A ball and a thin plate are made from different materials and have the same init

Question

A ball and a thin plate are made from different materials and have the same initial temperature. The ball does not fit through a hole in the plate, because the diameter of the ball is slightly larger than the diameter of the hole. However, the ball will pass through the hole when the ball and the plate are both heated to a common higher temperature. In each of the arrangements in the drawing the diameter of the ball is 1.0

A ball and a thin plate are made from different materials and have the same initial temperature. The ball does not fit through a hole in the plate, because the diameter of the ball is slightly larger than the diameter of the hole. However, the ball will pass through the hole when the ball and the plate are both heated to a common higher temperature. In each of the arrangements in the drawing the diameter of the ball is 1.0 ? 10-5 m larger than the diameter of the hole in the thin plate, which has a diameter of 0.10 m. The initial temperature of each arrangement is 21.0 °C. At what temperature will the ball fall through the hole in each arrangement?

Explanation / Answer

The idea for each arrangement is that:

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    expansion of diameter of hole - expansion of diameter of sphere = 1.0 x10-5

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             D ?1 ?T    - D?2 ?T    = 1.0 x10-5

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Or...        ?T = 1.0 x 10-5 / D(?1 - ?2)    

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If we express alphas as    x 10-6C-1    then we get:

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         ?T = 1.0 x 10-5 / 0.10 *(?1 - ?2 ) x10-6   = 100 / (?1 - ?2 )

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Now we just plug in numbers and get ?T for eachpair:

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(a)   gold, alpha is 14    lead is 29

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     ?T = 100 / (29- 14 )   =    6.67 sothe final temp is   21 + 6.67 =    27.67 degC

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(b)   steel, alpha is 11   aluminum is 24

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     ?T = 100 / (24- 11 )   = 7.692 sothe final temp is   21 + 7.692 =    28.692 degC

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(a)   silver, alpha is 18   quartz is 0.59

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     ?T = 100 / (18- 0.59 )   = 5.744 so the finaltemp is   21 + 5.744 =    26.744 degC

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