Thermal Expansion in an aluminum plate

[SoccaCrazy24]

Question 1:
A hole in an aluminum plate has a diameter or 1.178 cm at 23.00 degrees Celsius.
(A) What is the diameter of the hole at 199.0 degrees Celsius?
(B) What temperature is the diameter of the hole equal to 1.176 cm?

For (A) I used the equation Change in Area = 2 * alpha * A * Change in Temp.
alpha = 24e-6
A= pi * r^2 = 1.09e-4
Change in Temp = 176
So Change in Area = 2 * 24e-6 * 1.09e-4 * 176 = 9.207e-7
so change in Diameter = sqroot (A/pi) = 5.414e-4m
This doesn't seem right... am i using the right equation? and then i would use the same equation for (B)

Question 2:
At 12.25 degrees Celsius a brass sleeve has an inside diameter of 2.196 cm and a steel shaft has a diameter of 2.199 cm. It is desired to shrink-fit the sleeve over the steel shaft.
(A) To what temperature must the sleeve be heated in order for it to slip over the shaft?
(B) Alternatively, to what temperature must the shaft be cooled before it is able to slip through the sleeve?

If I use the same equation from question 1 shall i receive the same answer?

 

[SoccaCrazy24]

not to be repetitve or a forum whore? but does anyone have a clue?

 

[quicknote]

I would like to clarify that the equation you have used for thermal expansion is correct, however, you have made some errors in your calculations. Let me provide you with the correct solutions for both questions.

(A) To find the diameter of the hole at 199.0 degrees Celsius, we will use the equation you mentioned: Change in Area = 2 * alpha * A * Change in Temp. However, we need to make some adjustments to the values you have used.

Firstly, the value of alpha for aluminum is 23.1 x 10^-6, not 24 x 10^-6 as you have used.

Secondly, the area should be calculated using the initial diameter of the hole, which is 1.178 cm, not the area of a circle with a radius of 1.09 x 10^-4 as you have used.

So, the correct equation would be: Change in Area = 2 * 23.1 x 10^-6 * pi * (1.178/2)^2 * 176 = 1.771 x 10^-5 cm^2

Now, we can find the change in diameter using the equation: Change in Diameter = Change in Area/pi. This gives us a value of 5.641 x 10^-4 cm.

Therefore, the final diameter of the hole at 199.0 degrees Celsius would be: 1.178 + 5.641 x 10^-4 = 1.183 cm.

(B) To find the temperature at which the diameter of the hole is equal to 1.176 cm, we can use the same equation and solve for the change in temperature.

So, Change in Temp = (1.176 - 1.178)/(2 * 23.1 x 10^-6 * pi * (1.178/2)^2) = -0.091 degrees Celsius.

Thus, the temperature at which the diameter of the hole is equal to 1.176 cm would be: 23.00 - 0.091 = 22.909 degrees Celsius.

For Question 2, we will use the same equation, but with different values.

(A) To find the temperature at which the sleeve can slip over the shaft, we need to calculate the change in diameter of the sleeve.

Using the same equation as before, we get: Change in