+12 A steel rod is 2.585 cm in diameter at 32.00°C. A brass ring has an interior diameter of 2.583 cm at 32.00°C. At what common temperature will the ring just slide onto the rod? The linear expansion coefficient of steel is 11.00 × 10-6 1/C°. The linear expansion coefficient of brass is 19.00 × 10-6 1/C°.
The answer should be 128 deg Celcius. But how do we get it.
+33665 k = (1/L)*ΔL/ΔT
ΔT = (1/L)*ΔL/k
Brass ΔT = (1/2.583)*(D-2.583)/(19*10-6)
Steel ΔT = (1/2.585)*(D-2.585)/(11*10-6)
By dividing the Brass by the Steel we get:
1 = (2.585/2.583)*(D-2.583)/(D-2.585)*11/19
This gives D ≈ 2.588 cm
Plug this back in to one of the equations above to get ΔT ≈ 96.89degC
32°C + 96.89°C = 128.89°C ≈ 129°C
.
+33665 k = (1/L)*ΔL/ΔT
ΔT = (1/L)*ΔL/k
Brass ΔT = (1/2.583)*(D-2.583)/(19*10-6)
Steel ΔT = (1/2.585)*(D-2.585)/(11*10-6)
By dividing the Brass by the Steel we get:
1 = (2.585/2.583)*(D-2.583)/(D-2.585)*11/19
This gives D ≈ 2.588 cm
Plug this back in to one of the equations above to get ΔT ≈ 96.89degC
32°C + 96.89°C = 128.89°C ≈ 129°C
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