PHYS102Lec1

PHYS102Lec1 - ! " $ &% # # # % # # # % # $...

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Unformatted text preview: ! " $ &% # # # % # # # % # $ # $ $ ' ( )# % # * &% # * # % # # ' ( # % # % # % # # + # % ," , ( . / . # % # $ . $ # # 0 # # . /. # $ # # # 0 1 2 1 2 # . $ $ ! . . .0 3 4 4 567 8 92 : 567 8 92 . . 4 ; : 5 67 8 4 9 . 1 ! $ 4 , .0 3 #) , <9= > 0 $ # 0 & # 4 2 : =9? , .0 . . @ 67 8 @ A8 67 0 0 . . . B # 4 .# > , 0 + 0 4 862 862 # 6 62 ? . # 0 . 0 = − 67 8 9 < = + 86 9 9 = ( − 86 ) < < ∆=∆ 9 0 # ; 9 9 TF = TC + 32 = ( −273.15 ) + 32 = − 460°F 5 5 5 5 TC = ( TF − 32 ) = ( 98.6 − 32 ) = 37.0°C 9 9 9 9 9 TF = TC + 32 = ( T − 273.15) + 32 = ( −173.15) + 32 = − 280°F 5 5 5 0 # TF = 9 5 ; < 9 5 TC + 32 = ( 43 ) + 32 = ( 77 + 32 ) °F = 109°F The normal body temperature is 98.6°F Thus, this patient has a high fever and needs immediate attention . / $ $ . # # % .# . $ # % ( > ∆ =α α $ 0 . / $ ∆ − =α ( − ) # . , ( # / $ 8 ; We choose the radius as our linear ∆ L = α L0 ( ∆ T ) dimension. Then, from ∆ T = TC − 20.0°C = L − L0 2 .21 cm − 2 .20 cm = = 35.0°C −1 α L0 130 × 10 −6 ( °C ) ( 2 .20 cm ) TC = 5 5.0°C / $ 5* 0 + $ 0 . $ / $ $ ∆ = γ − =γ ∆ γ = 6α $ - / $ $ ∆V = β Vo ∆t for solids, β = 3α > $ . . # ( # / $ = ; (a) The diameter is a linear dimension, so we consider the linear expansion of steel: d = d0 1 + α( ∆T) = ( 2.540 cm) 1 + 11 × 10−6 ( °C) ( −1 ) (100°C − 25°C) = 2.542 cm b) If the volume increases by 1%, then Then, using ∆T = ∆V V0 ∆ V = βV0 ( ∆T ) 1.0 × 10 −2 3 11 × 10 −6 ∆V = (1.0 × 10 −2 ) V0 where β = 3α β = ( °C ) −1 = 3.0 × 102 °C , / $ $! 0 3 . . + * . 3 2 =2 $# $ "A 8 # . =2 $ ,$ 2 = ) ) ! )$ ) 04 . # ) ! # ) , # # $ . ) ! . / $ / " & 1 . #. 1 , )C # . . . $ mass mass n= molar mass & # 6 # 16 . # CB # B :@ 6 $ B 68 6 A # . # matom ; molar mass = NA . , # CB # . D $ # 6 # # 6 B ) -: ! E ( E E : ?8 FA E: ?( 6 ) A ) - ! - : B "* ( . "* ! "* : E AB : B 8?$ # 168 F A : B AB B # # ! # . # # # . B C 5 ! # 1 " 5 # ) # . . " ! 3N P= 2V 1 mv 2 2 # . # ) . ) ) . " # # . " , ) . " 123 mv = k BT 2 2 " # KE total 3 = nRT 2 ) ) / / 3 U = nRT 2 + ) # .# " # . ) # / = 8 3 U = nRT 2 The average translational kinetic energy per molecule in an ideal gas at absolute temperature T is KE molecule = 3k BT 2 kB = 1.38 × 10−23 J K T = TC + 273.15 = ( 77.0 + 273.15 ) K = 350 K 3 = 1.38 × 10 −23 J K ( 350 K ) = 7.25 × 10 −21 J 2 KE molecule ( ) 0 / $ GH , # # v rms 3 kB T 3RT = = m M . . . . ( 0 0 ,$ . $# . # ; , . ## ,$ > . . # I .. I . . # ...
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