Linear Expansion of Solid problem

[toddler]

Homework Statement

There is an aluminum ring, .07 m in diameter at 5 deg. C. There is also an aluminum shaft whose diameter is .07003 m in diameter at 5 deg. C. What temp. should the shaft be cooled such that the ring will fit over the shaft.

Homework Equations

/\L = Coefficient of Lin. Expansion (Original Length) /\T

The Attempt at a Solution

step 1) .07 - .07003 = 2.4 x 10^-5 (.07003) (5 - Tf)

* The original length is .07003 since is what is being cooled to eventually reach .07 m. Also, the Coefficient of Lin. Expansion for Aluminum is 2.4 x 10^ -5

step 2) -3.0 x 10^-5 = 1.681 x 10^-6 (5 - Tf)

step 3) -3.0 x 10^-5 = 8.405 x 10^-6 - 1.681 x 10^-6 Tf


now when i solve for Tf, it is not coming out to -12.85 deg. C, which the book states is the answer...so I'm assuming there is an error somewhere, or a few places...any help would be appreciated , thanks

 

[AlephZero]

Your step 3 is OK. Check you didn't make a mistake with the signs or the exponents. Multiplying everything by 10^6 might help.

-30 = 8.405 - 1.681 Tf

EDIT:

Oops, P3X-018 is right. I made two mistakes which canceled out and gave me the right answer. Need more coffee...

 

[P3X-018]

I assume you get a reversed sign? When you calculate the temperature difference you use "initial temperature - final temperature" you need to use the same order about the difference in the diameter, that is "inital diameter - final diameter". You do it reversed for the length, what is the initial diameter?

 

[mohanancsp]

Linear expansion

consider
for variation in lenth Delta L for any variation in temp Delta T is

Delt a L = Coeff of Linear expansion * Original Length * Delta T
or
Delta T = Delta L / ( C of L exp * Original Length)
in the present case


delta L = 0.07003 - 0.07000 = 0.00003
then delta T = 0.00003 / ( 2.4 * 10 ^-5 *0.07003)
= ( 3 * 10^5)/ (2.4 * 7003)
= 300000/(1687.02)

=17.849
since delta L is negative, delta T also will be negative , which means
delta T = - 17.849
or -17.85 = final temp -original temp = Ft - 5 or

FT= -12.85
hope this is clear