and length of aluminum is 4.0000cm assume both are heated and cooled together at what temperature will rod barely fits into tubefind mass of copper ,find N of copper.Find final temperature of system.
linear thermal expansion coef Cu 17e-6 /K linear thermal expansion coef Al 23.1e-6 /K ?L/L α?T, α is linear thermal expansion coef But what is D1 and D2 supposed to mean? You really have to supply all the needed info and drawings and not make us guessI'll assume the 1.0 cm is the ID and 1.1 cm the OD of the tube So you need the temperature at which the 1.001 cm of Cu will fit into the 1.000 cm of aluminumCu expands less than aluminum with temperature, so there is a temperature where the aluminum will have expanded enough for the numbers to be equal ?L/L α?T for the Cu, ?L? L?(17e-6)?T for the Al, ?L? L?(23e-6)?T ?L? 1.001(17e-6)?T ?L? 1.000(23e-6)?T 1.000 + ?L? 1.001 + ?L? now it's just algebra ?L? 0.001 + ?L? ?L? 1.000(23e-6)?T 0.001 + ?L? 1.000(23e-6)?T ?L? 1.000(23e-6)?T – 0.001 ?L? 1.001(17e-6)?T subtracting 0 1.000(23e-6)?T – 0.001 – 1.001(17e-6)?T (23e-6)?T – (17.017e-6)?T 0.001 5.983e-6 ?T 0.001 ?T 167?C T 187?C but check my math Mass of copper you can get by calculating the volume and using the density of copper/Don't know what N of copper means.
