Question
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
Answers
GPT 3.5
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.
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.