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meerkat18
In physics, certain metals elongate when it is heated. This is a consequence of the expansion of the molecules present in a metal tube, for example. This elongation is described by the equation:

ΔL = L0*α*ΔT, where

ΔL is the elongation. In other words, this is the difference between the original length and the elongated length.
L0 is the original leng
α is the coefficient of linear expansion. This is an empirical data for specific kind of materials. For brass, α = 18.9 x 10^6/°C
ΔT is the change in temperature

Rearranging the equation,

ΔL/L0 = α*ΔT, where ΔL/L0 is the percentage of length expansion which is equal to 0.018 (1.8^%)

0.018 = (18.9 x 10^-6)(T-30)
T = 982.4°C

We have to heat a brass rod at 982.4[tex]\rm ^\circ C[/tex] for it to be 1.8 % longer than it is at 30.

Given :

Brass rod is 1.8 % longer than it is at 30[tex]\rm ^\circ C[/tex].

Solution :

We know that,

[tex]\rm \dfrac{\Delta L}{L_0}=\alpha \Delta T[/tex]

[tex]\rm \dfrac{\Delta L}{L_0}=\alpha ( T_2-T_1)[/tex]      ---- (1)

Where, [tex]\rm \Delta L[/tex] is the elongation,

[tex]L_0[/tex] is the original length,

[tex]\alpha[/tex] is the coefficient of linear expansion. For brass,

[tex]\rm \alpha = 18.9\times 10^-^6/^\circ C[/tex],

[tex]\rm \Delta T[/tex] is the change in temperature.

1.8 % length expansion means:

[tex]\rm \dfrac{\Delta L}{L_0} = 0.018[/tex]

Now put the values of

[tex]\rm \alpha ,\;\dfrac{\Delta L}{L_0},\;and\;T_1[/tex]     in equation (1) we get:

[tex]\rm 0.018 = 18.9\times 10^-^6 \times(T_2 - 30)[/tex]

[tex]\rm T_2 = 982.4\; ^\circ C[/tex]

We have to heat a brass rod at 982.4[tex]\rm ^\circ C[/tex] for it to be 1.8 % longer than it is at 30.

For more information, refer the link given below

https://brainly.com/question/852985

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