A thermometer is calibrated on a custom scale, X, where the ice point (0^\circ\text{C}) is… — JAMB UTME

A thermometer is calibrated on a custom scale, $X$ , where the ice point ( $0^\circ\text{C}$ ) is marked as $-10^\circ\text{X}$ and the steam point ( $100^\circ\text{C}$ ) is marked as $110^\circ\text{X}$ . What is the reading on this custom scale when a standard Celsius thermometer reads $35^\circ\text{C}$ ?

Answer: 32 °X

Answer

32 °X

Using the linear temperature interpolation formula

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

, substituting

\text{LFP}_X = -10

\text{UFP}_X = 110

C = 35

\text{LFP}_C = 0

, and

\text{UFP}_C = 100

gives

\frac{X - (-10)}{110 - (-10)} = \frac{35 - 0}{100 - 0}

. Simplifying gives

\frac{X + 10}{120} = 0.35

, leading to

X + 10 = 42

, so

X = 32^\circ\text{X}

Step-by-Step Solution

Determine fundamental intervals for both temperature scales

Celsius fundamental interval =

100 - 0 = 100^\circ\text{C}

; Custom scale fundamental interval =

110 - (-10) = 120^\circ\text{X}

Linear temperature scale interpolation requires calculating the total interval between the lower fixed point (LFP) and upper fixed point (UFP).

Set up the ratio equation between the two thermometric scales

\frac{X - (-10)}{120} = \frac{35 - 0}{100}

The fractional position of any given temperature relative to its fixed points must be equal on all linear scales.

Solve the algebraic equation for

X

X + 10 = 120 \times 0.35 = 42 \implies X = 32^\circ\text{X}

Isolating

X

gives the corresponding reading on the custom temperature scale.

Key Concept

Linear Temperature Scale Conversion and Interpolation

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