Physics 231 Topic 13: Heat Alex Brown Dec 1,2312015 MSU Physics Fall 2015 1 8th 10 pm correction for 3rd exam 9th 10 pm attitude survey (1% for participation) 10th 10 pm concept test timed (50 min)) (1% for performance) 11th 10 pm last homework set 17th 8-10 pm final (Thursday) VMC E100 MSU Physics 231 Fall 2015 2 Clicker Quiz! What happens to the volume of a balloon if you put it in the a) it increases b) it does not change c) it decreases freezer? 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. PV = nRT MSU Physics 231 Fall 2015 3 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 MSU Physics 231 Fall 2015 4 Thermal equilibrium Thermal contact Low temperature Low kinetic energy Particles move slowly High temperature High kinetic energy Particles move fast Transfer of kinetic energy Thermal equilibrium: temperature is the same everywhere MSU Physics 231 Fall 2015 5 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 10C) MSU Physics 231 Fall 2015 6 MSU Physics 231 Fall 2015 7 MSU Physics 231 Fall 2015 8 Heat transfer to an object The amount of energy transfer Q to an object with mass m when its temperature is raised by T: Q = c m T Change in temperature Energy transfer (J or cal) Mass of object Specific heat J/(kgoC) or cal/(goC) MSU Physics 231 Fall 2015 9 Example Substance Specific Heat Specific Heat J/kg oC cal/g oC aluminum 900 0.215 copper 387 0.092 water 4186 1.00 A 1 kg block of Copper is raised in temperature by 10oC. 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 MSU Physics 231 Fall 2015 10 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? ccopper=387 J/(kgoC) Potential energy: mgh (Joules) All transferred into heat Q = cm T mgh = cm T T = gh/c = (10)(9.81)/387 = 0.25oC MSU Physics 231 Fall 2015 11 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: T (cold) T (hot) c h Energy flow into cold part = Energy flow out of hot part mc cc ( Tf - Tc) = mh ch (Th - Tf) the final temperature is: Tf = mc cc Tc + mh ch Th mc cc + mh ch Tf (final) note that T can be in Kelvin or Centigrade MSU Physics 231 Fall 2015 12 Clicker Question A block of iron that has been heated to 1000C is dropped in a glass of water at room temperature (200C). 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! MSU Physics 231 Fall 2015 13 Heating Water with a Ball of Lead A ball of Lead at T=100oC with mass 400 g is dropped in a glass of water (0.3 L) at T=200C. What is the final temperature of the system? cwater=1 cal/g oC Tfinal = clead=0.03 cal/g oC water=103kg/m3 mwatercwaterTwater + mleadcleadTlead mwatercwater + mleadclead = [(300) (1) (20) + (400)(0.03)(100)] / [(300)(1) + (400) (0.03)] = 7200/312 = 23.1 oC MSU Physics 231 Fall 2015 14 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 (ccopper=0.093 cal/g0C). After the system has reached thermal equilibrium the temperature T equals 290K. What is the specific heat of the unknown material in cal/goC? Cunknown = mcopperccopper(Tcopper-Tfinal ) munknown (Tfinal -Tunknown) Cunkown = (5000) (0.093) (320-290) = 0.17 cal/goC 8000 (290-280) MSU Physics 231 Fall 2015 copper ???? 15 Mixing 3 liquids Three different liquids are mixed together in a calorimeter. The masses, specific heats and initial temperatures of the liquids are: T1 = 24.5 °C, m1 = 475 g, c1 = 225 J/kgC, T2 = 53.5 °C, m2 = 355 g, c2 = 500 J/kgC, T3 = 81.5 °C. m3 = 795 g. c3 = 840 J/kgC. What will be the temperature of the mixture in Celsius? Like with two substances, the final temperature is a weighted average of T1,T2 and T3 with the c’s and m’s being the weights Tf = c1m1T1+c2m2T2+c3m3T3 c1m1+c2m2+c3m3 = 225x475x24.5 + 500x355x53.5 + 840x795x81.5 225x475 + 500x355 + 840x795 MSU Physics 231 Fall 2015 = 69.9oC 16 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: negative! PE MSU Physics 231 Fall 2015 r 17 phase changes gas (high T) Q=cgasmT Gas liquid Q=csolidmT solid (low T) liquid (medium T) liquid solid Q=cliquidmT MSU Physics 231 Fall 2015 18 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” MSU Physics 231 Fall 2015 19 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. MSU Physics 231 Fall 2015 20 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 Qgas-liquid = m Lv m = mass Lv=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 MSU Physics 231 Fall 2015 21 solid liquid Qliquid-solid = m Lf m = mass Lf=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 MSU Physics 231 Fall 2015 22 phase changes gas (high T) Q=cgasmT Gas liquid Q=m Lv Q=csolidmT solid (low T) liquid (medium T) Q=cliquidmT MSU Physics 231 Fall 2015 liquid solid Q=m Lf 23 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 100oC MSU Physics 231 Fall 2015 24 ice water 0 ice+water steam T (oC) water+steam 100 THE PHASE TRANSFORMATIONS OF WATER MSU Physics 231 Fall 2015 25 water ice steam 0 ice+water T (oC) water+steam 100 A) Ice from -30 to 0oC B) Ice to water C) water from 0oC to 100oC D) water to steam E) steam from 100oC to 1500C TOTAL Ice with T=-30oC is heated to steam of T=1500C. How much heat (in cal) has been added in total? cice=0.5 cal/goC cwater=1.0 cal/goC csteam=0.480 cal/goC Lf=540 cal/g Lv=79.7 cal/g m=1 kg=1000g Q=1000*0.5*30= 15000 cal Q=1000*540= 540000 cal Q=1000*1.0*100=100000 cal Q=1000*79.7= 79700 cal Q=1000*0.48*50=24000 cal Q= =758700 cal MSU Physics 231 Fall 2015 26 Question A block of gold (room temperature 200C) is found to just melt completely after supplying 4x103 J of heat. What was the mass of the gold block? Properties of gold oC), oC 4 J/kg oC cGiven: = 129LJ/(kg melting point = 1063 =6.44x10 T =1063 c =129 J/kg0C f melt specific Lf = 6.45x104 J/kg Q = c m T + m Lf = (129)(m)(1063-20) + (m)(6.45x104) = 2.0x105 m 4000 = 2.0x105 m m = 0.02 kg MSU Physics 231 Fall 2015 27 How can heat be transferred? MSU Physics 231 Fall 2015 28 Conduction Touching different materials: Some feel cold, others feel warm, but all are at the same temperature… MSU Physics 231 Fall 2015 29 Thermal conductivity metal T=200C wood T=200C The heat transfer in the metal is much faster than in the wood: (thermal conductivity) T=370C MSU Physics 231 Fall 2015 T=370C 30 Heat transfer via conduction Tc Th Conduction occurs if there is a temperature difference between two parts of a conducting medium Rate of energy transfer P A P = Q/t (unit Watt = J/s) P = k A (Th-Tc)/x = k A T/x k: thermal conductivity Unit: J/(m s oC) x Metals Gases Nonmetals MSU Physics 231 Fall 2015 k ~ 300 J/(m s oC) k ~ 0.1 J/(m s oC) k ~ 1 J/(m s oC) 31 Th Tc Example A glass window (A=4m2, x=0.5cm) separates a living room (T=20 oC) from the outside (T=0 oC). A A) What is the rate of heat transfer through the window?, kglass=0.84 J/(m s oC) B) By what fraction does it change if the surface becomes 2x smaller and the outside temperature drops to -20 oC? x A) P = k A T/x = (0.84)(4)(20)/0.005 = 13440 Watt B) Porig= k A T/x Pnew= k(0.5A)(2T)/x = Porig The heat transfer is the same MSU Physics 231 Fall 2015 32 Another one An insulated gold wire (i.e. no heat lost to the air) is at one end connected to a heat reservoir (T=1000C) and at the other end connected to a heat sink (T=200C). If its length is 1m and P=200 W what is its cross section (A)? kgold = 314 J/(m s oC). P = k A T/x = (314)(A)(80)/1 = 25120 A = 200 A = 8.0x10-3 m2 MSU Physics 231 Fall 2015 33 Clicker Quiz! Given your experience of what feels colder when you walk on it, a) a rug which of the surfaces would b) a steel surface have the highest thermal c) a concrete floor conductivity? d) has nothing to do with thermal conductivity The heat flow rate is k A (T1 − T2)/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°). MSU Physics 231 Fall 2015 34 Multiple Layers Th k 1 Tc k2 A L1 Q A(Th Tc ) P= = t ( Li / ki ) i L2 (x) MSU Physics 231 Fall 2015 35 Q A(Th Tc ) P= = t ( Li / ki ) i inside Insulation Th Tc L1 L2 L3 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 kwood=0.10 J/(ms0C) and kinsulation=0.02 J/(ms0C), by what factor does he reduce his heating bill? Pbefore = A T/[0.10/0.10] = A T Pafter = 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) ! MSU Physics 231 Fall 2015 36 Convection T high low MSU Physics 231 Fall 2015 37 Radiation (photons) Nearly all objects emit energy through radiation: energy radiated per second P = A e T4 : Stefan’s law = 5.6696x10-8 W/m2K4 A: surface area e: object dependent constant emissivity (0-1) T: temperature (K) MSU Physics 231 Fall 2015 38 Emissivity Ideal reflector e=0 no energy is absorbed Ideal absorber (black body) e=1 all energy is absorbed also ideal radiator! MSU Physics 231 Fall 2015 39 A Barbecue The coals in a BBQ cover an area of 0.25m2. If the emissivity of the burning coal is 0.95 and their temperature 5000C, how much energy is radiated every minute? P = A e T4 (J/s) = 5.67x10-8 * 0.25 * 0.95 * (773)4 = 4808 (J/s) 1 minute: Q = 2.9x105 J (enough to boil off one liter of water) MSU Physics 231 Fall 2015 40 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. PNET = Ae (T4-T04) where T: temperature of object T0: temperature of surroundings. MSU Physics 231 Fall 2015 41 Example The temperature of the human body is 370C. If the room temperature is 200C, 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 m2. P = A e (T4-T04) = 5.67x10-8 * 2 * 0.9 * (310.54 - 293.54) = = 185 J/s Q = P * t = 185 * 60 = 1.1x104 J MSU Physics 231 Fall 2015 42 MSU Physics 231 Fall 2015 43 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=T4 with =5.67x10-8 W/m2K4 b=2.90×10−3 m K Wien’s displacement constant MSU Physics 231 Fall 2015 44 Infrared Radiation The human body emits radiation in the infrared. With a body temperature of T=(273+37 K) = 310 K the wavelength (lmax) of the thermal emission is 9.4 x10-5 m MSU Physics 231 Fall 2015 45 An Example The contents of a can of soda (0.33 kg) which is cooled to 4 oC is poured into a glass (0.1 kg) that is at room temperature (200C). What will the temperature of the filled glass be after it has reached full equilibrium (glass and liquid have the same temperature)? cwater=4186 J/(kgoC) and cglass=837 J/(kg0C) Tfinal= mwatercwaterTwater + mglasscglassTglass mwatercwater + mglasscglass = (0.33*4186*4 + 0.1*837*20) / (0.33*4186 + 0.1*837) = 4.9oC MSU Physics 231 Fall 2015 46 Water 0.5L 1000C And another A = 0.03m2 thickness = 0.5cm. 1500C 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 1500C. 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? (Lv=540 cal/g, kpan=1 cal/(m s 0C) To boil away m = 500g of water: Q = Lv(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… MSU Physics 231 Fall 2015 47