A thermocouple thermometer produces an electromotive force (e.m.f.) of 2.0\,\text{mV} at t… — JAMB UTME

A thermocouple thermometer produces an electromotive force (e.m.f.) of $2.0\,\text{mV}$ at the ice point ( $0^\circ\text{C}$ ) and $18.0\,\text{mV}$ at the steam point ( $100^\circ\text{C}$ ). When placed in a liquid bath, the recorded e.m.f. is $14.0\,\text{mV}$ . What is the temperature of the liquid bath on the Celsius scale?

A
$66.7^\circ\text{C}$
B
$70.0^\circ\text{C}$
$75.0^\circ\text{C}$ Answer
D
$87.5^\circ\text{C}$

Answer

75.0^\circ\text{C}

The temperature on a linear scale is proportional to the fraction of the interval traversed between the fixed points. Subtracting the baseline reading of

2.0\,\text{mV}

gives an effective increase of

12.0\,\text{mV}

out of a total range of

16.0\,\text{mV}

. Multiplying this fraction (

0.75

) by

100^\circ\text{C}

yields

75.0^\circ\text{C}

Step-by-Step Solution

Identify the given thermometric values for the lower fixed point, upper fixed point, and unknown temperature reading.

E_0 = 2.0\,\text{mV}

E_{100} = 18.0\,\text{mV}

, and

E_\theta = 14.0\,\text{mV}

Linear temperature scales relate the change in thermometric property proportionally to temperature changes.

Apply the standard linear interpolation formula for a Celsius temperature scale.

\theta = \frac{E_\theta - E_0}{E_{100} - E_0} \times 100^\circ\text{C}

This accounts for the baseline reading at the ice point (

0^\circ\text{C}

) and normalizes it over the

100^\circ\text{C}

fundamental interval.

Substitute the known values and evaluate the expression.

\theta = \frac{14.0 - 2.0}{18.0 - 2.0} \times 100 = \frac{12.0}{16.0} \times 100 = 0.75 \times 100 = 75.0^\circ\text{C}

Carrying out the subtraction yields a proportional change of

\frac{3}{4}

of the full fundamental interval.

Key Concept

Temperature Scale Calibration and Thermocouple Interpolation

Estimated Time:1m 30s

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