At 20∘C, the hole in an aluminum ring is 2.800 cm in diameter. You need to slip this ring over a steel shaft that has a room-temperature diameter of 2.804 cm . 1. To what common temperature should the ring and the shaft be heated so that the ring will just fit onto the shaft? Coefficients of linear thermal expansion of steel and aluminum are 12×10−6 K−1 and23×10−6 K−1 respectively.

Respuesta :

Answer:

Explanation:

 To measure increase in length due to thermal expansion,  the formula is

L₂ =L₁ ( 1 + α t ) where L₁ length increases to L₁ due to rise in temperature by t.

L₂ in both the case is same so

for aluminium

L₂ = 2.8 ( 1 + 23 X 10⁻⁶ t )

for steel

L₂ = 2.804 ( 1 + 12 X 10⁻⁶ t )

2.8 ( 1 + .000023t) = 2.804 ( 1 + .000012 t )

Solving this equation

t = 122C

Final temperature of both = 122 +20 = 142 C

Answer:

Temperature on the shaft [tex]142 C[/tex]

Explanation:

Given:

Diameter of the ring [tex]2.8cm[/tex]

Temperature [tex]20^{o} c[/tex]

To measure increase in length due to thermal expansion, the formula is

[tex]$L_{2}=L_{1}(1+\alpha t)$[/tex] where [tex]$L_{1}$[/tex] length increases to [tex]$L_{1}$[/tex] due to rise in temperature by t.

[tex]$\mathrm{L}_{2}$[/tex] in both the case is same so

for aluminium

[tex]$\mathrm{L}_{2}=2.8\left(1+23 \times 10^{-6} \mathrm{t}\right)$[/tex]

for steel

[tex]$L_{2}=2.804\left(1+12 \times 10^{-6} t\right)$[/tex]

[tex]$2.8(1+.000023 t)=2.804(1+.000012 t)$[/tex]

Solving this equation

[tex]$\mathrm{t}=122 \mathrm{C}$[/tex]

Final temperature of both [tex]$=122+20=142 \mathrm{C}$[/tex]

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