A resistance thermometer has a resistance of 4.0\,\Omega at the ice point (0^\circ\text{C}… — JAMB UTME

A resistance thermometer has a resistance of $4.0\,\Omega$ at the ice point ( $0^\circ\text{C}$ ) and $6.0\,\Omega$ at the steam point ( $100^\circ\text{C}$ ). When immersed in a liquid bath, its resistance is measured to be $5.2\,\Omega$ . What is the temperature of the bath?

$60.0^\circ\text{C}$ Answer
B
$260.0^\circ\text{C}$
C
$20.0^\circ\text{C}$
D
$40.0^\circ\text{C}$

Answer

60.0^\circ\text{C}

The temperature of the bath is found by calculating the fraction of resistance change relative to the total change between the ice point and steam point:

\theta = \frac{5.2 - 4.0}{6.0 - 4.0} \times 100^\circ\text{C} = 60.0^\circ\text{C}

Step-by-Step Solution

Identify the given thermometric parameters

R_0 = 4.0\,\Omega

R_{100} = 6.0\,\Omega

, and

R_\theta = 5.2\,\Omega

These are the measured resistance values corresponding to the lower fixed point, upper fixed point, and unknown temperature.

Apply the linear temperature interpolation formula for a resistance thermometer

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

Temperature on the Celsius scale is proportional to the relative change in the thermometric property between fixed points.

Substitute the values and compute the temperature

\theta = \frac{5.2 - 4.0}{6.0 - 4.0} \times 100^\circ\text{C} = \frac{1.2}{2.0} \times 100^\circ\text{C} = 60.0^\circ\text{C}

Evaluating the expression yields the exact temperature of the liquid bath.

Key Concept

Linear interpolation on temperature scales using thermometric properties

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