A ball and a thin plate are made from different materials andhave the same initi

Question

A ball and a thin plate are made from different materials andhave the same initial temperature. The ball does not fit through ahole in the plate, because the diameter of the ball is slightlylarger than the diameter of the hole. However, the ball will passthrough the hole when the ball and the plate are both heated to acommon higher temperature. In each of the arrangements in thedrawing the diameter of the ball is 1.0 × 10-5 mlarger than the diameter of the hole in the thin plate, which has adiameter of 0.11 m. The initial temperature of each arrangement is23.0 °C. At what temperature will the ball fall through thehole in each arrangement?

Explanation / Answer

The idea for each arrangement is that: .     expansion of diameter of hole - expansion of diameter of sphere = 1.0 x10-5 .              D 1 T    - D2 T    = 1.0 x10-5 . Or...        T = 1.0 x 10-5 / D(1 - 2)     . If we express alphas as    x 10-6C-1    then we get: .          T = 1.0 x 10-5 / 0.11 *(1 - 2 ) x10-6   =    90.90909 / (1 - 2 ) . Now we just plug in numbers and get T for eachpair: . (a)   gold, alpha is 14    leadis 29 .      T =       90.90909 / (29- 14 )   =    6.061   sothe final temp is   23 + 6.061 =    29.061 degC . (b)   steel, alpha is 11   aluminum is 24 .      T =       90.90909 / (24- 11 )   =    6.993   sothe final temp is   23 + 6.993 =    29.993 degC . (a)   silver, alpha is 18   quartz is 0.59 .      T =       90.90909 / (18- 0.59 )   =    4.938 so the finaltemp is   23 + 4.938 =    27.938 degC . Or...        T = 1.0 x 10-5 / D(1 - 2)     . If we express alphas as    x 10-6C-1    then we get: .          T = 1.0 x 10-5 / 0.11 *(1 - 2 ) x10-6   =    90.90909 / (1 - 2 ) . Now we just plug in numbers and get T for eachpair: . (a)   gold, alpha is 14    leadis 29 .      T =       90.90909 / (29- 14 )   =    6.061   sothe final temp is   23 + 6.061 =    29.061 degC . (b)   steel, alpha is 11   aluminum is 24 .      T =       90.90909 / (24- 11 )   =    6.993   sothe final temp is   23 + 6.993 =    29.993 degC . (a)   silver, alpha is 18   quartz is 0.59 .      T =       90.90909 / (18- 0.59 )   =    4.938 so the finaltemp is   23 + 4.938 =    27.938 degC (b)   steel, alpha is 11   aluminum is 24 .      T =       90.90909 / (24- 11 )   =    6.993   sothe final temp is   23 + 6.993 =    29.993 degC . (a)   silver, alpha is 18   quartz is 0.59 .      T =       90.90909 / (18- 0.59 )   =    4.938 so the finaltemp is   23 + 4.938 =    27.938 degC (a)   silver, alpha is 18   quartz is 0.59 .      T =       90.90909 / (18- 0.59 )   =    4.938 so the finaltemp is   23 + 4.938 =    27.938 degC

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