Question

When the temperature of a thin silver [α = 19 × 10-6 (C°)-1] rod is increased, the length of the rod increases by 1.9 × 10-3 cm. Another rod is identical in all respects, except that it is made from gold [α = 14 × 10-6 (C°)-1]. By how much ΔL does the length of the gold rod increase when its temperature increases by the same amount as that for the silver rod?

Answer 1

Answer on Question #37958, Physics, Other

Question:

When the temperature of a thin silver $[\alpha = 19 \times 10^{-6} (C{}^{\circ})^{-1}]$ rod is increased, the length of the rod increases by $1.9 \times 10^{-3} \mathrm{~cm}$. Another rod is identical in all respects, except that it is made from gold $[\alpha = 14 \times 10^{-6} (C{}^{\circ})^{-1}]$. By how much $\Delta L$ does the length of the gold rod increase when its temperature increases by the same amount as that for the silver rod?

Answer:

The change in the units' length when temperature change can be expressed as:

$$\Delta l = l _ {0} \alpha \Delta T$$

where $l_{0}$ is initial length, $\alpha$ is linear expansion coefficient, $\Delta T$ is change of temperature.

The change of length for silver rod equals:

$$\Delta l _ {s} = l _ {0} \alpha_ {s} \Delta T$$

The change of length for gold rod equals:

$$\Delta l _ {g} = l _ {0} \alpha_ {g} \Delta T$$

Therefore:

$$\frac {\Delta l _ {s}}{\Delta l _ {g}} = \frac {\alpha_ {s}}{\alpha_ {g}}$$ $$\Delta l _ {g} = \Delta l _ {s} \frac {\alpha_ {g}}{\alpha_ {s}} = 1.4 * 10 ^ {- 3} \mathrm{~cm}$$

Answer: $1.4 * 10^{-3} \mathrm{~cm}$