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 × 10-5 m larger than the diameter of the hole in the thin plate, which has a diameter of 0.12 m. The initial temperature of each arrangement is 23.0 °C. At what temperature will the ball fall through the hole in each arrangement?

Arrangement A) Gold ball through a lead plate
Coefficient of Linear expansion:
Gold: 14x10^-6
Lead: 29x10^6
If you just show me how to do this first arrangement i'm sure i can figure out the other two. I think im just missing a step. Please help !

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 / 0.12 *(1 - 2 ) x10-6   = 83.3333325 / (1 - 2 ) . Now we just plug in numbers and get T for eachpair: . (a)   gold, alpha is 14    leadis 29 .      T =       83.3333325 / (29- 14 )   =   5.55555 sothe final temp is   23 +5.55555 =    28.55555 degC . (b)   steel, alpha is 11   aluminum is 24 .      T =       83.3333325 / (24- 11 )   =    6.4102 sothe final temp is   23 +6.4102 =    29.4102 degC . (a)   silver, alpha is 18   quartz is 0.59 .      T =       83.3333325 / (18- 0.59 )   =    4.7865 so the finaltemp is   23 + 4.7865 =   27.7865 degC 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 / 0.12 *(1 - 2 ) x10-6   = 83.3333325 / (1 - 2 ) . Now we just plug in numbers and get T for eachpair: . (a)   gold, alpha is 14    leadis 29 .      T =       83.3333325 / (29- 14 )   =   5.55555 sothe final temp is   23 +5.55555 =    28.55555 degC . (b)   steel, alpha is 11   aluminum is 24 .      T =       83.3333325 / (24- 11 )   =    6.4102 sothe final temp is   23 +6.4102 =    29.4102 degC . (a)   silver, alpha is 18   quartz is 0.59 .      T =       83.3333325 / (18- 0.59 )   =    4.7865 so the finaltemp is   23 + 4.7865 =   27.7865 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 / 0.12 *(1 - 2 ) x10-6   = 83.3333325 / (1 - 2 ) . Now we just plug in numbers and get T for eachpair: . (a)   gold, alpha is 14    leadis 29 .      T =       83.3333325 / (29- 14 )   =   5.55555 sothe final temp is   23 +5.55555 =    28.55555 degC . (b)   steel, alpha is 11   aluminum is 24 .      T =       83.3333325 / (24- 11 )   =    6.4102 sothe final temp is   23 +6.4102 =    29.4102 degC . (a)   silver, alpha is 18   quartz is 0.59 .      T =       83.3333325 / (18- 0.59 )   =    4.7865 so the finaltemp is   23 + 4.7865 =   27.7865 degC (b)   steel, alpha is 11   aluminum is 24 .      T =       83.3333325 / (24- 11 )   =    6.4102 sothe final temp is   23 +6.4102 =    29.4102 degC . (a)   silver, alpha is 18   quartz is 0.59 .      T =       83.3333325 / (18- 0.59 )   =    4.7865 so the finaltemp is   23 + 4.7865 =   27.7865 degC (a)   silver, alpha is 18   quartz is 0.59 .      T =       83.3333325 / (18- 0.59 )   =    4.7865 so the finaltemp is   23 + 4.7865 =   27.7865 degC

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