The length of the mercury column in an uncalibrated liquid-in-glass thermometer is 2.0\,\t… — JAMB UTME

The length of the mercury column in an uncalibrated liquid-in-glass thermometer is $2.0\,\text{cm}$ at the ice point ( $0^\circ\text{C}$ ) and $18.0\,\text{cm}$ at the steam point ( $100^\circ\text{C}$ ). What is the temperature when the length of the mercury column is $9.2\,\text{cm}$ ?

A
$40^\circ\text{C}$
$45^\circ\text{C}$ Answer
C
$55^\circ\text{C}$
D
$57.5^\circ\text{C}$

Answer

The temperature corresponding to a mercury column length of

9.2\,\text{cm}

45^\circ\text{C}

The temperature on the Celsius scale is given by the formula

\theta = \frac{L_\theta - L_0}{L_{100} - L_0} \times 100^\circ\text{C}

. Substituting

L_0 = 2.0\,\text{cm}

L_{100} = 18.0\,\text{cm}

, and

L_\theta = 9.2\,\text{cm}

yields

\theta = \frac{7.2}{16.0} \times 100^\circ\text{C} = 45^\circ\text{C}

Step-by-Step Solution

Identify the fixed points and the thermometric property values.

Ice point length

L_0 = 2.0\,\text{cm}

, steam point length

L_{100} = 18.0\,\text{cm}

, and observed length

L_\theta = 9.2\,\text{cm}

The linear scale equation requires establishing the reference points on the Celsius scale.

Calculate the fundamental interval length (

L_{100} - L_0

L_{100} - L_0 = 18.0\,\text{cm} - 2.0\,\text{cm} = 16.0\,\text{cm}

The fundamental interval represents the total change in length corresponding to

100^\circ\text{C}

Apply the linear interpolation formula

\theta = \frac{L_\theta - L_0}{L_{100} - L_0} \times 100^\circ\text{C}

\theta = \frac{9.2 - 2.0}{16.0} \times 100^\circ\text{C} = \frac{7.2}{16.0} \times 100^\circ\text{C} = 45^\circ\text{C}

This scales the fractional change in thermometric property above the ice point to degrees Celsius.

Key Concept

Linear interpolation on empirical temperature scales

Estimated Time:1m 30s

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