Answer:
[tex]\theta_{2} = 15400^0 C[/tex]
Explanation:
The formula for linear expansivity, [tex]\alpha = \frac{l_{2} - l_{1} }{l_{1} ( \theta_{2} - \theta_{1} )}[/tex]
original length, l₁ = 123 cm = 1.23 m
final length, l₁ = 92.6 cm =0.926 m
original temperature, θ₁ = 200°C
Linear expansivity, α = 2 * 10⁻⁵ °C⁻¹
Putting these values into the formula:
[tex]2 * 10^{-5} = \frac{1.23 - 0.926 }{l_{1} ( \theta_{2} -200 )}\\ \theta_{2} -200 = \frac{0.304}{2 * 10^{-5} } \\\theta_{2} = 15200 + 200\\\theta_{2} = 15400^0 C[/tex]
