---
title: "Thermocouple laws and emf calculation"
subject: "Instrumentation and Measurement"
chapter: 5
key: "semester_4/instrumentation_and_measurement:thermocouple-laws-emf-numerical"
source: https://www.bctengineer.com/q/semester_4/instrumentation_and_measurement/thermocouple-laws-emf-numerical
site: BCT Engineering
---

# Thermocouple laws and emf calculation

Instrumentation and Measurement · Chapter 5: Sensors

- Asked in: year not given, 6 marks
- Video: [Thermocouple working | Cold Junction Compensation | Selection | Peltier Effect](https://www.youtube.com/watch?v=R5ewGMtr8cM) — Asad (I&C Engineer)
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/instrumentation_and_measurement/thermocouple-laws-emf-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_4/instrumentation_and_measurement/chapter-5

## Question

State the laws of thermocouples. A chromel-alumel thermocouple gives the following emf against a 0 $^\circ$C reference: 30 $^\circ$C, 1.203 mV; 100 $^\circ$C, 4.096 mV; 200 $^\circ$C, 8.138 mV; 300 $^\circ$C, 12.209 mV. Find (a) the emf when the junctions are at 300 $^\circ$C and 100 $^\circ$C, and (b) the hot-junction temperature if an instrument whose cold junction is at 30 $^\circ$C reads 7.5 mV (use linear interpolation).

## Short answer

Thermocouple laws: homogeneous circuit, intermediate metals, intermediate temperatures. (a) E_100,300 = E_0,300 - E_0,100 = 12.209 - 4.096 = 8.113 mV. (b) Referred to 0°C: E_0,T = 7.5 + 1.203 = 8.703 mV; linearly interpolating between 200°C (8.138 mV) and 300°C (12.209 mV) gives T ≈ 213.9°C.

## Answer

A thermocouple produces an emf when two dissimilar metals joined at two junctions are at different temperatures (Seebeck effect).

#### Laws of thermocouples
1. **Law of homogeneous circuit:** a thermoelectric current cannot be sustained in a circuit of a single homogeneous metal by heat alone, however the section varies. A temperature gradient in a uniform wire adds no emf.
2. **Law of intermediate metals:** a third metal inserted in the circuit (e.g. the voltmeter leads) adds no net emf if its two junctions are at the same temperature.
3. **Law of intermediate temperatures:** if the emf between junction temperatures $T_1$ and $T_2$ is $E_{1,2}$ and between $T_2$ and $T_3$ is $E_{2,3}$, then $E_{1,3} = E_{1,2} + E_{2,3}$. This lets tables with a 0 $^\circ$C reference be used for any cold junction.

#### (a) Junctions at 300 $^\circ$C and 100 $^\circ$C
$$
E_{100,300} = E_{0,300} - E_{0,100} = 12.209 - 4.096 = 8.113\ \text{mV}
$$

#### (b) Hot junction temperature
The instrument reads 7.5 mV with cold junction at 30 $^\circ$C. Referred to 0 $^\circ$C:

$$
E_{0,T} = E_{30,T} + E_{0,30} = 7.5 + 1.203 = 8.703\ \text{mV}
$$

This lies between 200 $^\circ$C (8.138 mV) and 300 $^\circ$C (12.209 mV).

$$
T = 200 + \frac{8.703 - 8.138}{12.209 - 8.138}\times 100 = 200 + \frac{0.565}{4.071}\times 100 = 213.9\ ^\circ\text{C}
$$

**Answer: (a) 8.113 mV; (b) $T \approx 213.9\ ^\circ$C.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. The law of intermediate temperatures states that
   A. E13 = E12 + E23
   B. E13 = E12 × E23
   C. E13 = E12 - E23 always zero
   D. emf is independent of temperature
   Answer: A. It lets 0 °C reference tables be used for any cold junction. (Key point: Intermediate temperatures)

2. The law of intermediate metals says a third metal in the circuit
   A. Always adds emf
   B. Adds no net emf if its two junctions are at the same temperature
   C. Cancels the Seebeck effect
   D. Doubles the emf
   Answer: B. So voltmeter leads do not change the emf. (Key point: Intermediate metals)

3. Emf with junctions at 300 °C and 100 °C is
   A. 16.305 mV
   B. 12.209 mV
   C. 8.113 mV
   D. 4.096 mV
   Answer: C. 12.209 - 4.096 = 8.113 mV. (Key point: emf 300/100)

4. A 7.5 mV reading with cold junction at 30 °C corresponds to what emf referred to 0 °C, and what temperature?
   A. 8.703 mV, about 213.9 °C
   B. 6.297 mV, about 150 °C
   C. 7.5 mV, about 190 °C
   D. 8.703 mV, about 300 °C
   Answer: A. Add E(0,30) = 1.203: 8.703 mV, then interpolate between 200 and 300 °C giving 213.9 °C. (Key point: Hot junction)

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