1. MSU Physics 231 Fall 2015 1 Physics 231 Topic 13: Heat Alex Brown Dec 1, 2015

2. MSU Physics 231 Fall 2015 2 9 th 10 pm attitude survey (1% for participation) 11 th 10 pm last homework set 8 th 10 pm correction for 3 rd exam 10 th 10 pm concept test timed (50 min)) (1% for performance) 17 th 8-10 pm final (Thursday) VMC E100

3. MSU Physics 231 Fall 2015 3 when the temperature is reduced at constant pressurevolume is reduced According to the Ideal Gas Law, when the temperature is reduced at constant pressure, the volume is reduced as well. The volume of the balloon therefore decreases. Clicker Quiz! nRTPV  a) it increases b) it does not change c) it decreases What happens to the volume of a balloon if you put it in the freezer?

4. MSU Physics 231 Fall 2015 4 Key Concepts: Heat Heat and Thermal Energy Heat & its units Heat Capacity & Specific Heat Thermal Equilibrium Equipartition theorem Phase Changes Latent heat of fusion, vaporization Conduction, Convection, and Radiation Thermal Conductivity Stefan-Boltzman law Covers chapter 13 in Rex & Wolfson

5. MSU Physics 231 Fall 2015 5 Thermal equilibrium Low temperature Low kinetic energy Particles move slowly High temperature High kinetic energy Particles move fast Thermal contact Transfer of kinetic energy Thermal equilibrium: temperature is the same everywhere

6. MSU Physics 231 Fall 2015 6 Heat Heat: The transfer of thermal energy between objects because their temperatures are different. Heat: energy transfer Symbol: Q Units: Calorie (cal) or Joule (J) 1 cal = 4.186 J (energy needed to raise 1g of water by 1 0 C)

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9. 9 Heat transfer to an object Q = c m  T Energy transfer (J or cal) Specific heat J/(kg o C) or cal/(g o C) Mass of object Change in temperature The amount of energy transfer Q to an object with mass m when its temperature is raised by  T:

10. MSU Physics 231 Fall 2015 10 Example A 1 kg block of Copper is raised in temperature by 10 o C. What was the heat transfer Q? Answer: Q = c m  T = (387)(1)(10) = 3870 J Q = (0.092)(1000)(10) = 924.5 cal 1 cal = 4.186 J SubstanceSpecific Heat J/kg o C Specific Heat cal/g o C aluminum9000.215 copper3870.092 water41861.00

11. MSU Physics 231 Fall 2015 11 Problem A block of Copper is dropped from a height of 10 m. Assuming that all the potential energy is transferred into internal energy when it hits the ground, what is the raise in temperature of the block? c copper =387 J/(kg o C) Potential energy: mgh (Joules) All transferred into heat Q = cm  T mgh = cm  T  T = gh/c = (10)(9.81)/387 = 0.25 o C

12. MSU Physics 231 Fall 2015 12 Calorimetry If we connect two objects with different temperature energy will transferred from the hotter to the cooler one until their temperatures are the same. If the system is isolated: Energy flow into cold part = Energy flow out of hot part m c c c ( T f - T c ) = m h c h (T h - T f ) the final temperature is: T f = note that T can be in Kelvin or Centigrade m c c c T c + m h c h T h m c c c + m h c h T c (cold) T h (hot) T f (final)

13. MSU Physics 231 Fall 2015 13 Clicker Question A block of iron that has been heated to 100 0 C is dropped in a glass of water at room temperature (20 0 C). After the temperatures in the block and the water have become equal: a)The water has changed more in temperature than the iron block. b) The water has changed less in temperature than the iron block c) the temperatures of both have changed equally d) I need more info to say anything!

14. MSU Physics 231 Fall 2015 14 Heating Water with a Ball of Lead A ball of Lead at T=100 o C with mass 400 g is dropped in a glass of water (0.3 L) at T=20 0 C. What is the final temperature of the system? c water =1 cal/g o C c lead =0.03 cal/g o C  water =10 3 kg/m 3 T final = = [(300) (1) (20) + (400)(0.03)(100)] / [(300)(1) + (400) (0.03)] = 7200/312 = 23.1 o C m water c water T water + m lead c lead T lead m water c water + m lead c lead

15. MSU Physics 231 Fall 2015 15 And another A block of unknown substance with a mass of 8 kg, initially at T=280K is thermally connect to a block of copper (5 kg) that is at T=320 K (c copper =0.093 cal/g 0 C). After the system has reached thermal equilibrium the temperature T equals 290K. What is the specific heat of the unknown material in cal/g o C? ???? copper C unknown = m copper c copper (T copper -T final ) m unknown (T final -T unknown ) C unkown = (5000) (0.093) (320-290) = 0.17 cal/g o C 8000 (290-280)

16. MSU Physics 231 Fall 2015 16 Mixing 3 liquids Three different liquids are mixed together in a calorimeter. The masses, specific heats and initial temperatures of the liquids are: m 1 = 475 g, m 2 = 355 g, m 3 = 795 g. What will be the temperature of the mixture in Celsius? c 1 = 225 J/kgC, c 2 = 500 J/kgC, c 3 = 840 J/kgC. T 1 = 24.5 °C, T 2 = 53.5 °C, T 3 = 81.5 °C. T f = = c 1 m 1 T 1 +c 2 m 2 T 2 +c 3 m 3 T 3 c 1 m 1 +c 2 m 2 +c 3 m 3 225x475x24.5 + 500x355x53.5 + 840x795x81.5 225x475 + 500x355 + 840x795 = 69.9 o C Like with two substances, the final temperature is a weighted average of T 1,T 2 and T 3 with the c’s and m’s being the weights

17. MSU Physics 231 Fall 2015 17 Internal Energy In chapter 12: The internal (total) energy for an ideal gas is the total kinetic energy of the atoms/particles in a gas. For a non-ideal gas: the internal energy is due to kinetic and potential energy associated with the inter-molecular potential energy PE r PE: negative!

18. MSU Physics 231 Fall 2015 18 phase changes gas (high T) liquid (medium T) solid (low T) Q=c gas m  T Q=c liquid m  T Q=c solid m  T Gas  liquid liquid  solid

19. MSU Physics 231 Fall 2015 19 phase changes SOLID to Gas and GAS to LIQUID DURING THESE PHASE TRANSITIONS THE TEMPERATURE DOES NOT CHANGE AND SO THE KINETIC ENERGY DOES NOT CHANGE. ALL ADDED HEAT GOES TO CHANGING PE “breaking the chemical bonds”

20. MSU Physics 231 Fall 2015 20 phase changes Gas  liquid When heat is added to a liquid, potential energy goes to zero - the energy stored in the stickiness of the liquid is taken away. When heat is taken from a gas, potential energy goes into the stickiness of the fluid liquid  solid When heat is added to a solid to make a liquid, potential energy in the bonds between the atoms become less. When heat is taken from a liquid, the bonds between atoms becomes stronger - potential energy is more negative.

21. MSU Physics 231 Fall 2015 21 Okay, the Temperature does not change in a phase transition! But what is the amount of heat added to make the phase transition? Gas  liquid Q gas-liquid = m L v L v =latent heat of vaporization (J/kg or cal/g) depends on material.(energy required to vaporize) Gas to liquid Q flows out Liquid to gas Q flows in m = mass

22. MSU Physics 231 Fall 2015 22 solid  liquid Q liquid-solid = m L f L f =latent heat of fusion (J/kg or cal/g) depends on material (energy required to liquify) Liquid to solid Q flows out Solid to liquid Q flows in m = mass

23. MSU Physics 231 Fall 2015 23 phase changes gas (high T) liquid (medium T) solid (low T) Q=c gas m  T Q=c liquid m  T Q=c solid m  T Gas  liquid liquid  solid Q=m L v Q=m L f

24. MSU Physics 231 Fall 2015 24 Clicker Question! Ice is heated steadily and becomes liquid and then vapor. During this process: a)the temperature rises continuously. b)when the ice turns into water, the temperature drops for a brief moment. c) the temperature is constant during the phase transformations d) the temperature cannot exceed 100 o C

25. MSU Physics 231 Fall 2015 25 T ( o C) 0 100 ice ice+water water water+steam steam THE PHASE TRANSFORMATIONS OF WATER

26. MSU Physics 231 Fall 2015 26 Ice with T=-30 o C is heated to steam of T=150 0 C. How much heat (in cal) has been added in total? c ice =0.5 cal/g o C c water =1.0 cal/g o C c steam =0.480 cal/g o C L f =540 cal/g L v =79.7 cal/g m=1 kg=1000g A) Ice from -30 to 0 o C Q=1000*0.5*30= 15000 cal B) Ice to water Q=1000*540= 540000 cal C) water from 0 o C to 100 o C Q=1000*1.0*100=100000 cal D) water to steam Q=1000*79.7= 79700 cal E) steam from 100 o C to 150 0 C Q=1000*0.48*50=24000 cal TOTALQ= =758700 cal T ( o C) 0 100 ice ice+water water water+steam steam

27. MSU Physics 231 Fall 2015 27 Question Given: L f =6.44x10 4 J/kg T melt =1063 o C c specific =129 J/kg 0 C A block of gold (room temperature 20 0 C) is found to just melt completely after supplying 4x10 3 J of heat. What was the mass of the gold block? Properties of gold c = 129 J/(kg o C), melting point = 1063 o C L f = 6.45x10 4 J/kg Q = c m  T + m L f = (129)(m)(1063-20) + (m)(6.45x10 4 ) = 2.0x10 5 m 4000 = 2.0x10 5 m m = 0.02 kg

28. MSU Physics 231 Fall 2015 28 How can heat be transferred?

29. MSU Physics 231 Fall 2015 29 Conduction Touching different materials: Some feel cold, others feel warm, but all are at the same temperature…

30. MSU Physics 231 Fall 2015 30 Thermal conductivity metal wood T=37 0 C T=20 0 C The heat transfer in the metal is much faster than in the wood: (thermal conductivity)

31. MSU Physics 231 Fall 2015 31 Heat transfer via conduction Conduction occurs if there is a temperature difference between two parts of a conducting medium Rate of energy transfer P P = Q/  t (unit Watt = J/s) P = k A (T h -T c )/  x = k A  T/  x k: thermal conductivity Unit: J/(m s o C) Metals k ~ 300 J/(m s o C) Gases k ~ 0.1 J/(m s o C) Nonmetals k ~ 1 J/(m s o C) ThTh TcTc xx A

32. MSU Physics 231 Fall 2015 32 Example A glass window (A=4m 2,  x=0.5cm) separates a living room (T=20 o C) from the outside (T=0 o C). A) What is the rate of heat transfer through the window?, k glass =0.84 J/(m s o C) B) By what fraction does it change if the surface becomes 2x smaller and the outside temperature drops to -20 o C? A)P = k A  T/  x = (0.84)(4)(20)/0.005 = 13440 Watt B)P orig = k A  T/  x P new = k(0.5A)(2  T)/  x = P orig The heat transfer is the same ThTh TcTc A xx

33. MSU Physics 231 Fall 2015 33 Another one An insulated gold wire (i.e. no heat lost to the air) is at one end connected to a heat reservoir (T=100 0 C) and at the other end connected to a heat sink (T=20 0 C). If its length is 1m and P=200 W what is its cross section (A)? k gold = 314 J/(m s o C). P = k A  T/  x = (314)(A)(80)/1 = 25120 A = 200 A = 8.0x10 -3 m 2

34. MSU Physics 231 Fall 2015 34 Clicker Quiz! Given your experience of what feels colder when you walk on it, which of the surfaces would have the highest thermal conductivity? a) a rug b) a steel surface c) a concrete floor d) has nothing to do with thermal conductivity The heat flow rate is k A (T 1 − T 2 ) /  x. All things being equal, bigger k leads to bigger heat loss. From the book: Steel = 40, Concrete = 0.84, Human tissue = 0.2, Wool = 0.04, in units of J/(s.m.C°).

35. MSU Physics 231 Fall 2015 35 Multiple Layers ThTh TcTc A L 1 L 2 (  x) k 1 k 2

36. MSU Physics 231 Fall 2015 36 Insulation L1L1 L2L2 L3L3 inside ThTh TcTc A house is built with 10cm thick wooden walls and roofs. The owner decides to install insulation. After installation the walls and roof are 4cm wood + 2cm insulation + 4cm wood. If k wood =0.10 J/(ms 0 C) and k insulation =0.02 J/(ms 0 C), by what factor does he reduce his heating bill? P before = A  T/[0.10/0.10] = A  T P after = A  T/[(0.04/0.10) + (0.02/0.02) + (0.04/0.10)] = A  T/1.8 Almost a factor of 2 (1.81) !

37. MSU Physics 231 Fall 2015 37 Convection T high  low

38. MSU Physics 231 Fall 2015 38 Radiation (photons) Nearly all objects emit energy through radiation: energy radiated per second P =  A e T 4 : Stefan’s law  = 5.6696x10 -8 W/m 2 K 4 A: surface area e: object dependent constant emissivity (0-1) T: temperature (K)

39. MSU Physics 231 Fall 2015 39 Emissivity Ideal reflector e=0 no energy is absorbed Ideal absorber (black body) e=1 all energy is absorbed also ideal radiator!

40. MSU Physics 231 Fall 2015 40 A Barbecue The coals in a BBQ cover an area of 0.25m 2. If the emissivity of the burning coal is 0.95 and their temperature 500 0 C, how much energy is radiated every minute? P =  A e T 4 ( J/s) = 5.67x10 -8 * 0.25 * 0.95 * (773) 4 = 4808 (J/s) 1 minute: Q = 2.9x10 5 J (enough to boil off one liter of water)

41. MSU Physics 231 Fall 2015 41 Net Power Radiated If an object would only emit radiation it would eventually have 0 K temperature. In reality, an object emits AND receives radiation. P NET =  Ae (T 4 -T 0 4 ) where T: temperature of object T 0 : temperature of surroundings.

42. MSU Physics 231 Fall 2015 42 Example The temperature of the human body is 37 0 C. If the room temperature is 20 0 C, how much heat is given off by the human body to the room in one minute? Assume that the emissivity of the human body is 0.9 and the surface area is 2 m 2. P =  A e (T 4 -T 0 4 ) = 5.67x10 -8 * 2 * 0.9 * (310.5 4 - 293.5 4 ) = = 185 J/s Q = P *  t = 185 * 60 = 1.1x10 4 J

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44. MSU Physics 231 Fall 2015 44 Black body A black body is an object that absorbs all electromagnetic radiation that falls onto it. They emit radiation, depending on their temperature. If T<700 K, almost no visible light is produced (hence a ‘black’ body). The energy emitted from a black body: P=  T 4 with  =5.67x10 -8 W/m 2 K 4 b=2.90×10 −3 m K Wien’s displacement constant

45. MSU Physics 231 Fall 2015 45 Infrared Radiation The human body emits radiation in the infrared. With a body temperature of T=(273+37 K) = 310 K the wavelength ( max ) of the thermal emission is 9.4 x10 -5 m

46. MSU Physics 231 Fall 2015 46 An Example The contents of a can of soda (0.33 kg) which is cooled to 4 o C is poured into a glass (0.1 kg) that is at room temperature (20 0 C). What will the temperature of the filled glass be after it has reached full equilibrium (glass and liquid have the same temperature)? c water =4186 J/(kg o C) and c glass =837 J/(kg 0 C) T final = = (0.33*4186*4 + 0.1*837*20) / (0.33*4186 + 0.1*837) = 4.9 o C m water c water T water + m glass c glass T glass m water c water + m glass c glass

47. MSU Physics 231 Fall 2015 47 And another Water 0.5L 100 0 C 150 0 C A = 0.03m 2 thickness = 0.5cm. A student working for his exam feels hungry and starts boiling water (0.5L) for some noodles. He leaves the kitchen when the water just boils. The stove’s temperature is 150 0 C. The pan’s bottom has dimensions given above. Working hard on the exam, he only comes back after half an hour. Is there still water in the pan? (L v =540 cal/g, k pan =1 cal/(m s 0 C) To boil away m = 500g of water: Q = L v (500)=270000 cal Heat added by the stove: P = kA  T/  x = (1)(0.03)(50)/0.005 = 300 cal/s P=Q/  t  t = Q/P = 270000/300 = 900 s (15 minutes) He’ll be hungry for a bit longer…