A platinum resistance thermometer has a resistance of 10\,\Omega at the ice point (0^\circ… — JAMB UTME

A platinum resistance thermometer has a resistance of $10\,\Omega$ at the ice point ( $0^\circ\text{C}$ ) and $50\,\Omega$ at the steam point ( $100^\circ\text{C}$ ). When placed in a warm liquid bath, an uncalibrated digital ohmmeter reads $38\,\Omega$ . If the ohmmeter has a known positive zero error of $+4\,\Omega$ (indicating $4\,\Omega$ above the true resistance), what is the actual temperature of the liquid bath on the absolute scale in kelvins?

$333\,\text{K}$ Answer
B
$343\,\text{K}$
C
$353\,\text{K}$
D
$60\,\text{K}$

Answer

The actual temperature of the liquid bath on the absolute scale is

333\,\text{K}

Subtracting the zero error of

+4\,\Omega

from the raw meter reading of

38\,\Omega

yields the true thermometric resistance of

34\,\Omega

. Applying the temperature formula

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

gives

\theta = \frac{34 - 10}{50 - 10} \times 100 = 60^\circ\text{C}

. Converting to absolute temperature gives

T = 60 + 273 = 333\,\text{K}

Step-by-Step Solution

Correct the measured resistance for zero error

True resistance

R_\theta = 38\,\Omega - 4\,\Omega = 34\,\Omega

A positive zero error means the meter reads higher than the true value, so the zero error must be subtracted.

Calculate the temperature on the Celsius scale using linear interpolation

\theta = \frac{R_\theta - R_0}{R_{100} - R_0} \times 100^\circ\text{C} = \frac{34 - 10}{50 - 10} \times 100 = \frac{24}{40} \times 100 = 60^\circ\text{C}

The thermometric property varies linearly between the fixed points.

Convert the Celsius temperature to the thermodynamic (absolute) Kelvin scale

T = \theta + 273 = 60 + 273 = 333\,\text{K}

The conversion from Celsius to Kelvin requires adding

273

(or

273.15

Key Concept

Linear interpolation of thermometric properties with instrument zero error correction

Estimated Time:2m 0s

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