Asteel ring with a 2.5000-in. inside diameter at 20.0 ∘c is to be warmed and slipped over a brass shaft with a 2.5020 in. outside diameter at 20.0 ∘c.
(a) to what temperature should the ring be warmed?
(b) if the ring and the shaft together are cooled by some means such as liquid air, at what temperature will the ring just slip off the shaft?

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

i mean i gues

explanation: