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To what temperature (in °C) must a cylindrical rod of one metal 10.034 mm in diameter and a plate of second metal having a circular hole 9.984 mm in diameter have to be heated for the rod to just fit into the hole? Assume that the initial temperature is 28°C and that the linear expansion coefficient values for metals one and two are 3.5 x 10-6 (°C)-1 and 17 x 10-6 (°C)-1, respectively.

User Kmort
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1 Answer

6 votes

Answer:


T_f=399.446\ ^(\circ)C

Step-by-step explanation:

Given:

  • diameter of a rod,
    d_r=10.034\ mm
  • diameter of hole on the plate,
    d_h=9.984\ mm
  • initial temperature,
    T_i=28^(\circ)C
  • coefficient of linear expansion of rod,
    \alpha_r=3.5* 10^(-6)\ ^(\circ)C^(-1)
  • coefficient of linear expansion of plate,
    \alpha_p=17* 10^(-6)\ ^(\circ)C^(-1)

Now using the equation of linear expansion:


\pi.d_r+\pi.d_r.\alpha_r.(T_f-T_i)=\pi.d_h+\pi.d_h.\alpha_p.(T_f-T_i)


\pi* 10.034 +\pi* 10.034 * (3.5* 10^(-6))* (T_f-28)=\pi* 9.984 +\pi* 9.984 * (17* 10^(-6))* (T_f-28)


T_f=399.446\ ^(\circ)C

User Zappee
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7.6k points