A brass ring of diameter 10.00 cm at 18.5°C is heated and slipped over an alumin

Question

A brass ring of diameter 10.00 cm at 18.5°C is heated and slipped over an aluminum rod of diameter 10.01 cm at 18.5°C. Assume the average coefficients of linear expansion are constant.
(a) To what temperature must the combination be cooled to separate the two metals?
1-178 -181
°C
(b) Is that temperature attainable?

2
Yes
No
.

(c) What if the aluminum rod were 8.99 cm in diameter?
317200 39200

Your response differs from the correct answer by more than 10%. Double check your calculations.°C
(d) Is that temperature attainable?

Explanation / Answer

   As per given data , it is obivious that ,     expansion of brass must be  0.01 cm   greater than expansion of aluminum    Thus, it is known by the formual for expansion ,     we have ,        L T [ brass] = L T    [ aluminum ]    +   0.01      but , brass =  19 x 10-6 / o C ,            aluminium =  24 x 10-6 / o C ,              where T is the temperature must the      combination be cooled to separate the two metals       10.00 x 19 x 10-6 *T   =   10.01 x 24 x 10-6 *T    + 0.01          0.000190T      =   0.00024024 T     + 0.01       - 0.00005024T    =   0.01                             T =   -199.0 thus, final temp wouldbe   :    18.5 - 199.0      =   -180.5 deg C on Aprroxomination : it is 181 (2 )   Yes (3) If we did the same thing for the new length ofaluminum...        L T for brass = L T    foraluminum     +         10.00 * 19 x 10-6 *T   =   8.99 * 24 x 10-6 *T    +   (10- 8.99)                0.000190T      =   0.00021576 T     + 1.01              -0.00002576 T    =   0.96                                    T = - 37267.08 oC (4 )   No                            expansion of brass must be  0.01 cm   greater than expansion of aluminum    Thus, it is known by the formual for expansion ,     we have ,        L T [ brass] = L T    [ aluminum ]    +   0.01      but , brass =  19 x 10-6 / o C ,            aluminium =  24 x 10-6 / o C ,              where T is the temperature must the      combination be cooled to separate the two metals       10.00 x 19 x 10-6 *T   =   10.01 x 24 x 10-6 *T    + 0.01          0.000190T      =   0.00024024 T     + 0.01       - 0.00005024T    =   0.01                             T =   -199.0 thus, final temp wouldbe   :    18.5 - 199.0      =   -180.5 deg C on Aprroxomination : it is 181 (2 )   Yes (3) on Aprroxomination : it is 181 (2 )   Yes (3) If we did the same thing for the new length ofaluminum...        L T for brass = L T    foraluminum     +         10.00 * 19 x 10-6 *T   =   8.99 * 24 x 10-6 *T    +   (10- 8.99)                0.000190T      =   0.00021576 T     + 1.01              -0.00002576 T    =   0.96                                    T = - 37267.08 oC (4 )   No                        (4 )   No      

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