A steel ball has a diameter of 60mm and is 0,01mm too large to pass through a hole in a brass plate when the ball and the plate are at 303k CALCULATE common temperature

Answer 1

To calculate the common temperature at which the steel ball would pass through the hole in the brass plate, we can use the coefficient of linear expansion formula:

ΔL = αΔT

Where:
ΔL = change in size of the steel ball
α = coefficient of linear expansion for steel (from table)
ΔT = change in temperature

Since the steel ball is 0.01mm too large to pass through the hole at 303K, we can say that ΔL = 0.01mm = 0.01 x 10^-3 meters.
The coefficient of linear expansion for steel is approximately 11 x 10^-6 K^-1.

Substitute the values into the formula:

0.01 x 10^-3 = (11 x 10^-6)ΔT
ΔT = (0.01 x 10^-3) / (11 x 10^-6)
ΔT = 0.909 K

Therefore, the common temperature at which the steel ball would pass through the hole in the brass plate is 303 K - 0.909 K = 302.091 K.