Measurements · Module 8 of 10

Temperature Measurement

Temperature is measured indirectly, through a property that changes with it. This module covers the two workhorses: the thermocouple, which makes a voltage, and the resistance detector, whose resistance rises with temperature.

01

Readiness check

This module measures temperature. Tick only what you can do closed-notes.

  • Divide a voltage by a sensitivity.
  • Evaluate R = R0(1 + αΔT).
  • Recall that resistance can change with temperature.
  • Recall that a voltage needs a reference.
  • Add a temperature rise to a reference temperature.
0 or 1 weak itemsContinue with this module.
2 weak itemsRevisit sensors in Electrical Circuits, Module 8.
3 or more weak itemsRevisit bridges for resistance in Module 6.
02

The core idea

Temperature is read through a property that depends on it. A thermocouple's Seebeck effect makes a voltage proportional to the temperature difference between its junctions, so it needs a known reference. A resistance detector's resistance rises almost linearly, R = R0(1 + αΔT).

thermocouple: ΔT = V / SRTD: R = R0(1 + αΔT)true temperature: T = Tref + ΔT

No sensor feels temperature directly; each exploits a property that varies with it. The thermocouple relies on the Seebeck effect: joining two dissimilar metals produces a small voltage that depends on the temperature difference between the measuring junction and a reference junction. The sensitivity, tens of microvolts per degree, is small, and crucially the reading is a difference, so you must know the reference junction temperature to get an absolute temperature, T = Tref + ΔT. Thermocouples are cheap, rugged, and span enormous ranges, which suits furnaces and engines. The resistance temperature detector, or RTD, uses the near-linear rise of a metal's resistance, usually platinum, with temperature: R = R0(1 + αΔT), where R0 is the ice-point resistance and α the temperature coefficient. A standard Pt100 reads 100 Ω at 0 degrees and about 138.5 Ω at 100 degrees. RTDs are more accurate and stable than thermocouples but cost more, respond slower, and can self-heat if driven too hard. Thermistors, semiconductor devices, give a large but strongly nonlinear change for sensitive narrow-range work. Choosing among them trades range, accuracy, speed, and cost.

The skill works when: you add the reference temperature for a thermocouple and use the linear law for an RTD.
The skill breaks down when: the reference junction is forgotten, or a thermistor is treated as linear.
The concept. An RTD gives a nearly straight resistance-temperature line from a known ice-point value. A thermocouple instead gives a difference voltage needing a reference.
03

The skills, taught in order

Five skills cover how temperature is actually measured.

8.1 The measurement principle

Every thermometer maps temperature onto a measurable property: expansion, resistance, or voltage. Temperature standards, the fixed points of the international scale, are what all these devices are ultimately calibrated against.

8.2 Thermal expansion thermometers

Liquid-in-glass and bimetallic thermometers convert temperature into a length or deflection. Simple and self-powered, they suit local indication but are hard to read electronically.

8.3 The resistance temperature detector

An RTD uses R = R0(1 + αΔT), with platinum giving excellent linearity and stability. It is read with a bridge, and the excitation current must stay small to avoid self-heating error.

SensorPrincipleBest for
ThermocoupleSeebeck voltagewide range, rugged
RTDresistance riseaccuracy and stability
Thermistorlarge nonlinear resistancesensitive narrow range

The three electrical thermometers and where each wins. The application decides the choice.

8.4 The thermocouple

Two dissimilar metals make a Seebeck voltage set by the junction temperature difference, so a reference junction is essential. Different metal pairs, the standard types, cover different ranges and sensitivities.

8.5 Thermistors and errors

Thermistors give a large but nonlinear response for sensitive work. All contact thermometers share physical errors: conduction along leads, radiation, and incomplete immersion, which bias the reading from the true temperature.

Engineering connection: a Pt100 in a bridge measures a lab bath to a hundredth of a degree, while a thermocouple monitors a furnace at a thousand degrees where the RTD would fail.

04

Worked example 1: a thermocouple reading

A thermocouple with a sensitivity of 40 µV per degree Celsius reads 2.0 mV, with its reference junction held at 20 degrees Celsius. Find the measuring junction temperature.

Figure 1. The voltage gives the temperature difference; adding the reference junction temperature gives the absolute measuring temperature.
  1. ProblemFind the measuring junction temperature in Figure 1.
  2. Given / findSensitivity S = 40 µV/°C, reading 2.0 mV, Tref = 20 °C. Find T.
  3. AssumptionsConstant sensitivity over the range; known reference temperature.
  4. ModelΔT = V / S; T = Tref + ΔT.
  5. EquationsΔT = 2000 µV / 40 µV/°CT = 20 + ΔT
  6. SolveΔT = 50 °C; T = 20 + 50 = 70 °C.
  7. Check2.0 mV is 2000 µV, and at 40 µV per degree that is 50 degrees above the 20 degree reference, so 70 degrees.
  8. ConclusionThe measuring junction is at 70 °C; forgetting the reference would have wrongly reported only the 50 degree rise.
Result. Measuring temperature 70 °C.
05

Worked example 2: an RTD resistance

A platinum Pt100 RTD has R0 = 100 Ω and α = 0.00385 per degree Celsius. Find its resistance at 100 degrees Celsius.

Figure 2. The linear law raises the ice-point 100 ohms by 38.5 percent over 100 degrees, to 138.5 ohms.
  1. ProblemFind the RTD resistance at 100 °C in Figure 2.
  2. Given / findR0 = 100 Ω, α = 0.00385/°C, ΔT = 100 °C. Find R.
  3. AssumptionsLinear resistance-temperature law over this range.
  4. ModelR = R0(1 + αΔT).
  5. EquationsR = 100(1 + 0.00385 × 100)
  6. SolveR = 100 × 1.385 = 138.5 Ω.
  7. CheckThe coefficient 0.00385 means 0.385 Ω per degree for a 100 Ω sensor, and 0.385 × 100 = 38.5 Ω added, giving 138.5 Ω.
  8. ConclusionThe Pt100 reads 138.5 Ω at 100 °C, the standard value a bridge would convert back to temperature.
Result. R = 138.5 Ω at 100 °C.
06

Misconceptions and diagnostics

MistakeSymptomDiagnostic questionCorrection
Forgetting the reference junctionThermocouple reads a rise, not a temperature"What is the reference temperature?"T = Tref + ΔT; the reading is a difference.
RTD self-heatingReading biased high under high current"Is the excitation current small?"Keep the drive current low to avoid self-heating.
Assuming a thermistor is linearLarge errors away from calibration"Is this device linear?"Thermistors are strongly nonlinear; use their curve.
Ignoring immersion errorsReading biased by conduction or radiation"Is the probe fully immersed and shielded?"Immerse deeply and account for conduction and radiation.
07

Practice ladder

Level 1 · Direct skill

Find the resistance of a Pt100 (α = 0.00385/°C) at 50 degrees Celsius.

Show answer

R = 100(1 + 0.00385 × 50) = 100 × 1.1925 = 119.25 Ω.

Level 2 · Mixed concept

A thermocouple of 40 µV/°C reads 1.6 mV with a reference at 25 degrees. Find the measuring temperature.

Show answer

ΔT = 1600 / 40 = 40 °C; T = 25 + 40 = 65 °C.

Level 3 · Independent problem

A Pt100 reads 123.4 Ω. Find the temperature (α = 0.00385/°C).

Show answer

ΔT = (123.4 − 100)/(100 × 0.00385) = 23.4 / 0.385 = 60.8 °C.

Transfer task | Real engineering

Choose a sensor for (a) a 900 degree furnace and (b) a laboratory bath held near 37 degrees to a hundredth of a degree. Justify each.

What good work looks like

(a) A thermocouple, for its wide range and ruggedness at high temperature; (b) a Pt100 RTD, for its accuracy and stability at modest temperature, read with a bridge and low excitation. A good answer ties range, accuracy, and speed to each choice.

08

Working with AI, and proving it yourself

Use AI as an examiner, not a solver

"Check that I added the reference junction temperature for this thermocouple."
"Give me three RTD readings; I will find each temperature."
"What temperature is this?" Apply the sensor law yourself.
"Pick a sensor for me." Weigh range, accuracy, and speed yourself.

Portfolio task

Convert a real thermocouple voltage or RTD resistance into a temperature, stating your reference or ice-point value.

Must include: the sensor law used, a reference or R0, and a temperature with a sanity check.
09

Retrieval and spaced review

Closed notes. Answer out loud, then reveal.

1. What makes a thermocouple's voltage?

The Seebeck effect, from the junction temperature difference.

2. Why does a thermocouple need a reference?

Because it measures a difference, so T = Tref + ΔT.

3. Write the RTD law.

R = R0(1 + αΔT).

4. Thermocouple versus RTD strength?

Thermocouple for wide rugged range; RTD for accuracy and stability.

5. What causes RTD self-heating?

Excessive excitation current dissipating power in the element.

TodayFinish this quiz and Levels 1 and 2 of the ladder.
+1 dayRe-derive a thermocouple and an RTD reading.
+3 daysConvert one real sensor output to temperature.
+7 daysMove on to pressure and flow in Module 9.
+30 daysAlways state the reference or ice-point with a temperature reading.
10

Textbook mapping

This module follows Figliola and Beasley, Theory and Design for Mechanical Measurements, 5th edition. Use these references to read further.

Topic in this moduleWhere to read more
Resistance temperature detectorsFigliola and Beasley, Section 8.4, Electrical Resistance Thermometry
Thermocouples and the Seebeck effectFigliola and Beasley, Section 8.5, Thermoelectric Temperature Measurement
Thermal expansion and physical errorsFigliola and Beasley, Sections 8.3 and 8.7

Section numbers refer to Figliola and Beasley, 5th edition. Any edition with the same chapter titles is equivalent for study.