!U = Q " P!V. Q = mc!t. Vocabulary, 3 Kinds of Energy. Chapter 11. Energy in Thermal Processes. Example Temperature and Specific Heat

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1 Vocabulary, 3 Kinds of Energy Chapter 11 Energy in Thermal Processes Internal Energy U = Energy of microscopic motion and intermolucular forces Work W = -F!x = -P!V is work done by compression (next chapter) Heat Q = Energy transfer from microscopic contact!u = Q " P!V next chapter Temperature and Specific Heat Add energy -> T rises Q = mc!t Example 11.1 Bobby Joe drinks a 130 calorie can of soda. If the efficiency for turning energy into work is 20%, how many 4 meter floors must Bobby Joe ascend in order to work off the soda and maintain her 55 kg mass? Mass Property of material c H20 = 1.0 cal/(gºc) 1 calorie = J N floors = 50.4 Example 11.2 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? 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 /transition T = 87.3 ºC

2 Example 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 MW-hr, what was the cost of heating and boiling the water? a) Q = 8.8x10 4 cal = 3.68x10 5 J b) Q = 5.4x10 5 cal = 2.26x10 6 J c) 5.5 Example 11.4 Consider Bobby Joe from the previous example. If the 80% 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) = 193 g A can of soda has ~ 325 g of H 2 0 Some fluid drips away Three Kinds of Heat Transfer Conduction Shake your neighbor - pass it down Examples: Heating a skillet, losing heat through the walls 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 P = ka! "T / "x Conduction Power depends on area A, thickness!x, temperature difference!t and conductivity of material Conductivity is property of material Example 11.5 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 m=1.43 kg Conductivities and R-values Conductivity (k) Property of Material SI units are W/(m ºC) " P = ka!t % # $!x & ' = A!T R R =!x / k R-Value Property of material and thickness!x. Measures resistance to heat Useful for comparing insulation products Quoted values are in AWFUL units

3 Conducitivities and R-values What makes a good heat conductor? ARGH! Free electrons (metals) Easy transport of sound (lattice vibrations) Stiff is good Low Density is good Pure crystal structure Diamond is perfect! Example 11.6a An large pipe carries steam at 224 ºC across a large industrial plant. The outside of the pipe is at room temperature, 24 ºC. The pipe is 120 m long and has a diameter of 70 cm. The pipe is constructed of an insulating material of conductivity k= 2.62 W/mºC. In order to reduce the rate of heat loss through the pipe by a factor of 1/2, an engineer could: a) Reduce the length of the pipe by a factor of 1/2 b) Reduce the diameter of the pipe by a factor of 1/2 c) Increase the thickness of the pipe by a factor of 2 d) All of the above e) None of the above Example 11.6b An large pipe carries steam at 224 ºC across a large industrial plant. The outside of the pipe is at room temperature, 24 ºC. The pipe is 120 m long and has a diameter of 70 cm. The pipe is constructed of an insulating material of conductivity k= 2.62 W/mºC. In order to reduce the rate of heat loss through the pipe by a factor of 1/2, an engineer could: a) Make the pipe using a new material with twice the conductivity, 5.24 W/m ºC b) Re-design the pipe to double the R-value c) All of the above d) None of the above Example 11.6c An large pipe carries steam at 224 ºC across a large industrial plant. The outside of the pipe is at room temperature, 24 ºC. The pipe is 120 m long and has a diameter of 70 cm. The pipe is constructed of an insulating material of conductivity k= 2.62 W/mºC. In order to reduce the rate of heat loss through the pipe by a factor of 1/2, an engineer could: a) Reduce the density of steam in the pipe by a factor of 1/2 b) Reduce the temperature of the steam to 124 ºC c) Reduce the velocity of the steam through the pipe by a factor of 1/2 d) All of the above e) None of the above R-values for layers Consider a layered system, e.g. glass-air-glass P = A!T R!T =!T 1 +!T 2 +!T 3... = PR 1 A + PR 2 A + PR 3 A +... = P A (R 1 + R 2 + R ) R = R 1 + R 2 + R

4 Example 11.7 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 W/mºC, k air = W/m ºC Convection P = 55.8 W 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 Emissivity, 0<e<1, usually near 1 P = e! AT 4 " = x10-8 W/(m 2 ºK 4 ) is the Stefan-Boltzmann constant Example 11.8 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? Example 11.9 DATA: The sun radiates 3.74x10 26 W Distance from Sun to Earth = 1.5x10 11 m Radius of Earth = 6.36x10 6 m - 34% a) What is the intensity (power/m 2 ) of sunlight when it reaches Earth? a) 1323 W/m 2 b) How much power is absorbed by Earth in sunlight? (assume that none of the sunlight is reflected) b) 1.68x10 17 W c) What average temperature would allow Earth to radiate an amount of power equal to the amount of sun power absorbed? c) T = 276 K = 3 ºC = 37 ºF

5 What is neglected in estimate? Earth is not at one single temperature Some of Sun s energy is reflected 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 Example 11.10a Two Asteroids A and B orbit the Sun at the same radius R. Asteroid B has twice the surface area of A. (Assume both asteroids absorb 100% of the sunlight and have emissivities of 1.0) a) (1/4)T A b) (1/2)T A c) T A d) 2T A e) 4T A Example 11.10b Two identical asteroids A and B orbit the sun. Asteroid B is located twice as far from sun as Asteroid A. R B =2R A (Assume both asteroids absorb 100% of the sunlight and have emissivities of 1.0) Example 11.10c Two Asteroids A and B orbit the Sun at the same radius R. Asteroid B is painted with reflective paint which reflects 3/4 of the sunlight, while asteroid A absorbs 100% of the sunlight. Both asteroids have emissivities of 1.0. a) (1/4)T A b) (1/2)T A c) (2-1/2 )T A d) (2-1/4 )T A e) T A a) (1/4)T A b) (1/2)T A c) (2-1/2 )T A d) (2-1/4 )T A e) (2-3/4 )T A Example 11.10d Two Asteroids A and B orbit the Sun at the same radius R. Asteroid B has an emissivity of 0.25, while the emissivity of asteroid A is 1.0. Both asteroids absorb 100% of the sunlight. a) 4T A 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 Barely affects incoming sunlight Reduces emissivity, e, of re-radiated heat b) 2T A c) 2 1/2 T A d) 2 1/4 T A e) 2 3/4 T A

6 Global warming Mercury and Venus T mercury = 700 K (day) & 90 K (night) T venus = 740 K T earth has risen ~ 1 ºF ~ consistent with greenhouse effect Other gases, e.g. S0 2, could cool Earth