A thermometric property X of a system has a value of 15.0\,\text{units} at the ice point (… — JAMB UTME

A thermometric property $X$ of a system has a value of $15.0\,\text{units}$ at the ice point ( $0^\circ\text{C}$ ) and $75.0\,\text{units}$ at the steam point ( $100^\circ\text{C}$ ). If the thermometric property is measured as $33.0\,\text{units}$ when immersed in a chemical bath, what is the temperature of the bath on the absolute thermodynamic scale?

$303\,\text{K}$ Answer
B
$317\,\text{K}$
C
$297\,\text{K}$
D
$30\,\text{K}$

Answer

303\,\text{K}

The temperature of the bath on the Celsius scale is found by taking the ratio of the change in thermometric property from the ice point to the total fundamental interval:

\frac{33.0 - 15.0}{75.0 - 15.0} \times 100^\circ\text{C} = \frac{18}{60} \times 100 = 30^\circ\text{C}

. Converting this temperature to the absolute (Kelvin) scale requires adding

273\,\text{K}

, yielding

30 + 273 = 303\,\text{K}

Step-by-Step Solution

Identify the given thermometric values for the fixed points and the unknown state

X_0 = 15.0\,\text{units}

(ice point,

0^\circ\text{C}

X_{100} = 75.0\,\text{units}

(steam point,

100^\circ\text{C}

), and

X_T = 33.0\,\text{units}

Linear thermometric interpolation requires defining the fixed reference points and the measured property value.

Calculate the temperature on the Celsius scale using the linear scale formula

\theta = \frac{X_T - X_0}{X_{100} - X_0} \times 100^\circ\text{C} = \frac{33.0 - 15.0}{75.0 - 15.0} \times 100 = \frac{18.0}{60.0} \times 100 = 30^\circ\text{C}

The temperature change relative to the fundamental interval determines the position on the Celsius scale.

Convert the temperature from degrees Celsius to Kelvin

T = \theta + 273 = 30 + 273 = 303\,\text{K}

The absolute thermodynamic scale (Kelvin) is shifted from the Celsius scale by adding

273\,\text{K}

Key Concept

Linear interpolation of temperature using thermometric properties and conversion to the absolute scale

Estimated Time:2m 0s

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