Question

To what temperature (in °C) must a cylindrical rod of one metal 10.083 mm in diameter and a plate of second metal having a circular hole 9.987 mm in diameter have to be heated for the rod to just fit into the hole? Assume that the initial temperature is 27°C and that the linear expansion coefficient values for metals one and two are 4.0 x 10-6 (°C)-1 and 16 x 10-6 (°C)-1, respectively.

Answer 1

834°C

Step-by-step explanation:

By setting the equation of final diameter of one metal and another metal equal to one another in order to determine final temperature, the equation for final diameter is given by,

`d_(f)` =
`d_{o`(1 + α(
`t_(f)`-
`t_{o`))

`d_{f1`=
`d_{f2`

9.987(1 + 16 x
`10^(-6)`(
`t_(f)`-27)) = 10.083( 1 + (4 x
`10^(-6)`)(
`t_(f)`-27))

9.987(1 + 16 x
`10^(-6)`
`t_(f)` - 432 x
`10^(-6)`) = 10.083( 1 + 4 x
`10^(-6)`
`t_(f)` - 108 x
`10^(-6)`)

9.987 + 1.59 x
`10^{-4`
`t_(f)` - 4.31 x
`10^{-3` = 10.083 + 4.033 x
`10^{-5`
`t_(f)` - 1.088 x
`10^{-3`

1.59 x
`10^{-4`
`t_(f)` - 4.033 x
`10^{-5`
`t_(f)` = 10.083-1.088 x
`10^{-3`- 9.987+4.31 x
`10^{-3`

1.187 x
`10^{-4`
`t_(f)`= 0.099

`t_(f)` = 0.099/ 1.187 x
`10^{-4`

`t_(f)` = 834°C