e final temperatureT1 = the initial temperature = the coefficient of linear expansion of the materialThe following equation 2 was changed by equation 1, which was used to calculate the = L . L1 ( T2 V T1)The coefficient of linear expansion () of five metals were calculated by equation 2. The accepted value of five metals in the text book (pg. 646) was shown in the following:Metal [ (XC)-1 x 10-6(Pyrex) glass3.3Brass20.0Copper17.0Stainless steel16.0Aluminum23.6GROUP #5Calculations for (Pyren) glass:L = (0.15)/1000(m), L1 = 0.6m, T2 = 95XC,T1 =27XC = (0.15/1000) . = 3.68 (XC)-1 x 10-6 0.6(95 V 27) for Brass:L = (0.88)/1000(m), L1 = 0.6m, T2 = 98XC,T1 =27XC = (0.88/1000) . = 20.7 (XC)-1 x 10-6 0.6(98 V 27) for Copper:L = (0.79)/1000(m), L1 = 0.6m, T2 = 98XC,T1 =27XC = (0.79/1000) . = 18.6 (XC)-1 x 10-6 0.6(98 V 27) for Stainless steel:L = (0.74)/1000(m), L1 = 0.6m, T2 = 98XC,T1 =27XC = (0.74/1000) . = 17.4 (XC)-1 x 10-6 0.6(98 V 27) for Aluminum:L = (1.09)/1000(m), L1 = 0.6m, T2 = 98XC,T1 =27XC = (1.09/1000) . = 25.6 (XC)-1 x 10-6 0.6(98 V 27) GROUP #5The percent deviation of the was shown in the following:% of Glass= (3.6 x 10-6) - (3.3 x 10-6) x 100 3.3 x 10-6= 9.1%% of Brass= (20.7 x 10-6) - (20 x 10-6) x 100 20 x 10-6= 3.5 %% of Copper= (18.6 x 10-6) - (17 x 10-6) x 100 17 x 10-6= 9.4 %% of Stainless = (17.4 x 10-6) - (1.6 x 10-6) x 100steel 1.6 x 10-6= 8.75 %% of Aluminum= (25.6 x 10-6) - (23.6 x 10-6) x 100 23.6 x 10-6= 8.5 %DiscussionThe result of the experiment turned out to be fairly close to the accepted value. The percentage deviation of the coefficient of linear expansion was in a range of 3.5% to 9.4% which was less than what it was expected due to many...
