A brass plug is to be placed in a ring made of iron. At room temperature, the diameter of the plug is 8.753 cm and that of the inside of the ring is 8.743 cm. They must both be brought to what common temperature in order to fit?

Apr 9, 2016

If you start at 20 °C, they must both be cooled to -150 °C.

Explanation:

Some observations:

1. The brass plug is larger than the ring.
2. The coefficient of thermal expansion of brass is greater than that of iron.
3. We will have to cool both the plug and the ring for them to fit.

The formula for linear expansion is

color(blue)(|bar(ul(color(white)(a/a) l = l_0 + l_0αΔTcolor(white)(a/a)|)))" "

where

• $l$ is the linear dimension (diameter) after the temperature change
• ${l}_{0}$ is the original linear dimension
• α is the coefficient of thermal expansion for the material
• ΔT is the change in temperature.

At some temperature, ${l}_{\text{brass" = l_"iron}}$

l = underbrace( l_0 + l_0αΔT)_color(red)("brass") = underbrace( l + l_0 + l_0αΔT)_color(red)("iron")

8.753 color(red)(cancel(color(black)("cm"))) + 8.753 color(red)(cancel(color(black)("cm"))) × 18.7 × 10^"-6" "°C"^"-1" × ΔT
= 8.743 color(red)(cancel(color(black)("cm"))) + 8.743 color(red)(cancel(color(black)("cm"))) × 12.0 × 10^"-6" "°C"^"-1" × ΔT

8.753 + 1.636 × 10^"-4"color(white)(l)"°C"^"-1"ΔT = 8.743 + 1.049× 10^"-4"color(white)(l)"°C"^"-1" ΔT

5.87 × 10^"-5"color(white)(l)"°C"^"-1"ΔT = "-0.010"

ΔT = -0.010/(5.87 × 10^"-5"color(white)(l)"°C"^"-1") = "-170 °C"

Both brass and ring must be cooled by 170°C.

If you start at 20 °C, you must cool them to -150 °C.