A faulty liquid-in-glass thermometer registers a reading of 5^\circ\text{C} at the melting… — JAMB UTME

A faulty liquid-in-glass thermometer registers a reading of $5^\circ\text{C}$ at the melting ice point and $95^\circ\text{C}$ at the steam point of pure water under standard atmospheric pressure. What is the actual temperature in degrees Celsius when this thermometer registers a reading of $41^\circ\text{C}$ ?

A
$36^\circ\text{C}$
$40^\circ\text{C}$ Answer
C
$41^\circ\text{C}$
D
$45.6^\circ\text{C}$

Answer

The actual temperature is

40^\circ\text{C}

The correct answer is obtained by setting up the linear interpolation formula for thermometric property values:

\theta = \frac{X_\theta - X_0}{X_{100} - X_0} \times 100

. Substituting

X_0 = 5

X_{100} = 95

, and

X_\theta = 41

gives

\theta = \frac{36}{90} \times 100 = 40^\circ\text{C}

Step-by-Step Solution

Determine the fundamental interval of the faulty thermometer.

Fundamental interval

= 95^\circ\text{C} - 5^\circ\text{C} = 90\,\text{divisions}

The total interval between the lower fixed point and upper fixed point represents

100^\circ\text{C}

on the standard Celsius scale.

Calculate the measured change from the ice point.

Measured difference

= 41^\circ\text{C} - 5^\circ\text{C} = 36\,\text{divisions}

The zero error of

+5^\circ\text{C}

must be subtracted from the observed reading.

Apply the linear scale interpolation formula to find the actual temperature

\theta

\theta = \frac{41 - 5}{95 - 5} \times 100^\circ\text{C} = \frac{36}{90} \times 100^\circ\text{C} = 40^\circ\text{C}

The ratio of the measured interval to the total fundamental interval equals the true fraction of

100^\circ\text{C}

Key Concept

Linear interpolation on non-standard or faulty thermometer scales using fixed points.

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