1. Chapter 11 Energy in Thermal Processes

2. Vocabulary, 3 Kinds of Energy  Internal Energy U = Energy of a system due to microscopic motion and inter-molucular forces  Work W = -F  x = -P  V is work done by expansion (next chapter)  Heat Q = Energy transfer from microscopic contact next chapter

3. Temperature and Specific Heat  Add energy -> T rises Property of the material mass of material c H20 = 1.0 cal/(gºC) 1 calorie = 4.186 J

4. 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! 10% of Q gets converted to PE N floors = 25

5. 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 T = 87.3 ºC Solve for T

6. 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 Property of substance and type of transition

7. 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?

8. Solution a) Given m=1000 g, c=1.0 cal/g,  T=88 Find Q Q = 8.8x10 4 cal = 3.68x10 5 J

9. Solution continued b) Given L=540 cal/g, m=1000g Find Q Q = 5.4x10 5 cal = 2.26x10 6 J c) Given Q = 2.26x10 6 +3.68x10 5 J Rate = $75/(1000 kW-hr) Find cost First, find rate in dollars/J Then find net cost = Q multiplied by rate = 5.5 ¢

10. 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

11. 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 = 0.9x1.3x10 5 cal, L = 540 cal/g Find: m evap m evap =Q/L = 217 g A can of soda has ~ 350 g of H 2 0 Some fluid drips away

12. Three Kinds of Heat Transer  Conduction oShake your neighbor - pass it down oExamples: Heating a skillet Losing heat through the walls of a house  Convection oMove hot region to a different location oExamples: Hot-water heating for buildings Circulating air Unstable atmospheres  Radiation oLight is emitted from hot object oExamples: Stars Incandescent bulbs

13. Conduction  Power depends on area, length, temperature difference and conductivity of material Conductivity is property of material

14. 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: k Cu = 397 W/mºC Solution Given:  x = 0.005 m, A =  r 2 (r=.12 m), T h = 115 ºC, T c = 100 ºC time = 60 sec, k Cu = 397 W/mºC, L = 540 cal/g First, find the power, P = 5.39x10 4 Watts Next, find Q = P·time = 3.23x10 6 J Finally, find m of vaporized water Remember (L=4.186·540 J/g) m=1.43 kg

15. Conductivities and R-values  Conductivity oProperty of Material oSI units are W/(m ºC)  R-Value o Property of material and thickness  x. o Measures resistance to heat o Useful for comparing insulation products o Quoted values are in AWFUL units

16. Conducitivities and R-values ARGH!

17. 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!

18. R-values for layers Consider a layered system, e.g. glass-air-glass

19. 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 m 2 glass door if the temperature outside is -40 ºC and the temperature inside is 20 ºC? DATA: k glass = 0.84 WmºC, k air = 0.0234 Wm ºC

20. Solution Known: k glass =0.84,  x glass =0.005, k air =0.0234,  x air =0.025, A=2.0,  T=60 Find: P First, find R glass and R air for one layer of each R glass =0.00595, R air =1.068 Next, find R for all the layers Finally, find the power P = 55.7 W

21. 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

22. Why are windows triple paned? To stop convection!

23. 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  = 5.6696x10 -8 W/(m 2 ºK 4 ) is the Stefan-Boltzmann constant Emissivity, 0 < e < 1, usually near 1

24. 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? Solution - 34%

25. Example: Power of the Sun DATA: The sun radiates 3.74x10 26 W Distance from Sun to Earth = 1.5x10 11 m Radius of Earth = 6.36x10 6 m a)What is the intensity (power/m 2 ) 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?

26. Solution a) Find I=P/A of sunlight at the Earth’s orbit Given: P sun, R earth-sun, R earth = 1320 W/m 2 b) Find power absorbed by earth = 1.67x10 17 W c) Find average T of earth T = 276 ºK = 3 ºC = 37 ºF

27. 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

28. 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  CO 2 in Earth’s atmosphere reflects in the infrared oBarely affects incoming sunlight oReduces emissivity, e, of re-radiated heat

29. Global warming T earth has risen ~ 1 ºF ~ consistent with greenhouse effect Other gases, e.g. S0 2, could cool Earth