Phy 212: General Physics II 1 Chapter 18 Worksheet 3/20/2008

Transcription

1 Phy 1: General Physics II 1 hapter 18 rksheet 3/0/008 Thermal Expansin: 1. A wedding ring cmpsed f pure gld (inner diameter = 1.5 x 10 - m) is placed n a persn s finger (diameter = 1.5 x 10 - m). Bth the ring and the finger are at 7.0. a. The finger and ring are placed in cld running water (5.0 ). Determine the cntractin f the ring and the finger, respectively. Determine the rati f L finger / L ring. Based n yur calculatin, is the ring lser r tighter? b. The finger and ring are placed in warm running water (0.0 ). Determine the rati f L finger / L ring. Based n yur calculatin, is the ring lser r tighter? c. Based n (a) and (b), which apprach, warm vs. cld water, wuld be the mst effective fr remving a tight ring frm a finger? hy? Heat apacity & Heat Transfrmatin:. A lead bullet (m=0.050 kg & V= m 3 ) at 0.0 impacts a blck, made f an ideal thermal insulatr, and cmes t rest at its center. After impact, the temperature f the bullet is 37. a. Hw much heat was needed t raise the bullet t its final temperature? Q = cm T = ( 18 ) 0.050kg 307 =1965 b. hat is the cefficient f vlume expansin (β) fr the bullet? β = 3 α = = c. hat is the vlume f the bullet after it cmes t rest? β V=V T= m = m V =V + V= m m = m d. Hw much additinal heat wuld be needed t melt the bullet? Q = ml f = ( 0.050kg) ( ) 3 3 = e. Hw fast was the bullet traveling prir t hitting the blck? Assume that all mechanical energy in the bullet is translatinal kinetic energy, prir t cntact with the blck and n energy is lst t the blck. Assuming that all heat gain by bullet is due t lss f kinetic energy, K=1965, and the final kinetic energy fr the bullet is K f =0, then K i = K- K f =1965-0=1965

2 Phy 1: General Physics II hapter 18 rksheet 3/0/ A kg piece f cpper at an initial temperature f 0.0 is placed in a water bath and the temperature f the metal is raised t a. Hw much heat was required t raise the temperature f the cpper? Q = cm T = ( 386 ) 0.500kg 80 = b. Hw much mre heat wuld be required t raise the cpper t its melting pint? The melting pint fr cpper is T melt =1083 : ( ) Q = cm T = kg 983 = c. Hw much heat wuld be required t cmpletely melt the piece f cpper, frm an initial temperature f 100.0? Q = Q +Q = kg tt T melt kg Q tt = =.9 10 d. The piece f cpper in (a) is then placed in a thermally islated cntainer, called a calrimeter, cntaining 1.00 kg f water initially at 0.0. hat is the equilibrium temperature f the cpper/water system? Q = -Q c m T = -c m T c m T-0.0 = c m T HO u HO HO HO u u u HO HO u u T = cum c m u HO HO c m u u 1+ c HO m HO = 3.5 e. Suppse the piece f cpper frm (a) were placed in a calrimeter cntaining kg f an unknwn liquid initially at 0.0. The equilibrium temperature f the cpper/liquid system is hat is the specific heat capacity f the unknwn liquid? ( 386 ) ( 0.500kg) ( 69.1 ) c m T c = = = 50 u u u? m? T? ( 0.500kg) ( 10.9 ) f. hat is the identity f the unknwn liquid? Prbably ethyl alchl (verify using yur textbk). Suppse a 0.70 kg piece f irn and a 0.50 kg piece f cpper (bth at initial temperature f ) were placed tgether in a calrimeter cntaining 1.00 kg water (initially at 0.0 ). hat is the final temperature f the water in the calrimeter?

3 Phy 1: General Physics II 3 hapter 18 rksheet 3/0/008 Q = -Q -Q c m T = -c m T -c m T HO Al u HO HO HO Al Al Al u u u T = T = c m T-0.0 = c m +c m T HO HO Al Al u u calm Al+cumu chomh O c m +c m 1+ c HO m HO Al Al u u ( 900 ) ( 0.70kg ) kg ( ) ( 0.50kg) ( 186 ) ( 1.00kg) kg ( 900 ) ( 0.70kg ) + ( 386 ) ( 0.50kg) 1 ( 186 ) ( 1.00kg) kg = A kg glass (c=80 /kg ) cntaining 1.00 L f water (at 0.0 ) is filled with kg f ice (at -5.0 ). a. hat is the mass f the liquid water initially in the glass? m = V = x10 m =1.0 kg kg -3 3 ρ ( 3 ) HO HO HO m b. hat is the equilibrium temperature f the water and glass when all f the ice has melted? Ignre any heat gained frm r lst t the surrundings. Q +Q +Q = -Q -Q ice melt melted HO glass ice micel ice+cicemice ( ) +ch Omice T-0.0 = ch Om HO+cglassmglass T ( ch ) Om ice+chom HO+cglassmglass T = ch Om HO+cglassmglass 0.0 -micel ice-cicemice 5.0 ( ch ) Om HO+cglassmglass 0.0 -micel ice-cicemice 5.0 T = = 11.5 ( ch Om ice+ch Om H O+cglassmglass ) c. A persn then drinks all f the water in the glass. Hw much heat des the water gain as it is warmed up t 37.0 in the digestive tract f the persn? 5 Q HO= chomh O T HO = kg T = ( )

4 Phy 1: General Physics II hapter 18 rksheet 3/0/008 1 st Law f Thermdynamics: 6. An enclsed gas perfrms the fllwing 3 step clsed cycle. a) alculate the wrk perfrmed by the system fr: i) A t B Since P is linear frm A t B: =P V A B avg x10 Pa+1.0x10 Pa A B= 1.0m 5 A B=1.11x10 N m ii) B t P is linear frm B t : =P V B avg B 5 3 = 1.0x10 Pa -1.0m 5 B =-1.0x10 N m 3 B iii) t A P is linear frm t A: =P V A avg A =0 N m A b) hat is the ttal wrk perfrmed by the clsed cycle? net is the area enclsed in the PV graph r alternatively: = + + Net A B B A 3 Net=-9.x10 N m c) Is the net wrk perfrmed by this system ver a cmplete cycle psitive r negative? Explain. net is negative. Net wrk is perfrmed ON the system, since the clsed cycle perates cunter-clckwise. d) alculate the ttal heat absrbed by the system during 1 cycle. Since this is a clsed cycle, E int =0, therefre Q = net =-9x10 3. Heat is released(lst) by the system during ne cycle.

5 Phy 1: General Physics II 5 hapter 18 rksheet 3/0/ An enclsed gas perfrms the fllwing 3 step clsed cycle. a) alculate the wrk perfrmed by the system fr: i) A t B A B=Pavg V=0 N m ii) B t =P V B avg B 5 3 =.00x10 Pa 0.5m 5 B =1.00x10 N m iii) t D D=Pavg V=0 N m ii) D t A =P V D A avg D A 5 3 = 1.00x10 Pa 0.5m D A=-5.0x10 N m b) hat is the ttal wrk perfrmed by the clsed cycle? net is the area enclsed in the PV graph r alternatively: = Net A B B D D A Net=5.0x10 N m B A D c) Is the net wrk perfrmed by this system ver a cmplete cycle psitive r negative? Explain. net is psitive, net wrk is perfrmed BY the system (the clsed cycle perates clckwise). d) alculate the ttal heat absrbed by the system during 1 cycle. Since this is a clsed cycle, E int =0, therefre Q = net =+5.x10. Heat is gained (absrbed) by the system during ne cycle.

6 Phy 1: General Physics II 6 hapter 18 rksheet 3/0/008 Heat Transfer & king: 8. A 0. cm irn pan is used t fry a cylindrical piece f meat (mass = 0.1 kg, 1.5 cm thickness and 10.0 cm diameter). The temperature f the pan is heated t a temperature f 350 F. The thermal cnductivity f the meat is 0.0 /(m. K) and the emissivity f the meat is 0.9. Assume the specific heat capacity f the meat is 3500 /(kg. K). a) hat is the temperature f the pan in elsius & Kelvin? F 9 F ( ) ( ) T = T 3 F = 350 F 3 F =177. F T = K 1K K= 50K b) hat is the area f the frying surface f the meat in m? 0.10m A= π r = π = m c) hen the meat placed n the pan its temperature is raised t a cnstant temperature f 100 and the ppsite face f the meat initially is 5. hat is the rate f cnductive heat flw thrugh the meat? dq A = k dt = 19.9 y s d) hy des the heat f the ht surface f the meat never get warmer than 100 when the meat is mist? The water n the surface limits the surface temperature until it has all evaprated. e) hen enugh heat has passed thrugh the meat, the ppsite face f the meat will reach the temperature f the rm (5 ), what is the temperature at the center f the meat? Assuming the temperature prfile thrugh the meat is linear: T bttm + Ttp T middle= = 37.5 f) hat is the rate f cnductive heat flw thrugh the meat in (e)? dq A = k dt = 15.8 y s g) Hw much ttal heat (energy) has the meat absrbed (cmpared t when the whle piece was at 5 ) when the uter surface is 5? Assuming the temperature prfile thrugh the meat is linear: Q = c m T - 5 = meat meat middle h) hat is the net rate f radiative heat flw frm the meat when the uter surface is at 5 and 5 respectively? hen meat surface temperature is T=5 :

7 Phy 1: General Physics II 7 hapter 18 rksheet 3/0/008 dq = εσ A T - T = m 73.15K K meat rm m K dq = ( ) π hen meat surface temperature is T=5 : dq = -8 εσ A T - T = m 98.15K K meat rm m K π dq = 0 i) If the meat were left n the pan (same side) what wuld be the final equilibrium temperature f the uter surface f the meat? {Hint: Try using the equatin slver feature f yur (r yur neighbr s) TI85/89 calculatr} cnd This is a tugh ne! = net rad ka T -T = A T - T y ( bttm tp ) εσ ( tp rm ) k k k T - T = ( T -T ) = T - T εσ y εσ y εσ y tp rm bttm tp bttm tp k k T = - T + T + T εσ y εσ y tp tp bttm rm T tp 357 K r 8 { I slved this graphically, check it w/ yur calculatr... } j) hat is the net rate f radiative/cnductive heat flw frm the meat when the meat is in thermal equilibrium? The net rate f thermal transfer int the meat is zer, since it is in thermal equilibrium. This can be checked by calculating the radiative heat lss, when meat surface temperature is T=357 K r 137 : dq = -8 εσ A ( T - T ) = ( 0.9 ) ( ) π ( 0.05m ) ( 357K ) - ( 98.15K ) surface rm m K dq = 3. mpare this t the cnductive heat flw acrss the meat: cnd ka = T -T y ( bttm tp ) ( 0. ) π ( 0.05m) dq cnd 0.015m m K = K-357K = 3.

8 Phy 1: General Physics II 8 hapter 18 rksheet 3/0/ The Greenhuse Effect can be analyzed using the radiatin mdel f h 18. Assume that radiatin is the primary mechanism f heat transfer between the earth and the universe, where bth the earth and sun are perfect emitters (ε = 1.0). Use the fllwing values (all ther values can be fund in the back f the textbk): T sun = 5350 K (the average temperature f the surface f the sun) T earth = 88 K (the average temperature f the surface f the earth) a. Determine the rate at which energy is radiated by the sun. dq = -8 8 εσ AT = ( 1.0 ) ( ) m K π m 5350K dq = b. Hw much f the sun s radiant energy reaches the earth? Hint: hat is the intensity (pwer per unit area) at a distance frm the sun f d earth-sun = 1.50x10 11 m? The intensity f sun s radiant energy at the earth is: I = = = m A π m The incident pwer is related t intensity x the effective crss sectinal area f the earth: ( ) dq = IA = crss sectin π m = m c. Hw much f the sun s pwer is absrbed by the earth? Nte: The earth s atmsphere absrbs ~30% f the ttal radiant energy frm the sun. dq absrbed = = d. Hw much pwer (at what energy rate) des the earth radiate? radiated radiated ( ) π ( ) -8 6 = εσat = m 95K m K 17 = e. hat fractin f the earth s radiant energy is reflected back t the surface and reabsrbed? This is the Greenhuse Effect! Assuming the earth is in thermal equilibrium: = - + = 0 earth frm sun radiated re-absrbed = - = re-absrbed radiated frm sun re-absrbed 16 =

9 Phy 1: General Physics II 9 hapter 18 rksheet 3/0/008 f. If there were n reflectin and re-absrptin f radiant energy by the earth, estimate the average temperature f the earth. The temperature (earth surface) fr thermal equilibrium: = - = 0 earth frm sun radiated = AT = ( 1.0) ( ) ( m) T = m K εσ π radiated T = 30K g. If the earth had n atmsphere at all, estimate the average temperature f the surface f the earth. The temperature (earth surface) fr thermal equilibrium: radiated = εσ AT = earth T earth = =5 K εσa 1