Question

An iron rod is 2.58m long at 0°C. Calculate the length of a brass rod at 0°C, if the difference between the lengths of the two rods must remain the same at all temperature. Linear expansivity of iron = 1.2 × 10^-5k^-1. Linear expansivity of brass = 1.9 × 10^-5k^-1

Answer 1

By the definition of the linear thermal expansion we have:

$$\Delta L_{iron} = \alpha_{iron} L_{0, iron} \Delta T_{iron},$$ $$\Delta L_{brass} = \alpha_{brass} L_{0, brass} \Delta T_{brass},$$

here, $\Delta L_{iron}$, $\Delta L_{brass}$ are the difference in the lengths of the copper and aluminum rods after the change in the temperature, respectively; $L_{0, iron}$, $L_{0, brass}$ are the length of the copper and aluminum rods before the change in the temperature (at 0°C), respectively; $\alpha_{iron}$, $\alpha_{brass}$ are the coefficients of linear expansion for the copper and aluminum rods, respectively; $\Delta T_{iron}$, $\Delta T_{brass}$ are the change in temperature, respectively.

Since the difference between the lengths of the two rods must remain the same at all temperature, we can write:

$$\Delta T_{iron} = \Delta T_{brass} = \Delta T,$$ $$\Delta L_{iron} = \Delta L_{brass}.$$

Then, we get:

$$\alpha_{iron} L_{0, iron} \Delta T = \alpha_{brass} L_{0, brass} \Delta T,$$ $$L_{0, brass} = \dfrac{\alpha_{iron} L_{0, iron}}{\alpha_{brass}} = \dfrac{1.2 \cdot 10^{-5} K^{-1} \cdot 2.58 m}{1.9 \cdot 10^{-5} K^{-1}} = 1.63 m.$$

Answer:

$L_{0, brass} = 1.63 m.$