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Physics 101 Lecture 10. Thermal Physics Applications of Newton’s Laws to Large Number of Particles Can’t apply Newton’s Laws to large number of particles.

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Presentation on theme: "Physics 101 Lecture 10. Thermal Physics Applications of Newton’s Laws to Large Number of Particles Can’t apply Newton’s Laws to large number of particles."— Presentation transcript:

1 Physics 101 Lecture 10

2 Thermal Physics

3 Applications of Newton’s Laws to Large Number of Particles Can’t apply Newton’s Laws to large number of particles There would be too many equations, initial positions, initial velocities and forces for even a modern computer to handle

4 Large Numbers - 1 Ten atoms Six have 0 units of energy Four have 1 unit of energy How many distinct arrangements 10! / (6! x 4!) = 210 arrangements

5 Large Numbers - 2 100 atoms 60 have 0 units of energy 40 have 1 unit of energy How many distinct arrangements  10 28 arrangements

6 Large Numbers - 3 1000 atoms 600 have 0 units of energy 400 have 1 unit of energy How many distinct arrangements  10 290 arrangements

7 Large Numbers - 4 6 x 10 23 atoms 3.6 x 10 23 have 0 units of energy 2.4 x 10 23 have 1 unit of energy How many distinct arrangements  Humongous

8 Temperature

9 What is Temperature ? -1 We can measure temperature This does not tell us what temperature is

10 What is Temperature? - 2 Observations  Put together two objects with same measured temperature  Temperatures do not change  Put together two objects with different measured temperatures  High temperature decreases and lower temperature increases until they are equal This only tells us that temperature is a property of matter

11 Temperature Measurement - 1 Bodily sensation of temperature is arbitrary Defining temperature as hotness or coldness is arbitrary

12 Temperature Measurement - 2 Temperature is measured using a physical property that changes as temperature changes Thermometer is a device for measuring temperature  Materials expand / contract with changing temperature  Electrical properties change  Glowing color changes

13 Temperature Scales Temperature scales are man-made There are a number of temperature scales:  Celsius (centigrade)  Fahrenheit  Reamur, Rankin, and others The scales were defined by assigning two temperature points on the scale  Ice melting point  Water boiling point Then the distance between two points was divided into equally spaced intervals

14 Celsius / Fahrenheit

15 Temperature Scale - Kelvin Kelvin temperature scale has greater scientific significance than Celsius or Fahrenheit 1 Kelvin is identical to that of 1 Celsius degree There exists a lowest possible temperature below which no substance can be cooled Determined using a constant-volume gas thermometer Lowest temperature is defined to be the zero point on the Kelvin scale This is absolute zero Ice point (0 0 C) is 273.15 K

16 Constant Volume Thermometer Absolute Zero

17 Celsius / Kelvin

18 Temperature Conversions  T C = (5/9)(T F – 32)  T F = (9/5)(T C ) + 32  T K = T C + 273.15

19 Temperature Conversion Example Temperature of the Universe is 2.726 K What is that temperature in Celsius?  T K = T C + 273.15  2.726 = T C + 273.15  T C = -270.43 0 C

20 Heat

21 Heat is energy that flows from a higher- temperature object to a lower-temperature object because of the difference in temperatures Symbol: Q SI Unit of Heat: joule (J)

22 Internal Energy Heat that flows from higher to lower temperature originates in the internal energy of the higher temperature substance The internal energy of a substance is the sum of:  Molecular kinetic energy (random motion of molecules)  Molecular potential energy (due to forces that act between atoms and molecules)  Other kinds of molecular energy When heat flows, where the work done is negligible, the internal energy of the higher temperature substance decreases and the internal energy of the lower temperature substance increases

23 Heat and Temperature Change Solids and Liquids Heat Q that must be supplied or removed to change the temperature of a substance of mass m by an amount  T is  Q = cm  T c is specific heat capacity of the substance SI Unit of Specific Heat Capacity: J/(kg C 0 )

24 Heat and Temperature Change Gases Value of specific heat capacity depends on whether the pressure or volume is held constant while energy in the form of heat is added to or removed

25 Heat and Temperature Change Example How much heat does 0.025 kg of aluminum give off as it cools from 100 0 C to 20 0C? For aluminum, c = 880 J/kg 0 C  Q = cm(T f – T i )  Q = 880(0.025)(20 – 100) = -1760 J

26 Temperature Scales If you are using an individual temperature, T, then you must use Kelvin temperature If you are using a temperature difference, (T f – T i ), then you can use Celsius or Kelvin temperature

27 Heat and Phase Change There are three phases of matter:  solid(for example, water ice)  liquid(for example, water)  gas(for example, steam) Heat Q that must be supplied or removed to change the phase of a mass m is  Q = mL  L is the latent heat of the substance SI Unit of Latent Heat: J/kg

28 Latent Heat Latent heat of fusion, L f  refers to change between solid and liquid phase Latent heat of vaporization, L v  applies to change between liquid and gas phases Latent heat of sublimation, L s  refers to change between solid and gas phases

29 Temperature during Phase Change During a phase change, the temperature remains constant at the freezing temperature or boiling temperature Example:  Block of solid ice starts at 0 0 C  Add heat  Some of the ice melts  You have a mixture of ice and water  Temperature remains 0 0 C  Add more heat  All the ice melts  Temperature is still 0 0 C  Add more heat  Temperature rises above 0 0 C

30 Temperature during Phase Change

31 Calorimetry Example A thermos bottle contains 0.250 kg of coffee at 90 0 C To this is added 0.020 kg of milk at 5 0 C After equilibrium is established, what is the temperature of the liquid?  Water, coffee, and milk all have c = 4186.  (heat change of coffee) + (heat change of milk) = 0  [cm(T f – T i )] coffee + [cm(T f – T i )] milk = 0  c cancels out.  (0.250)(T f – 90) + (0.020)(T f – 5) = 0  0.250T f – 22.5 + 0.020T f – 0.1 = 0  0.27T f = 22.6  T f = 83.7 0 C


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