Wenchao LouPHY151 T1 Specific Heat of Metals by the Method of MixtruesCopper Sample mass: 161.7gCalorimeter cup mass: 25.5gMc+Mw = 73.6gSection V-2m:s t(s)T(℃)1:00 60 19.22:00 120 19.13:00 180 19.14:00 240 19.25:00 300 19.36:00 360 19.37:00 420 19.48:00 480 19.49:00 540 19.510:00 600 19.5Section V-4m:s t(s)T (℃)11:30 690 13.412:00 720 15.512:30 750 16.113:00 780 16.513:30 810 16.714:00 840 16.914:30 870 17.115:00 900 17.115:30 930 17.316:00 960 17.316:30 990 17.417:00 1020 17.517:30 1050 17.518:00 1080 17.518:30 1110 17.619:00 1140 17.619:30 1170 17.620:00 1200 17.720:30 1230 17.721:00 1260 17.721:30 1290 17.80 200 400 600 800 1000 1200 14000510152025Temperature vs. TimeTime (seconds)Temperature (°C)Mw=48.1 gThe 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◦CUsing 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:Q1=1 calg ℃*Mw(Ti-Tf)= 1cal/g℃*(48.1g)*(19.5-13.4)→293.41 calHeat lost by cup:Q2=.22 calg ℃*Mc(Ti-Tf)= .22cal/g℃*(25.5)*( 19.5-13.4)→ 34.221calHeat 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.Ccu*2166.78 cal=[1 calg ℃*Mw(Ti-Tf)]+[.22 calg ℃*Mc(Ti-Tf)]Ccu*2166.78 cal=293.41+34.221Ccu*2166.78 cal=327.631Ccu=0.15 cal/g℃Aluminum Sample mass:52.5gCalorimeter cup mass:25.5gMc+Mw = 70.8gSection V-2m:s t (s) T(°C)1:00 60 19.82:00 120 19.83:00 180 19.84:00 240 19.85:00 300 19.96:00 360 19.97:00 420 19.98:00 480 19.99:00 540 19.910:00 600 20Section V-4m:s t (s) (°C)10:30 630 1411:00 660 15.611:30 690 17.112:00 720 17.312:30 750 17.413:00 780 17.513:30 810 17.514:00 840 17.614:30 870 17.715:00 900 17.715:30 930 17.716:00 960 17.716:30 990 17.817:00 1020 17.817:30 1050 17.818:00 1080 17.818:30 1110 17.919:00 1140 17.919:30 1170 17.920:00 1200 17.90 200 400 600 800 1000 1200 14000510152025Temperature vs. Time (Sample 2)Time (seconds)Temperature (°C) Mw=45.3gUsing the graph I got the final and initial temperature for Al.Ti= 20◦C and Tf= 14◦CUsing 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:Q1=1 calg ℃*Mw(Ti-Tf)= 1cal/g℃*(45.3)*(20-14)→271.8 calHeat lost by cup:Q2=.22 calg ℃*Mc(Ti-Tf)= .22cal/g℃*(25.5g)*(20-14)→ 33 calHeat 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.CAl*735cal=[1 calg ℃*Mw(Ti-Tf)]+[.22 calg ℃*Mc(Ti-Tf)]CAl*735 cal=271.8+33CAl*735 cal=304.8CAl=0.41 cal/g℃VI-2Using 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% differenceUsing 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%