Question

When the temperature of a thin silver [α = 19 × 10-6 (C°)-1] rod is increased, the length of the rod increases by 3.1 × 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#37928, 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 $3.1 \times 10^{-3} \mathrm{~cm}$. 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 increasement of length of the rod is given by formula

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

where $\alpha$ – is linear expansion coefficient of the rod, $L_{0}$ – initial length of the rod, $\Delta T$ – increasement of temperature of the rod.

The initial length and the increasement of temperature of the silver and the gold rods are the same, so we can write an equation

$$\frac {\Delta L _ {s}}{\alpha_ {s}} = \frac {\Delta L _ {g}}{\alpha_ {g}}$$

where $\Delta L_{s}, \alpha_{s}$ – the increasement of length and the linear expansion coefficient of the silver rod and $\Delta L_{g}, \alpha_{g}$ – the increasement of length and the linear expansion coefficient of the gold rod.

So we can find $\Delta L_{g}$

$$\Delta L _ {g} = \frac {\alpha_ {g}}{\alpha_ {s}} \Delta L _ {s}$$ $$\Delta L _ {g} = \frac {1 4 \times 1 0 ^ {- 6}}{1 9 \times 1 0 ^ {- 6}} \cdot 3. 1 \times 1 0 ^ {- 3} = 2. 3 \times 1 0 ^ {- 3} c m$$

The answer is: $\Delta \mathbf{L}_{\mathbf{g}} = 2.3\times 10^{-3}\mathbf{cm}$