A resistance thermometer registers a resistance of 5.0\,\Omega at the ice point (0^\circ\t… — JAMB UTME

A resistance thermometer registers a resistance of $5.0\,\Omega$ at the ice point ( $0^\circ\text{C}$ ) and $25.0\,\Omega$ at the steam point ( $100^\circ\text{C}$ ). When placed in a heated liquid bath, the resistance measured is $30.0\,\Omega$ . What is the temperature of the bath on the Kelvin scale?

A
$125\,\text{K}$
$398\,\text{K}$ Answer
C
$423\,\text{K}$
D
$373\,\text{K}$

Answer

The temperature of the bath on the Kelvin scale is

398\,\text{K}

Using the linear interpolation formula for a thermometric property

\theta = \frac{R_\theta - R_0}{R_{100} - R_0} \times 100^\circ\text{C}

, substituting

R_0 = 5.0\,\Omega

R_{100} = 25.0\,\Omega

, and

R_\theta = 30.0\,\Omega

yields

\theta = \frac{25.0}{20.0} \times 100 = 125^\circ\text{C}

. Converting to absolute thermodynamic temperature gives

T = 125 + 273 = 398\,\text{K}

Step-by-Step Solution

Calculate the temperature on the Celsius scale using linear interpolation of thermometric property.

\theta = \frac{R_\theta - R_0}{R_{100} - R_0} \times 100^\circ\text{C} = \frac{30.0 - 5.0}{25.0 - 5.0} \times 100^\circ\text{C} = \frac{25.0}{20.0} \times 100^\circ\text{C} = 125^\circ\text{C}

The change in resistance is directly proportional to the temperature change between fixed points.

Convert the temperature from degrees Celsius to Kelvins.

T = \theta + 273 = 125 + 273 = 398\,\text{K}

Absolute temperature in Kelvin is obtained by adding 273 to the temperature in degrees Celsius.

Key Concept

Temperature Scale Interpolation and Kelvin Conversion

Estimated Time:1m 30s

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