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Chapter 11 Energy in Thermal Processes Vocabulary, 3 Kinds of Energy Internal Energy U = Energy of a system due to microscopic motion and inter-molucular forces Work W = -FDx = -PDV is work done by expansion (next chapter) Heat Q = Energy transfer from microscopic contact DU Q PDV next chapter Temperature and Specific Heat Add energy -> T rises Q mcDT mass of material Property of the material •cH20 = 1.0 cal/(gºC) •1 calorie = 4.186 J Example: Converting calories Bobby Joe drinks a 130 “calorie” can of soda. If the efficiency for turning energy into work is 10%, how many 4 meter floors must Bobby Joe ascend in order to work off the soda and maintain her 55 kg mass? Solution •First, note that the “calories” listed in food are actually kilocalories! J Q 130 kcal 1.3 10 cal 4.186 5.44 105 J cal 5 •10% of Q gets converted to PE Q / 10 mgh, h 101 m Nfloors = 25 Example (Calorimetry) Aluminum has a specific heat of .0924 cal/gºC. If 110 g of hot water at 90 ºC is added to an aluminum cup of mass 50 g which is originally at a temperature of 23 ºC, what is the final temperature of the equilibrated water/cup combo? Solution •Equate heat loss of water with heat gain of cup DU Q mcDT mcupc Al (T 23 C ) mH 20cH 20 (90 C T ) •Solve for T T = 87.3 ºC Phase Changes and Latent Heat T does not rise when phases change (at constant P) Examples: solid --> liquid (fusion), liquid --> vapor (vaporization) Latent heat = energy required to change phases Q mL Property of substance and type of transition Example: Boiling water 1.0 liters of water is heated from 12 ºC to 100 ºC, then boiled away. a) How much energy is required to bring the water to boiling? b) How much extra energy is required to vaporize the water? c) If electricity costs $75 per 1000 kW-hrs, what was the cost of boiling the water? Solution a) Given m=1000 g, c=1.0 cal/g, DT=88 Find Q Q mcDT Q = 8.8x104 cal = 3.68x105 J Solution continued b) Given L=540 cal/g, m=1000g Find Q Q mL f Q = 5.4x105 cal = 2.26x106 J c) Given Q = 2.26x106+3.68x105 J Rate = $75/(1000 kW-hr) Find cost •First, find rate in dollars/J $75 Rate dollars/J 2.08 10 8 dollars/J 1000 1000 3600 •Then find net cost = Q multiplied by rate = 5.5 ¢ Announcements •Midterms graded on scale of 11 •Who wants extra review/recitation sessions? •You can pick up your exams (not bubble sheet) in Friday helproom Example: Body cooling Consider Bobby Joe from the previous example. If the 90% of the 130 kcals from her soda went into heat which was taken from her body from radiation, how much water was perspired to maintain her normal body temperature? (Assume a latent heat of vaporization of 540 cal/g even though T = 37 ºC) Solution Given: Q = 0.9x1.3x105 cal, L = 540 cal/g Find: mevap Q mL mevap=Q/L = 217 g A can of soda has ~ 350 g of H20 Some fluid drips away Three Kinds of Heat Transer Conduction o Shake your neighbor - pass it down o Examples: Heating a skillet Losing heat through the walls of a house Convection o Move hot region to a different location o Examples: Hot-water heating for buildings Circulating air Unstable atmospheres Radiation o Light is emitted from hot object o Examples: Stars Incandescent bulbs Conduction Power depends on area, length, temperature difference and conductivity of material DT P kA Dx Conductivity is property of material Example A copper pot of radius 12 cm and thickness 5 mm sits on a burner and boils water. The temperature of the burner is 115 ºC while the temperature of the inside of the pot is 100 ºC. What mass of water is boiled away every minute? DATA: kCu = 397 W/mºC Solution Given: Dx = 0.005 m, A = pr2 (r=.12 m), Th = 115 ºC, Tc = 100 ºC time = 60 sec, kCu = 397 W/mºC, L = 540 cal/g DT P kA Dx Q mL f •First, find the power, P = 5.39x104 Watts •Next, find Q = P·time = 3.23x106 J •Finally, find m of vaporized water m Q / Lf Remember (L=4.186·540 J/g) m=1.43 kg Conductivities and R-values Conductivity o Property of Material o SI units are W/(m ºC) DT DT P kA A R Dx R Dx / k R-Value o Property of material and thickness Dx. o Measures resistance to heat o Useful for comparing insulation products o Quoted values are in AWFUL units Conducitivities and R-values ARGH! What makes a good heat conductor? •“Free” electrons (metals) •Easy transport of sound (lattice vibrations) •Stiff is good •Low Density is good •Pure crystal structure Diamond is perfect! R-values for layers Consider a layered system, e.g. glass-air-glass DT PA R DT DT1 DT2 DT3 ... PR1 PR2 PR3 ... A A A P ( R1 R2 R3 ...) A R R1 R2 R3 ... Example: Glass Door Consider three panes of glass, each of thickness 5 mm. The panes trap two 2.5 cm layers of air in a large glass door. How much power leaks through a 2.0 m2 glass door if the temperature outside is -40 ºC and the temperature inside is 20 ºC? DATA: kglass= 0.84 WmºC, kair= 0.0234 Wm ºC Solution Known: kglass=0.84, Dxglass=0.005, kair=0.0234, Dxair=0.025, A=2.0, DT=60 Find: P •First, find Rglass and Rair for one layer of each R Dx / k Rglass=0.00595, Rair=1.068 •Next, find R for all the layers R R1 R2 R3 ... R 3Rglass 2 Rair 2.155 •Finally, find the power DT PA R P = 55.7 W Convection If warm air blows across the room, it is convection If there is no wind, it is conduction Can be instigated by turbulence or instabilities Why are windows triple paned? To stop convection! Transfer of heat by radiation All objects emit light if T > 0 Colder objects emit longer wavelengths (red or infra-red) Hotter objects emit shorter wavelengths (blue or ultraviolet) Stefan’s Law give power of emitted radiation P AeT 4 Emissivity, 0 < e < 1, usually near 1 = 5.6696x10-8 W/(m2ºK4) is the Stefan-Boltzmann constant Example If the temperature of the Sun fell 5%, and the radius shrank 10%, what would be the percentage change of the Sun’s power output? P AeT 4 Solution 4 2 4 P AeT 4pR eT 4 2 4 P0 A0 eT0 4pR 0 eT0 0.9 0.95 0.660 2 4 - 34% Example: Power of the Sun DATA: The sun radiates 3.74x1026 W Distance from Sun to Earth = 1.5x1011 m Radius of Earth = 6.36x106 m a) What is the intensity (power/m2) of sunlight when it reaches Earth? b) How much power is absorbed by Earth in sunlight? (assume that none of the sunlight is reflected) c) What average temperature would allow Earth to radiate an amount of power equal to the amount of sun power absorbed? Solution Given: Psun, Rearth-sun, Rearth a) Find I=P/A of sunlight at the Earth’s orbit I sun Psun 2 4pRearth -sun3 = 1320 W/m2 b) Find power absorbed by earth P Ia, a pR 2 earth = 1.67x1017 W c) Find average T of earth P AeT 4 Aearth 4pR 2 earth P T Ae 4 , e 1 T = 276 ºK = 3 ºC = 37 ºF Why is the Earth warmer? •Earth is not at one single temperature •Emissivity lower at Earth’s thermal wavelengths than at Sun’s wavelengths •Radioactive decays inside Earth •Hot underground (less so in Canada) •Most of Jupiter’s radiation Greenhouse Gases Sun is much hotter than Earth so sunlight has much shorter wavelengths than light radiated by Earth (infrared) Emissivity of Earth depends on wavelength CO2 in Earth’s atmosphere reflects in the infrared oBarely affects incoming sunlight oReduces emissivity, e, of re-radiated heat Global warming •Tearth has risen ~ 1 ºF • ~ consistent with greenhouse effect •Other gases, e.g. S02, could cool Earth