To what temperature would an aluminum ring have to be heated to make its diameter 1.2% larger than it is at 20 oC?
We need to increase the diameter of the aluminum ring from Lo to 1.012Lo. That is an increase of
L = 0.012Lo
we find that the coefficient of thermal expansion for aluminum is
= 24 x 10 - 6 (Co) - 1
L = LoT
T = [L ] / [ Lo ]
T = 0.012Lo/ [ ( 24 x 10 - 6 (Co) - 1 ) Lo]
T = 500 Co
Tf = 520 oC
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