A metal rod is 40.125cm long at 20.0°C and 40.148cm long at 45.0°C.calculate the average coefficient of linear expansion of the rod for this temperature range.

2.3×10?5

(40.148-40.125)=coefficint of linear expantion*40.125*(45-20) (0.023)/(40.125*25)=coefficint of linear expantion coefficint of linear expantion=0.000022928 therefore coefficint of linear expantion=0.000022928

given that Initial length(L1)=40.125 Final length (L2)=40.148 (t1)=20 t2=45 we know that coefficient of linear expansion = (L2-L1)/L1*(t2-t1) = (40.148-40.125)/40.125*(25) = (0.023)/1003.125 =2.29*10^-5

given Initial length (l1) = 40.125 final length (l2) = 40.148 temp (t1) =20 temp (t2) =45 we had know that coeffecient of l.e =(L2-L1)/L1*(T2-T1) =(40.148-40125)/40.125*(25) =(0.023)/1003.125 =2.29*10^-5