Answer to Question #84297 in Molecular Physics | Thermodynamics for Pearl Valencia

Question

A copper rod ( α = 0.000017 /°C ) is 20 cm longer than an aluminum rod ( α = 0.000022 /°C ). How long should the copper rod be if the difference in their lengths is to be independent of temperature? Hint: For their difference in length not to change with temperature, ΔL must be the same for both rods under the same temperature change.

Answer 1

By the definition of the linear thermal expansion we have:

$$\Delta L_{copper} = \alpha_{copper} L_{0, copper} \Delta T_{copper},$$ $$\Delta L_{aluminum} = \alpha_{aluminum} L_{0, aluminum} \Delta T_{aluminum},$$

here,

$$\Delta L_{copper}$$

,

$$\Delta L_{aluminum}$$

are the difference in the lengths of the copper and aluminum rods after the change in the temperature, respectively;

$$L_{0, copper}$$

,

$$L_{0, aluminum}$$

are the length of the copper and aluminum rods before the change in the temperature, respectively;

$$\alpha_{copper}$$

,

$$\alpha_{aluminum}$$

are the coefficients of linear expansion for the copper and aluminum rods, respectively;

$$\Delta T_{copper}$$

,

$$\Delta T_{aluminum}$$

are the change in temperature, respectively.

From the definition of the question we know that the difference in the lengths of the rods is independent of temperature. Therefore, we can write:

$$\Delta T_{copper} = \Delta T_{aluminum},$$ $$\Delta L_{copper} = \Delta L_{aluminum}.$$

Also, we know that the copper rod is 20 cm longer than the aluminum rod. So, we can write the expression for the length of the aluminum rod:

$$L_{0, aluminum} = L_{0, copper} - 0.2 m.$$

Finally, we can substitute this expression into the previous equation and find the length of the copper rod:

$$\alpha_{copper} L_{0, copper} \Delta T_{copper} = \alpha_{aluminum} ( L_{0, copper} - 0.2 m) \Delta T_{aluminum},$$ $$\alpha_{copper} L_{0, copper} = \alpha_{aluminum} ( L_{0, copper} - 0.2 m),$$ $$L_{0, copper} = \frac{0.2 m \cdot \alpha_{aluminum}}{(\alpha_{aluminum} - \alpha_{copper})}.$$

Let's substitute the numbers:

$$L_{0, copper} = \frac{0.2 m \cdot 2.20 \cdot 10^{-5} ℃^{-1}}{(2.20 \cdot 10^{-5} ℃^{-1} - 1.70 \cdot 10^{-5} ℃^{-1})} = 0.88 m.$$

Answer:

$$L_{0, copper} = 0.88 m$$

.