Answer:
a) 87 C
b) -94 C
Explanation:
The linear thermal expansion equation is:
D1 = D0 * (1 + a*(t1 - to))
Where
D1: diameter after heating
D: diameter before heating
a: thermal dilation coefficient 12e-6/K for steel
t1: heating temperature
t0: ambient temperature (20 C is considered standard)
Then
1 + a*(t1 - t0) = D1/D0
a * (t1 - t0) = D1/D0 - 1
t1 - t0 = ((D1/D0) - 1)/a
t1 = ((D1/D0) - 1)/a + t0
t1 = ((2.502/2.5) - 1)/12e-6 + 20 = 87 C
The thermal dilation coefficient for brass is: 19e-6/K
For the steel ring to slip off the brass it must have the same diameter
Dsteel = Dbrass
D0s * (1 + as*(t1 - t0)) = D0b * (1 + ab*(t1 - t0))
D0s + D0s*as*t1 - D0s*as*t0 = D0b + D0b*ab*t1 - D0b*ab*t0
D0s*as*t1 - D0b*ab*t1 = D0b * (1 - ab*t0) - D0s * (1 - as*t0)
t1 * (D0s*as - D0b*ab) = D0b * (1 - ab*t0) - D0s * (1 - as*t0)
t1 = (D0b * (1 - ab*t0) - D0s * (1 - as*t0))/(D0s*as - D0b*ab)
t1 = (2.502 * (1 - 19e-6*20) - 2.5 * (1 - 12e-6*20))/(2.5*12e-6 - 2.502*19e-6) = -94 C
In this exercise we have to use the knowledge of the coefficient of expansion to calculate the values, like this:
a)The temperature should be is 87 C
b)The temperature will be -94 C
The linear thermal expansion equation is:
[tex]D_1 = D_0 * (1 + a*(t_1 - t_0))[/tex]
Where:
Then we have:
[tex]1 + a*(t_1 - t_0) = D_1/D_0\\a * (t_1 - t_0) = D_1/D_0 - 1\\t_1 - t_0 = ((D_1/D_0) - 1)/a\\t_1 = ((D_1/D_0) - 1)/a + t_0\\t_1 = ((2.502/2.5) - 1)/12e^{-6} + 20 = 87 C[/tex]
For the steel ring to slip off the brass it must have the same diameter
[tex]t_1 * (D0s*as - D0b*ab) = D0b * (1 - ab*t_0) - D0s * (1 - as*t_0)\\t1 = (D0b * (1 - ab*t_0) - D0s * (1 - as*t_0))/(D0s*as - D0b*ab)\\t1 = (2.502 * (1 - 19e-6*20) - 2.5 * (1 - 12e-6*20))/(2.5*12e-6 - 2.502*19e-6) \\= -94 C[/tex]
See more about thermal expanion at brainly.com/question/3022807
