Question

A metal wire 4 m long is at 50°C, when it is cooled to 0°C, it contracts by 5 mm. Calculate the coefficient of linear expansion of the metal.​

Answer 1

Answer:

Coefficient of linear expansion of metal = $\rm{2\times 10^{-5}\;K^{-1}}$

Explanation:

Given:-

 Initial length of the metal wire, $\rm{L_1}$ = 4 m

 Initial temperature, $\rm{T_1}$ = 50° C

(On cooling the wire it contracts by 5 mm means 0.005 m, so)

New length of the wire, $\rm{L_2}$ = 4 - 0.005 m = 3.995 m

 New temperature, $\rm{T_2}$ = 0° C

To find:-

Coefficient of linear expansion of metal, $\alpha$ =?

Formula required:-

• Formula for linear expansion

       $\qquad \rm{L_2 = L_1 + \alpha\; L_1 (T_2 - T_1)}$

Where,

$\rm{L_1}$ = Initial length (original length)

$\rm{L_2}$ = new length (expaned length)

$\rm{\alpha}$ = coefficient of linear expansion

$\rm{T_1}$ = Initial temperature

$\rm{T_2}$ = new temperature

Solution:-

Converting the temperatures into Kelvin for the correctness of solution because in the expression of linear expansion temperature must be in Kelvin unit.

$\rm{T_1}$ = 50° C = 50 + 273 K = 323 K

$\rm{T_2}$ = 0° C = 0 + 273 K = 273 K

Using the formula for the linear expansion of solid

$\longmapsto \rm{L_2 = L_1 + \alpha\; L_1 (T_2 - T_1)}$

$\longmapsto \rm{3.995 = 4 + \alpha\; (4) \;(273 - 323)}$

$\longmapsto \rm{3.995 - 4 = \alpha\; (4) \;(-50)}$

$\longmapsto \rm{-0.005 = -200\;\alpha}$

$\longmapsto \rm{\alpha = 2\times10^{-5}\;\;K^{-1}}$

Therefore,

Coefficient of linear expansion for the metal is $\rm{2\times 10^{-5}\;K^{-1}}$