A thermometer calibrated on an arbitrary scale X registers a lower fixed point of -10^\cir… — JAMB UTME

A thermometer calibrated on an arbitrary scale $X$ registers a lower fixed point of $-10^\circ\text{X}$ and an upper fixed point of $110^\circ\text{X}$ . What is the true temperature in degrees Celsius ( $^\circ\text{C}$ ) when this thermometer reads $20^\circ\text{X}$ ?

Answer: 25 °C

Answer

The true temperature on the Celsius scale is

25^\circ\text{C}

Using the relation

\frac{X - \text{LFP}_X}{\text{UFP}_X - \text{LFP}_X} = \frac{\theta}{100}

, substituting

X = 20

\text{LFP}_X = -10

, and

\text{UFP}_X = 110

gives

\frac{20 - (-10)}{110 - (-10)} = \frac{30}{120} = 0.25

. Multiplying

0.25

100

gives

25^\circ\text{C}

Step-by-Step Solution

Set up the linear relationship between the arbitrary temperature scale

X

and the Celsius scale

\frac{X - \text{LFP}_X}{\text{UFP}_X - \text{LFP}_X} = \frac{\theta - \text{LFP}_C}{\text{UFP}_C - \text{LFP}_C}

Thermometric properties vary linearly with temperature between fixed points.

Substitute the given numerical values into the formula

\frac{20 - (-10)}{110 - (-10)} = \frac{\theta - 0}{100 - 0} \implies \frac{30}{120} = \frac{\theta}{100}

The lower fixed point on scale

X

-10^\circ\text{X}

and the upper fixed point is

110^\circ\text{X}

Solve for the unknown temperature

\theta

in degrees Celsius

\theta = \frac{30}{120} \times 100 = 25^\circ\text{C}

Simplifying the fraction

\frac{30}{120}

yields

\frac{1}{4}

, and

\frac{1}{4} \times 100 = 25

Key Concept

Linear interpolation and conversion between thermometric temperature scales

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