The length of the mercury column in an uncalibrated thermometer is 4.0\,\text{cm} at the i… — JAMB UTME

The length of the mercury column in an uncalibrated thermometer is $4.0\,\text{cm}$ at the ice point ( $0^\circ\text{C}$ ) and $24.0\,\text{cm}$ at the steam point ( $100^\circ\text{C}$ ). What is the temperature in degrees Celsius when the length of the mercury column is $19.0\,\text{cm}$ ?

Answer: 75 °C

Answer

The temperature corresponding to a mercury column length of

19.0\,\text{cm}

75^\circ\text{C}

The temperature on the Celsius scale is determined by the ratio of the length change above the ice point to the total length change between the ice and steam points:

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

. Substituting

L_0 = 4.0\,\text{cm}

L_{100} = 24.0\,\text{cm}

, and

L_T = 19.0\,\text{cm}

gives

T = \frac{15.0}{20.0} \times 100 = 75^\circ\text{C}

Step-by-Step Solution

Identify given thermometric length values at fixed points and at the unknown temperature

L_0 = 4.0\,\text{cm}

L_{100} = 24.0\,\text{cm}

, and

L_T = 19.0\,\text{cm}

These represent the length at the lower fixed point (

0^\circ\text{C}

), upper fixed point (

100^\circ\text{C}

), and intermediate temperature

T

respectively.

Set up the linear interpolation equation on the Celsius scale

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

Thermometric expansion is assumed to vary linearly with temperature over the operational range.

Substitute the given values and perform arithmetic calculation

T = \frac{19.0 - 4.0}{24.0 - 4.0} \times 100 = \frac{15.0}{20.0} \times 100 = 75^\circ\text{C}

Simplifying

\frac{15.0}{20.0}

gives

0.75

, which multiplied by

100

equals

75

Key Concept

Temperature measurement using linear variation of thermometric properties

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