Question

The electrical resistance in ohms of a certain thermometer varies with temperature accoding to the approximate law : $R=R_0[1=a(T-T_0\left)\right]$

The resistance is $101.6\Omega$ at the triple -point of water $273.16K,$ and $165.6\Omega$ at the normal melting point of lead $600.5K.$ What is the temperature when the resistance is $123.4\Omega ?$

Answer 1

Given relation ,
$R=R_0[1+a(T-T_0\left)\right]....\left(i\right)$

Here,
$R\text{and}{T}_0$ initial resistance and temperature, respectively.

$R\text{and}T$ are the final resistance and temperature respectively and $\alpha$ is coefficient of linear expansion.

At triple point of water, $T_0=273.16K,$
Resistance of lead, $R=101.6\Omega$

At normal melting point of lead, $T=600.5K$

Resistance of lead, $R=165.5\Omega$

Using thess values in equation (i)

$165.5=101.6[1+\alpha (600.5-273.16\left)\right]$

$\frac{165.5}{101.6}=1+\alpha \left(327.34\right)$

$1.629-1=\alpha \left(327.34\right)$

$\alpha =\frac{0.629}{327.34}$

$\alpha =1.92\times 10-3/K$

when resistance $R=123.4\Omega$

Then from equation (i)

$123.4=101.6[1+1.92\times {10}^{-3}\times (T-273.16\left)\right]$

$\frac{123.4}{101.6}=1+1.92\times {10}^{-3}(T-273.16)$

$1.214-1=1.92\times {10}^{-3}(T-273.16)$

$\frac{0.214}{1.92\times {10}^{-3}}=T-273.16$

$T=384.61K$