Question

A silver wire has a resistance of $ 2.1\Omega $ at $ 27.5^\circ C $ and resistance of $ 2.7\Omega $ at $ 100^\circ C $ . Determine the temperature coefficient of expansion of silver.

Answer 1

Hint : We need to take the lower temperature as the reference temperature, and the corresponding resistance as the reference resistance. The temperature coefficient is positive and can be calculated by the formula, $ R = {R_0}\left[ {1 + \alpha \left( {T - {T_0}} \right)} \right] $ .

Formula used: In this solution we will be using the following formula;
 $\Rightarrow R = {R_0}\left[ {1 + \alpha \left( {T - {T_0}} \right)} \right] $ where $ R $ is the resistance of the substance at any particular temperature $ T $ , $ {R_0} $ is the reference resistance of the substance at any particular reference temperature $ {T_0} $ , $ \alpha $ is the coefficient of resistivity of the substance.

Complete step by step answer
It is known that when the temperature of a substance changes, the resistance of the same substance also changes. For most conductors, the resistance (or resistivity) increases as the temperature increases. This has been understood to be mostly due to the thermal agitation (violent vibration) of the free electrons, which makes them harder to be directed in a particular direction. This changes in resistance is approximately given as
 $\Rightarrow R = {R_0}\left[ {1 + \alpha \left( {T - {T_0}} \right)} \right] $ where $ R $ is the resistance of the substance at any particular temperature $ T $ , $ {R_0} $ is the reference resistance of the substance at any particular reference temperature $ {T_0} $ , $ \alpha $ is the coefficient of resistivity of the substance. Usually, the reference temperature is $ 20^\circ C $ and sometimes $ 0^\circ C $ . But any known temperature can be used if the resistance at the temperature is known.
Making $ 27.5^\circ C $ our reference, and thus inserting all known values, we have that
 $\Rightarrow 2.7 = 2.1\left[ {1 + \alpha \left( {100 - 27.5} \right)} \right] $ . Dividing both sides by $ 2.1 $ and computing, we have
 $\Rightarrow 1.2857 = 1 + 72.5\alpha $ . Making $ \alpha $ subject we have,
 $\Rightarrow 72.5\alpha = 1.2857 - 1 $
 $\Rightarrow \alpha = \dfrac{{0.2857}}{{72.5}} = 0.00394 $
Hence, $ \alpha = 0.00394 $.

Note
Usually the temperature, for any scientific measurement, is in Kelvin, but the Celsius is allowed in this question because only temperature difference is involved, and the difference in temperature on a Kelvin scale is equal to that on a Celsius scale, as in:
 $\Rightarrow {T_{1K}} = {T_{1C}} + 273 $
 $\Rightarrow {T_{2K}} = {T_{2C}} + 273 $ where the $ k's $ and $ c's $ are temperatures in Kelvin and Celsius respectively. Hence,
 $\Rightarrow {T_{2K}} - {T_{1K}} = {T_{2C}} + 273 - {T_{1C}} + 273 = {T_{2C}} - {T_{1C}} $
 $\Rightarrow \Delta {T_K} = \Delta {T_C} $
Hence, it can be substituted.