2 kt = 3 2 (1.38x10 23 )(310) = 6.42x10 21 J

61 Example: What is the rms speed of air molecules (O 2 and N 2 ) at room temperature (20 0 C)? The molecular masses of O 2 and N 2 are 32 u and 28 u respectively. Also 1u=1.67x10-27 kg. Solution: m(o 2 )= (32)(1.67x10-27 kg)= 5.3x10-26 kg m(n 2 )= 4.7x10-26 kg Thus for Oxygen: For Nitrogen: υ rms = 3kT m = (3)(1.38x10 23 )(293) 5.3x10 26 = 480 m /s υ rms = 510 m /s

62 Distribution of Molecular Speeds These two graphs show the distribution of speeds of molecules in a gas, as derived by Maxwell. The most probable speed, v p, is not quite the same as the rms speed. As expected, the curves shift to the right with temperature.

63 The Laws of Thermodynamics

64 The First Law of Thermodynamics The change in internal energy of a closed system will be equal to the energy added to the system minus the work done by the system on its surroundings. This is the law of conservation of energy, written in a form useful to systems involving heat transfer. *On the AP formula sheet W is the work done on the system by the environment. This is why the formula is ΔU = Q +W on the formula sheet.

65 The sum total of all the energy of all the molecules in a substance is its internal (or thermal) energy U. ΔU is the change in internal energy of the system Temperature: measures molecules average kinetic energy. The internal energy U of a system, is directly correlated to the system s temperature. Q= Heat: a transfer of energy due to difference in temperature W= work. See next slide.

66 If the pressure is constant, the work done (W= work done by system) is the pressure multiplied by the change in volume: In general, the work done is the area under the curve of a P versus V diagram. In an isometric (isochoric) process, the volume does not change, so the work done is zero.

67 For processes where the pressure varies, the work done is the area under the P-V curve.

68 Thermodynamic Processes An isothermal process is one where the temperature does not change.

69 In order for an isothermal process to take place, we assume the system is in contact with a heat reservoir. In general, we assume that the system remains in equilibrium throughout all processes.

70 An adiabatic process is one where there is no heat flow into or out of the system. Question: Based on the 1st Law, how can you show that AB must be isothermal, and AC is adiabatic?

71 An isobaric process (a) occurs at constant pressure; an isometric (isochoric) one (b) at constant volume.

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73 The Second Law of Thermodynamics The absence of the process illustrated above indicates that conservation of energy is not the whole story. If it were, movies run backwards would look perfectly normal to us!

74 The second law of thermodynamics is a statement about which processes occur and which do not. There are many ways to state the second law; here is one: Heat can flow spontaneously from a hot object to a cold object; it will not flow spontaneously from a cold object to a hot object. Entropy is another way of stating the law. The entropy (state of disorder) of the universe can only increase.

75 Heat Engines It is easy to produce thermal energy using work, but how does one produce work using thermal energy? This is a heat engine; mechanical energy can be obtained from thermal energy only when heat can flow from a higher temperature to a lower temperature.

76 We will discuss only engines that run in a repeating cycle; the change in internal energy over a cycle is zero, as the system returns to its initial state. The high temperature reservoir transfers an amount of heat Q H to the engine, where part of it is transformed into work W and the rest, Q L, is exhausted to the lower temperature reservoir. Note that all three of these quantities are positive.

77 A steam engine is one type of heat engine.

78 The internal combustion engine is a type of heat engine as well.

79 The efficiency of a heat engine is the ratio of the work done to the heat input: Using conservation of energy to eliminate W, we find:

80 The Carnot engine was created to examine the efficiency of a heat engine. It is idealized, as it has no friction. Each leg of its cycle is reversible. The Carnot cycle consists of: Isothermal expansion Adiabatic expansion Isothermal compression Adiabatic compression An example is on the next slide.

81 Carnot Cycle

82 For an ideal reversible engine, the efficiency can be written in terms of the temperature: From this we see that 100% efficiency can be achieved only if the cold reservoir is at absolute zero, which is impossible. Real engines have some frictional losses; the best achieve 60-80% of the Carnot value of efficiency.