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

Answer:

834°C

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