key-eng-fall07-e8059x03

key-eng-fall07-e8059x03 - Exam 3 answer key The test...

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Exam 3 answer key 5 Nov 07 The test version is listed after problem 23. PART I # pts alpha beta gamma-2-a 1 3.25 e e d 2 3.25 e a e 3 3.25 e e b 4 3.25 a e a 5 3.25 e a d 6 3.25 d d e 7 3.25 e e b 8 3.25 b b e 9 3.25 a b e 10 3.25 d a c 11 3.25 c c b 12 3.25 c e d 13 3.25 c d a 14 3.25 a c d 15 3.25 d e c 16 3.25 e b a 17 3.25 a a b 18 3.25 bc d e 19 3.25 e e d 20 3.25 d d e 21 3.25 d bc d 22 3.25 d d c 23 3.25 a a bc These are worth 3.5 pts each: alpha#5, beta#4, and gamma-2-a#6. There were TWO (b)’s listed. The 2 nd (b) is correct. Anyone putting (b) OR (c) will get credit. PART II 24. (9 pts) find the energy of a photon E h hc Js m s m J photon = = = × × × = × - - - ν λ ( . )( . / ) . 6 626 10 300 10 633 10 314 10 34 8 9 19 find the energy of the pulses E pulses photons pulse J photon J pulses = × × × × = - 1120 500 10 1 314 10 1 176 17 19 . . The pulses are emitted, so the energy is negative. -176 J 0 = q cold + q hot + E light 0 = s cold ·m cold · Δ T cold + s cold ·m cold · Δ T cold + E light 0 = (4.184 J/g·ºC)(50.0g)(T f – 13.6 ºC) + (4.184 J/g·ºC)(85.0g)(T f – 93.7 ºC) – 176 J Solve for T f = 64.3 ºC 1 pt for writing the equation for the heat of the cold using specific heat of water
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