A steel bridge originally 100 m long expands by 5 cm. Find the temperature change. The question omits α; use the chapter’s steel value 1.2×10⁻⁵ °C⁻¹.

Solution & explanation

1. Given: L₀ = 100 m; ΔL = 5 cm; assumed steel α = 1.2×10⁻⁵ °C⁻¹.

2. Convert 5 cm to metres before substitution.

   $${\Delta }_L=\frac{5}{100}=0.05$$

3. Start with ΔL = αL₀ΔT.

   $${\Delta }_L={\Delta }_T\times L_0\times \alpha$$

4. Divide by αL₀ to isolate the interval.

   $${\Delta }_T=\frac{{\Delta }_L}{L_0\times \alpha }$$

5. Insert the values.

   $${\Delta }_T=\frac{0.05}{\frac{120.0}{100000}}$$

6. Calculate denominator, then divide.

   $${\Delta }_T=\frac{0.05}{0.0012}=\frac{125}{3}$$

7. State the result with units: 41.67°C increase.

ΔT ≈ 41.67°C