Question

The following observations were recorded on a platinum resistance thermometer. Resistance at melting point of ice ${{ = 3}}{{.70 ohm}}$, resistance at boiling point of water at normal pressure $4.71{{ }}\Omega $ and resistance at $t^{\circ}C {{ = 5}}{{.29 ohm}}$. Calculate temperature coefficient of resistance of platinum.

Answer 1

Hint: The relation between resistance of the platinum and temperature is ${{{R}}_{{t}}}{{ = }}{{{R}}_{{0}}}{{(1 + \alpha t)}}$and temperature coefficient, ${{\alpha = }}\dfrac{{{{{R}}_{{t}}}{{ - }}{{{R}}_{{0}}}}}{{{{{R}}_{{0}}}{{ \times t}}}}$. On substituting the given values in above formulas, temperature coefficient of resistance of platinum can be determined.

Complete step by step solution:
The platinum thermal resistance thermometer uses platinum in order to determine temperature. Temperature coefficient of resistance is the resistance-change factor per degree Celsius of temperature. It is represented by Greek letter alpha (${{\alpha }}$).

For pure metals, the value of ${{\alpha }}$ is positive. Having a positive value of temperature coefficient of resistance means that the value of resistance increases with the increase in temperature. For the elements like carbon, silicon, and germanium ${{\alpha }}$ is a negative. It means that the value of resistance decreases with increase in temperature. For metal alloys, the value of ${{\alpha }}$ is very close to zero, which means that the resistance hardly changes at all with variations in temperature.

Given: Resistance at temperature ${{{0}}^{{0}}}{{C}}$, ${R_0} = 3.70{{ }}\Omega $
Resistance at temperature ${100^0}C$, ${R_{100}} = 4.71{{ }}\Omega $
Resistance at temperature $t°C$, ${{{R}}_{{t}}}{{ = 5}}{{.29 \Omega }}$
To find: Temperature coefficient of resistance of platinum.
Formula for temperature, ${{t = }}\left( {\dfrac{{{{{R}}_{{t}}}{{ - }}{{{R}}_{{0}}}}}{{{{{R}}_{{{100}}}}{{ - }}{{{R}}_{{0}}}}}} \right){{ \times 100}}$
Resistance at temperature ${{{t}}^{{0}}}{{C}}$
${{{R}}_{{0}}}$= Resistance at temperature ${{{0}}^{{0}}}{{C}}$
${{{R}}_{{{100}}}}$= Resistance at temperature ${{10}}{{{0}}^{{0}}}{{C}}$
On substituting values in above formula, we get
$
  {{t = }}\left( {\dfrac{{{{5}}{{.29 - 3}}{{.68}}}}{{{{4}}{{.71 - 3}}{{.68}}}}} \right){{ \times 100}} \\
   \Rightarrow {{t = }}\dfrac{{{{1}}{{.68}}}}{{{{1}}{{.03}}}}{{ \times 100}} \\
   \Rightarrow {{t = 16}}{{{3}}^{{0}}}{{C}} \\
$
Now, the value of temperature is ${{t = 16}}{{{3}}^{{0}}}{{C}}$.
Finding the value of temperature was the first step in order to calculate the temperature coefficient of resistance of platinum. Now, the relation between resistance of the platinum with temperature is given by:
$
  {{{R}}_{{t}}}{{ = }}{{{R}}_{{0}}}{{(1 + \alpha t)}} \\
$
Where $R_t$ = Resistance at temperature t°C
$R_0$ = Resistance at temperature 0°C
$\alpha$ = temperature coefficient of resistance

On rearranging terms for temperature coefficient, we get
${{\alpha = }}\dfrac{{{{{R}}_{{t}}}{{ - }}{{{R}}_{{0}}}}}{{{{{R}}_{{0}}}{{ \times t}}}}$
On substituting values in the above formula, we get
$
  {{\alpha = }}\left( {\dfrac{{{{5}}{{.29 - 3}}{{.68}}}}{{{{3}}{{.68 \times 163}}}}} \right) \\
  \Rightarrow {{\alpha = }}\dfrac{{{{1}}{{.68}}}}{{{{599}}{{.84}}}} \\
  \Rightarrow {{\alpha = 0}}{{.002800}} \\
  {{Or\ \alpha = 2}}{{.8 \times 1}}{{{0}}^{{{ - 3}}}}{{ }}{}^{{0}}{{{C}}^{{{ - 1}}}} \\
$
Temperature coefficient of resistance of platinum is ${{\alpha = 2}}{{.8 \times 1}}{{{0}}^{{{ - 3}}}}{{ }}{}^{{0}}{{{C}}^{{{ - 1}}}}.$

Note: Melting point of ice is ${{{0}}^{{0}}}{{ C}}{{.}}$ Boiling point of water is ${{10}}{{{0}}^{{0}}}{{C}}$. The temperature coefficient of resistance of platinum means the change in the value of resistance as a function of the ambient temperature.