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Thermal Energy Processes and Heat Transfer in Materials

Learn about internal energy, work, heat transfer, phase changes, latent heat, conductivity, R-values, and heat transfer mechanisms like conduction, convection, and radiation. Explore examples and solutions to thermodynamics problems.

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Thermal Energy Processes and Heat Transfer in Materials

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  1. Chapter 11Energy in Thermal Processes

  2. next chapter Vocabulary, 3 Kinds of Energy • Internal Energy U = Energy of a system due to microscopic motion and inter-molucular forces • WorkW = -FDx = -PDV is work done by expansion (next chapter) • HeatQ = Energy transfer from microscopic contact

  3. Property of the material mass of material • cH20 = 1.0 cal/(gºC) • 1 calorie = 4.186 J Temperature and Specific Heat • Add energy -> T rises

  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 Nfloors = 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 • Solve for T T = 87.3 ºC

  6. Property of substance and type of transition 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

  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, DT=88 Find Q Q = 8.8x104 cal = 3.68x105 J

  9. Solution continued b) Given L=540 cal/g, m=1000gFind Q 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 • 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 = 0.9x1.3x105 cal, L = 540 cal/g Find: mevap mevap=Q/L = 217 g A can of soda has ~ 350 g of H20 Some fluid drips away

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

  13. Conductivity is property of material Conduction • Power depends on area, length, temperature difference and conductivityof 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: 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 • First, find the power, P = 5.39x104 Watts • Next, find Q = P·time = 3.23x106 J • Finally, find m of vaporized water Remember (L=4.186·540 J/g) m=1.43 kg

  15. Conductivities and R-values • Conductivity • Property of Material • SI units are W/(m ºC) • R-Value • Property of material and thickness Dx. • Measures resistance to heat • Useful for comparing insulation products • Quoted values are in AWFUL units

  16. ARGH! Conducitivities and R-values

  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 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

  20. 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 Rglass=0.00595, Rair=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. Emissivity, 0 < e < 1, usually near 1 s = 5.6696x10-8 W/(m2ºK4) is the Stefan-Boltzmann constant 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

  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 3.74x1026 W Distance from Sun to Earth = 1.5x1011 m Radius of Earth = 6.36x106 m • What is the intensity (power/m2) of sunlight when it reaches Earth? • How much power is absorbed by Earth in sunlight? (assume that none of the sunlight is reflected) • What average temperature would allow Earth to radiate an amount of power equal to the amount of sun power absorbed?

  26. Solution Given: Psun, Rearth-sun, Rearth a) Find I=P/A of sunlight at the Earth’s orbit = 1320 W/m2 b) Find power absorbed by earth = 1.67x1017 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 • CO2 in Earth’s atmosphere reflects in the infrared • Barely affects incoming sunlight • Reduces emissivity, e, of re-radiated heat

  29. Global warming • Tearth has risen ~ 1 ºF • ~ consistent with greenhouse effect • Other gases, e.g. S02, could cool Earth

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