A metal rod has initial length L0 = 2.50 m and a linear expansion coefficient alpha = 2.5 × 10^-5 per °C. If the temperature increases by delta T = 75 °C, what is the change in length delta L?

A metal rod has initial length L0 = 2.50 m and a linear expansion coefficient alpha = 2.5 × 10^-5 per °C. If the temperature increases by delta T = 75 °C, what is the change in length delta L?

• A. 0.00469 m
• B. 0.000469 m
• C. 0.046875 m
• D. 0.0470 m

Answer: (A)

When a metal heats up, its length increases in proportion to how much the temperature changes. The change in length is given by ΔL = α L0 ΔT. Here, α = 2.5 × 10^-5 per °C, L0 = 2.50 m, and ΔT = 75 °C. Compute αΔT: (2.5 × 10^-5)(75) = 1.875 × 10^-3. Then ΔL = (1.875 × 10^-3)(2.50 m) = 4.6875 × 10^-3 m, which rounds to 0.00469 m. So the rod lengthens by about 0.00469 m (4.69 mm).

When a metal heats up, its length increases in proportion to how much the temperature changes. The change in length is given by ΔL = α L0 ΔT. Here, α = 2.5 × 10^-5 per °C, L0 = 2.50 m, and ΔT = 75 °C.

Compute αΔT: (2.5 × 10^-5)(75) = 1.875 × 10^-3. Then ΔL = (1.875 × 10^-3)(2.50 m) = 4.6875 × 10^-3 m, which rounds to 0.00469 m.

So the rod lengthens by about 0.00469 m (4.69 mm).