Answer:
T' = 865.15 °C
Explanation:
In order for the copper collar to just slip on the steel shaft the, assuming are heated simultaneously, we must find the final parameters of both and equate them. Because the final diameters of both must be same for the slipping to occur.
FOR COPPER COLLAR:
dc' = dc(1 + ∝c*ΔT)
where,
dc' = final diameter of copper ring
dc = initial diameter of copper ring = 5.98 cm
∝c = coefficient of linear expansion for copper = 16 x 10⁻⁶ °C⁻¹
ΔT = Change in Temperature
Therefore,
dc' = (5.98 cm)[1 + (16 x 10⁻⁶ °C⁻¹)ΔT] ------------- equation (1)
FOR STEEL SHAFT:
ds' = ds(1 + ∝s*ΔT)
where,
ds' = final diameter of steel shaft
ds = initial diameter of steel shaft = 6 cm
∝s = coefficient of linear expansion for steel = 12 x 10⁻⁶ °C⁻¹
ΔT = Change in Temperature
Therefore,
dc' = (6 cm)[1 + (12 x 10⁻⁶ °C⁻¹)ΔT] ------------- equation (2)
Comparing equation (1) and equation (2):
(5.98 cm)[1 + (16 x 10⁻⁶ °C⁻¹)ΔT] = (6 cm)[1 + (12 x 10⁻⁶ °C⁻¹)ΔT]
(5.98 cm/6 cm)[1 + (16 x 10⁻⁶ °C⁻¹)ΔT] = [1 + (12 x 10⁻⁶ °C⁻¹)ΔT]
0.9967 + (1.59 x 10⁻⁵ °C⁻¹)ΔT = 1 + (12 x 10⁻⁶ °C⁻¹)ΔT
1 - 0.9967 = [(15.9 -12) x 10⁻⁶ °C⁻¹]ΔT
0.0033/3.9 x 10⁻⁶ °C⁻¹ = ΔT
ΔT = 846.15 °C
but,
ΔT = T' - T = T' - 19°C = 846.15°C
T' = 846.15 °C + 19 °C
T' = 865.15 °C
