Question

The electrical resistance (in ohms) of a thermometer varies with temperature roughly as shown. The resistance is 101.6 $\omega$ at the triple point of water (273.16 K) and 165.5 $\omega$ at the normal melting point of lead (600.5 K). What is the temperature when the resistance is 123.4 $\omega$.​

Answer 1

★ Concept

This question is based on the concept of Thermal properties of matter or more specifically on (Temperature scales and Thermal expansion). Here we are provided with a relation stated as $\sf\: R_T = R_O \: [1 + 0.005 (T - T_O)]$ where, $\sf \: \red{R_T}$ is the resistance at the triple point of water and it's value is 101.6 Ω (given). First we'll substitute this value in the given eqⁿ, also the resistance at the normal point of lead is given as 165.5 Ω. In this way, we have formed two eqⁿ using a mathematical operation - division , we'll able to calculate the value for $\sf \:\purple{T_O}$. Now, substitute this value in the above stated relation and we would able to calculate the value for $\sf \: \red{R_O}$. Now, for calculating T when resistance 123.4 Ω is given we will substitute all above values of Triple resistance, electrical resistance and electrical temperature.

Let's proceed with Calculation !!

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The resistance $\sf \: \red{R_T}$ at the triple point of water (T = 273.16 K) is 101.6 Ω. Substituting this in the given equation, we have

$\sf101.6 \: \omega = R_O[1 + 0.005(273.16 - T_O)] \qquad \: ( \red1)$

Also, the resistance at the normal melting point of lead (600.5 K) is 165.5 Ω. Therefore,

$\sf165.5 \: \omega = R_O[1 + 0.005(600.5 - T_O)] \qquad \: ( \red2)$

Dividing eq (2) by (1), we have

$\sf \dfrac{1 + 0.005(600.5 -T_O )}{1 + 0.005(273.16 -T_O )} = \dfrac{165.5}{101.6} = \purple{ 1.629}$

$\sf or, \: 1.629(1 + 1.366 - 0.005 \: T_O) = 1 + 3.0025 - 0.005 \: T_O$

$\sf\: or, \:1.629(2.366 - 0.005 \: T_O) = 4.0025 - 0.005 \: T_O$

$\sf or, \: 3.854 - 0.008 \: T_O = 4.0025 - 0.005 \: T_O$

$\sf or, - 0.003 \: T_O = 4.0025 - 3.854 = 0.1485$

$\rm \: T_O = \dfrac{0.1485}{0.003} = { - 49.5} \quad \underline{\boxed{ \red{ \rm \: T_O = - 49.5}}}$

Substituting this value of $\sf \: {T_O}$ in eq (1), we have

$\sf \: R_O[1 + 0.005(273.16 - ( - 49.5))] = 101.6 \omega$

$\sf \: 2.613 \: R_O = 101.6 \omega$

$\sf \: R_O = \dfrac{101.6 \omega}{2.613} = 38.9 \omega \quad \underline{ \boxed{ \blue{ \sf R_O = 38.9 \omega}}}$

Now, let T be the Temperature when resistance $\sf \: \red{R_T}$ is 123.4 Ω. Substituting for $\sf \: {R_T}$, $\sf \: {R_O}$ and $\sf \: {T_O}$ in the given equation, we have

$\sf \: 123.4 \omega = 38.9 \omega[1 + 0.005(T - ( - 49.5))]$

$\sf{ \huge( }\dfrac{123.4}{38.9} - 1 {\huge)} \dfrac{1}{0.005} = T + 49.5$

$\sf434.4 = T + 49.5$

$\rm \: T \: = 434.4 - 49.5 = 384.9 \quad \underline{\boxed{\red{ \sf T = 384.9K}}}$

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Additional Information

Temperature

The temperature is that property of a body by which we know whether the body is in thermal equilibrium with another given body or not.

Thermometery

The branch of physics which deals with the measurement of temperature is called 'thermometery'.

Thermal Expansion

The expansion of a substance on heating is called 'thermal expansion' of substance.