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Consider a 20 mm long brass wire aligned with a 10 mm long aluminium wire. The wires are...

Question:

Consider a 20 mm long brass wire aligned with a 10 mm long aluminium wire. The wires are separated by a distance of 30 mm at a temperature of 22°C. If the far ends of the wires are restrained, at what temperature will the two wires touch?

Take the following values to be the coefficients of linear expansion:

i) Brass = 19 x {eq}10^{-6} {/eq}

ii) Al = 22 x {eq}10^{-6} {/eq}

Linear Expansion:

When temperature of any substance is increased it expands. We assume the expansion to be directly proportional to the change in temperature and initial length for small changes in temperature. i.e {eq}\Delta l = kl\Delta T {/eq} where l is the initial length, {eq}\Delta l {/eq} is the increase in length, k is the coefficient of linear expansion and {eq}\Delta T {/eq} is the temperature change.

Answer and Explanation:

GIven:

Length of brass wire = 20m

Length of aluminium wire = 10m

Distance between them = 30mm

{eq}K_{brass} = 19 \times 10^{-6}\\ K_{Al} = 22 \times 10^{-6} {/eq}

Let the final temperature be T.

{eq}\Delta l_{brass} = K_{brass} l_{brass} \Delta T = 19 \times 10^{-6} \times 20 \times ( T - 22 ) \\ \Delta l_{Al} = K_{Al} l_{Al} \Delta T = 22 \times 10^{-6} \times 10 \times (T -22) {/eq}

For the wires to touch:

{eq}\Delta l_{brass} + \Delta l_{Al} = Initial ~Separation = 30 mm\\ => 19 \times 10^{-6} \times 20 \times ( T - 22 ) + 22 \times 10^{-6} \times 10 \times (T -22) = 30 \times 10^{-3}\\ => 0.0006 (T - 22) = 30 \times 10^{-3} \\ => (T - 22) = 50\\ => T = 50 + 22 = 72° C {/eq}

The temperature at which the wires touch = 72 {eq}° {/eq}C


Learn more about this topic:

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Thermal Expansion: Definition, Equation & Examples

from General Studies Science: Help & Review

Chapter 16 / Lesson 3
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