Answer to Question #209839 in Molecular Physics | Thermodynamics for Nkanyezi

Question

A  brass  plug has  a  diameter  of 10,000 cm  at  20°C.  To what  temperature  must  it  be  cooled in order to fit  into  a  hole  A  brass  plug has  a  diameter  of 10,000 cm  at  20°C.  To what  temperature  must  it  be  cooled in order to fit  into  a  hole  with a  diameter  of 9,997 cm?  (α  =19×  10–  6)

Answer 1

By the definition of the linear thermal expansion we have:

$$\Delta L = \alpha L_0 \Delta T,$$

here, $\Delta L = L - L_0$ is the increace in diameter of the brass plug when it is heated, $L_0 = 10 \ m$ is the initial diameter of the brass plug at some temperature $T_0$, $L = 10 \ m$ is the diameter of the brass plug at $T = 20 \ ^{\circ}C$, $\alpha = 19 \cdot 10^{-6} \ ^{\circ}C^{-1}$ is the linear expansion coefficient for steel, $\Delta T = T - T_0$ is the change in temperature.

From this formula we can find the change in temperature:

$$\Delta T = \dfrac{L - L_0}{\alpha L_0},$$ $$\Delta T = \dfrac{10 \ m - 9.997 \ m}{19 \cdot 10^{-6} \ ^{\circ}C^{-1} \cdot 9.997\ m} = 15.79 \ ^{\circ}C.$$

Finally, we can calculate the temperature at which the brass plug will fit exactly into a hole of constant diameter:

$$T_0 = T - \Delta T = 20 \ ^{\circ}C - 15.79 \ ^{\circ}C = 4.21 \ ^{\circ}C.$$

Answer:

$T_0 = 4.21 \ ^{\circ}C.$