A thermistor has an electrical resistance of 800\,\Omega at the melting point of ice (0^\c… — JAMB UTME

A thermistor has an electrical resistance of $800\,\Omega$ at the melting point of ice ( $0^\circ\text{C}$ ) and $200\,\Omega$ at the boiling point of water ( $100^\circ\text{C}$ ). Assuming the thermometric property varies linearly with temperature, what is the temperature in degrees Celsius ( $^\circ\text{C}$ ) when the resistance of the thermistor is $500\,\Omega$ ?

Answer: 50 °C

Answer

The temperature corresponding to a resistance of

500\,\Omega

50^\circ\text{C}

Applying the standard thermometric interpolation relation

\theta = \frac{R_\theta - R_0}{R_{100} - R_0} \times 100^\circ\text{C}

with

R_0 = 800\,\Omega

R_{100} = 200\,\Omega

, and

R_\theta = 500\,\Omega

gives

\theta = \frac{500 - 800}{200 - 800} \times 100 = \frac{-300}{-600} \times 100 = 50^\circ\text{C}

Step-by-Step Solution

Identify the values of the thermometric property at the ice point and steam point.

R_0 = 800\,\Omega

and

R_{100} = 200\,\Omega

These established values represent the fixed points of

0^\circ\text{C}

and

100^\circ\text{C}

respectively.

Set up the linear relationship formula for temperature conversion on the Celsius scale.

$\theta = \frac{R_\theta - R_0}{R_{100} - R_0} \times 100^\circ\text{C}$

The temperature scale is defined linearly between the two fixed calibration points.

Substitute the unknown resistance

R_\theta = 500\,\Omega

into the equation and compute the result.

$\theta = \frac{500 - 800}{200 - 800} \times 100^\circ\text{C} = \frac{-300}{-600} \times 100^\circ\text{C} = 50^\circ\text{C}$

Dividing the change from the lower fixed point by the total interval between fixed points gives the fraction of

100^\circ\text{C}

Key Concept

Linear relationship between a thermometric property and temperature

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