To what temperature would you have to heat a brass rod for it to be 1.8 % longer than it is at 30 ∘c?

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

Answer8.0m;