UB PHY 151 - phy151 T2 (5 pages)

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phy151 T2

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phy151 T2

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School:
University at Buffalo, The State University of New York
Course:
Phy 151 - College Physics Lab 1
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Wenchao Lou PHY151 T2 Phase Transition of the First Order The Melting of Ice Calorimeter cup mass 7 5g Mc Mw 60g Section V 2 m s 1 00 2 00 3 00 4 00 5 00 t s 60 120 180 240 300 T 34 4 34 1 33 8 33 5 33 3 Section V 4 m s t s T 6 00 360 11 4 7 00 420 7 6 8 30 480 6 8 9 00 540 6 0 10 30 600 5 8 11 00 660 5 8 12 30 720 5 8 13 00 780 5 7 14 30 840 5 7 15 00 900 5 7 Calorimeter Cup Water Melted ice mass Mc Mw Mi 78 8g Temperature vs Time 25 Temperature C 20 15 10 5 0 0 200 400 600 800 1000 1200 1400 Time seconds Mw 48 1 g The graph is on a separate sheet of paper Using the graph I got the final and initial temperature for Cu Ti 19 5 C and Tf 13 4 C Using these values I can now calculate the heat for each component which will allow me to calculate the specific heat for copper Heat lost by water 1 cal Q1 Mw Ti Tf 1cal g 48 1g 19 5 13 4 293 41 cal g Heat lost by cup 22 cal Q2 Mc Ti Tf 22cal g 25 5 19 5 13 4 34 221cal g Heat gained by copper Q3 Ccu Mcu Ti Tf Ccu 161 7g 13 4 0 Ccu 2166 78 To get Ccu we use equation 3 from the manual 1 cal 22 cal Ccu 2166 78 cal Mw Ti Tf Mc Ti Tf g g Ccu 2166 78 cal 293 41 34 221 Ccu 2166 78 cal 327 631 Ccu 0 15 cal g Aluminum Sample mass 52 5g Calorimeter cup mass 25 5g Mc Mw 70 8g Section V 2 m s t s 1 00 2 00 3 00 4 00 5 00 6 00 7 00 8 00 9 00 10 00 T C 60 120 180 240 300 360 420 480 540 600 19 8 19 8 19 8 19 8 19 9 19 9 19 9 19 9 19 9 20 Section V 4 m s t s 10 30 11 00 11 30 12 00 12 30 13 00 13 30 14 00 14 30 15 00 15 30 16 00 16 30 17 00 17 30 18 00 18 30 19 00 19 30 20 00 C 630 660 690 720 750 780 810 840 870 900 930 960 990 1020 1050 1080 1110 1140 1170 1200 14 15 6 17 1 17 3 17 4 17 5 17 5 17 6 17 7 17 7 17 7 17 7 17 8 17 8 17 8 17 8 17 9 17 9 17 9 17 9 Temperature vs Time Sample 2 25 Temperature C 20 15 10 5 0 0 200 400 600 800 1000 1200 1400 Time seconds Mw 45 3g Using the graph I got the final and initial temperature for Al Ti 20 C and Tf 14 C Using these values I can now calculate the heat for each component which will allow me to calculate the specific heat for Aluminum Heat lost by water 1 cal Q1 Mw Ti Tf 1cal g 45 3 20 14 271 8 cal g Heat lost by cup 22 cal Q2 Mc Ti Tf 22cal g 25 5g 20 14 33 cal g Heat gained by copper Q3 CAl MAl Ti Tf CAl 52 5 g 14 0 CAl 735 To get CAl we use equation 3 from the manual 1 cal 22 cal CAl 735cal Mw Ti Tf Mc Ti Tf g g CAl 735 cal 271 8 33 CAl 735 cal 304 8 CAl 0 41 cal g VI 2 Using this source I got the specific heat of copper The actual specific heat of copper is Ccu 0 0919 cal g My calculated value was Ccu 0 15 cal g The percent difference is 0 15 cal g 0 0919 cal g 0 019 100 3 difference Using this source I got the specific heat of aluminum The actual specific heat of aluminum is CAl 0 2157 cal g My calculated value was CAl 0 41 cal g The percent difference is 0 41 cal g 0 2157 cal g 0130 100 14 9 difference

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