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hemetite
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Homework Statement
An aluminium sleeve of internal diameter 4.000cm at 10 degrees Celsius is heated and slipped over a steel rod of diameter 4.040cm (at 10 degrees Celsius. When the aluminium sleeve and steel rod return to room temp they are impossible to separate. Assume that the coeffiecients of thermal expansion are constant with temperature.
1) To what temperature must the aluminium sleeve be heated before it can slipped over the steel rod?
2) Can the aluminium sleeve and steel rod be separated by heating or cooling form together to some common temperature? If so, what temperature?
Homework Equations
delta L = alpha x L initial x delta temperature
The Attempt at a Solution
This problem should be quite straightforward and simple...but i am not confident of my answer i got for part 1)...care to spot my mistake?
Let the room temperature be 295 K.
At 10 degrees = 273.15 K + 10 = 283.15 K
Firstly i check the Diameter for both aluminum and steel at room temperature.
The aluminum ring have expanded by
delta L = (25x10^-6)(4x10^-2)(295-283.15)
= 1.185 x 10 ^ (-5) m
The diameter of aluminium at room temp
= (4x10^-2) + = 1.185 x 10 ^ (-5)
= 0.04001185 m
Okay i cut short...i just do the same for the steel rod and it is 0.040405744m.
To get the temperature of the aluiminium so that it can be slipped over the rod ..
Diameter of steel - Diameter aluminium = 0.000393894 m
so i put back in the thermal expansion equation
0.000393894 = (25x10^-6)(0.04001185)(Tempearture - 295 K)
solving the equation:
The Temperature the aluminium need to be heated = 683.8 K = 410 degree celsius
I felt that the number is a bit big...any mistakes?