i need some help in this questions

Question

A machinist wishes to insert a brass rod with a diameter of 7 mm into a hole with a diameter of 6.998 mm. By how much would the machinist have to lower the temperature (in °C) of the rod to make it fit the hole?

Answer 1

Note that the Coefficient Of Linear Expansion For Brass is α equal to 1.93×10^-5°C^-1.

Use the formula L_cooled-rod − L₀ = αL₀(T_cooled-rod − T₀).

Here, take Room Temperature as T₀ = 23°C. Take Initial Rod Diameter or L₀ as 0.007.

Take L_cooled-rod as 0.006998.

Place values to obtain (0.006998 − 0.007) = [(1.93×10^-5) × (0.007)](T_cooled-rod − 23) which gives

T_cooled-rod as (0.006998 − 0.007) divided by [(1.93×10^-5) × (0.007)] plus 23 equal to 8.196151°C.

(8.196151 − 23) or -14.803849°C is then the drop in rod temperature sought.