Unit 15 · Electricity and Magnetism · 3 teaching hours · 16 min read
Thermoelectric effects
What the syllabus asks for 2
- Seebeck effect; Thermocouples
- Peltier effect: Variation of thermoelectric e.m.f. with temperature; Thermopile
Heat and electricity are usually separate chapters, but at the junction of two different metals they are directly linked: a temperature difference produces an e.m.f., and a current produces a temperature difference. These are the Seebeck and Peltier effects. They are the basis of the thermocouple thermometer used in every industrial furnace, of the thermopile that detects heat radiation, and of the solid-state coolers in portable fridges.
Three things, every year: the direction of the thermoelectric current (Seebeck series), the parabola E = \alpha\theta + \frac{1}{2}\beta\theta^{2} with its neutral and inversion temperatures, and the table of differences between Peltier and Joule heating. The numericals are almost always about \theta_n and \theta_i.
15.1 Seebeck effect; Thermocouples
When two wires of different metals are joined to form a closed circuit and the two junctions are kept at different temperatures, an e.m.f. is set up in the circuit and a current flows round it. The effect is called the Seebeck or thermoelectric effect, the e.m.f. is the thermo-e.m.f., and the pair of metals with its two junctions is a thermocouple.
Thomas Seebeck discovered it in 1821. The thermo-e.m.f. is small — of the order of tens of microvolts per degree — and it depends on only two things:
- the nature of the two metals, and
- the temperature difference between the two junctions.
It does not depend on the thickness of the wires, the length of the wires, or the shape of the junctions.
Direction of the thermoelectric current: the thermoelectric series
Arrange metals in the Seebeck (thermoelectric) series:
Sb, Fe, Cd, Zn, Ag, Au, Cr, Sn, Pb, Hg, Mn, Cu, Pt, Co, Ni, Bi
Two rules follow:
- Direction. In the thermocouple the current flows through the cold junction from the metal that comes earlier in the series to the metal that comes later. (Through the hot junction it therefore flows the other way.)
- Magnitude. The further apart the two metals are in the series, the larger the thermo-e.m.f. Antimony and bismuth stand at the two ends, so an Sb–Bi couple gives the biggest e.m.f. of all — which is why thermopiles are built from it.
"A–B–C" — Antimony to Bismuth through the Cold junction. "hot coffee" — Copper to Fe (iron) through the hot junction, and therefore iron to copper through the cold junction.
Why it happens. The free-electron concentration is different in the two metals. Where they touch, electrons diffuse from the metal of higher concentration to the metal of lower concentration until a contact potential difference stops the flow. This contact p.d. depends on temperature. If both junctions are at the same temperature the two contact p.d.s are equal and opposite and cancel round the loop; if the junctions are at different temperatures they no longer cancel, and the difference is the thermo-e.m.f.
Thermocouple thermometers
To use a thermocouple as a thermometer, keep the cold junction in melting ice (a fixed 0\,^{\circ}\text{C} reference) and put the hot junction where the temperature is wanted. The galvanometer or potentiometer reading is converted to temperature from a calibration curve.
| Thermocouple | Useful range |
|---|---|
| Copper–constantan | -200\,^{\circ}C to 400\,^{\circ}C |
| Iron–constantan | -200\,^{\circ}C to 750\,^{\circ}C |
| Chromel–alumel | -200\,^{\circ}C to 1100\,^{\circ}C |
| Platinum–platinum/rhodium | 0\,^{\circ}C to 1600\,^{\circ}C |
Advantages: very wide range; the junction is tiny, so its heat capacity is small and it responds in a fraction of a second; it measures the temperature at a point; the reading can be taken hundreds of metres away, which is why the kilns of a brick factory or a cement plant in Hetauda are monitored this way.
Disadvantages: the e.m.f. is small, so a sensitive instrument is needed; the calibration is not linear; a reference junction must be maintained.
The Seebeck e.m.f. depends on the difference of junction temperatures. If the "cold" junction is allowed to drift to the temperature of the hot one, the e.m.f. falls to zero even though the furnace is white hot. Always check the reference junction first.
15.2 Peltier effect: Variation of thermoelectric e.m.f. with temperature; Thermopile
The Peltier effect
When an electric current is passed through the junction of two dissimilar metals, heat is evolved at one junction and absorbed at the other. If the direction of the current is reversed, the two junctions exchange roles. This is the converse of the Seebeck effect.
The Peltier coefficient \pi of a junction is the heat absorbed or evolved when unit charge crosses it. Since a current I carries a charge q = It in time t,
so \pi is measured in joule per coulomb, i.e. volts; for metal junctions it is of the order of millivolts.
| Peltier heat | Joule heat |
|---|---|
| Only at a junction of two different metals | In any conductor |
| H = \pi I t — proportional to I | H = I^{2}Rt — proportional to I^{2} |
| Reversible: reversing I turns evolution into absorption | Irreversible: heat is evolved whichever way I flows |
| Can cool a junction below room temperature | Always warms the conductor |
A junction that absorbs Peltier heat is one at which the current flows in the direction it would take through the hot junction of a Seebeck circuit of the same pair — for copper and iron, from copper to iron. The reason is simply that a working thermocouple is a heat engine: it takes in heat at the hot junction and rejects it at the cold one.
Thomson effect (for completeness). Heat is also absorbed or evolved when a current flows along a single conductor whose two ends are at different temperatures. It is positive for copper, silver and zinc, negative for iron, cobalt and nickel, and zero for lead — which is why lead is taken as the reference metal.
Variation of thermo-e.m.f. with temperature
Keep the cold junction at 0\,^{\circ}\text{C} and raise the hot junction to \theta. Experiment shows that the thermo-e.m.f. is not proportional to \theta; it follows a parabola:
where \alpha and \beta are constants of the pair of metals. For most couples \alpha is positive and \beta is small and negative, so the curve rises, turns over and comes back down.
Thermoelectric power. Differentiating E = \alpha\theta + \frac{1}{2}\beta\theta^{2},
This is the thermoelectric power (or Seebeck coefficient) — the e.m.f. per degree. It falls linearly as \theta rises.
Neutral temperature \theta_n. The e.m.f. is a maximum where dE/d\theta = 0:
\theta_n contains only \alpha and \beta, so it depends only on the pair of metals — not on the temperature of the cold junction. The maximum e.m.f. is E_{max} = -\alpha^{2}/2\beta.
Temperature of inversion \theta_i. Beyond \theta_n the e.m.f. falls, and at some higher temperature it becomes zero and then reverses direction. Putting E = 0 with the cold junction at 0\,^{\circ}\text{C}:
If the cold junction is at \theta_c instead of 0\,^{\circ}\text{C}, the neutral temperature stays put but the inversion temperature shifts, because \theta_n always lies midway between \theta_c and \theta_i:
A very common error is to say that both temperatures depend on the cold junction. Only \theta_i does. Raise the cold junction by 10\,^{\circ}\text{C} and \theta_i falls by 10\,^{\circ}\text{C}, while \theta_n does not move at all.
Problem. For a thermocouple with its cold junction at 0\,^{\circ}\text{C}, the thermo-e.m.f. in microvolts is E = 25\theta - 0.05\theta^{2}. Find (a) \alpha and \beta, (b) the neutral temperature, (c) the temperature of inversion, (d) the maximum e.m.f., (e) the thermoelectric power at 100\,^{\circ}\text{C}, and (f) the two hot-junction temperatures at which E = 3000\ \mu\text{V}.
Solution.
(a) Comparing with E = \alpha\theta + \frac{1}{2}\beta\theta^{2}: \alpha = 25\ \mu\text{V}\,^{\circ}\text{C}^{-1} and \frac{1}{2}\beta = -0.05, so \beta = -0.1\ \mu\text{V}\,^{\circ}\text{C}^{-2}.
(b) \theta_n = -\alpha/\beta = -25/(-0.1) = 250\,^{\circ}\text{C}.
(c) \theta_i = 2\theta_n - \theta_c = 2(250) - 0 = 500\,^{\circ}\text{C}.
(d) E_{max} = 25(250) - 0.05(250)^{2} = 6250 - 3125 = 3125\ \mu\text{V} = 3.125\ \text{mV}.
(e) S = \alpha + \beta\theta = 25 - 0.1(100) = 15\ \mu\text{V}\,^{\circ}\text{C}^{-1}.
(f) 25\theta - 0.05\theta^{2} = 3000 \Rightarrow \theta^{2} - 500\theta + 60000 = 0,
so \theta = 200\,^{\circ}\text{C} or 300\,^{\circ}\text{C} — one on each side of \theta_n, as the parabola requires.
Problem. The neutral temperature of a copper–iron thermocouple is 270\,^{\circ}\text{C}. Find its temperature of inversion when the cold junction is kept at (a) 0\,^{\circ}\text{C} and (b) 20\,^{\circ}\text{C}. If \alpha = 13.5\ \mu\text{V}\,^{\circ}\text{C}^{-1}, find \beta and the e.m.f. at \theta = 100\,^{\circ}\text{C} with the cold junction at 0\,^{\circ}\text{C}.
Solution.
(a) \theta_i = 2\theta_n - \theta_c = 540 - 0 = 540\,^{\circ}\text{C}.
(b) \theta_i = 540 - 20 = 520\,^{\circ}\text{C}.
From \theta_n = -\alpha/\beta:
Problem. The Peltier coefficient of a junction is 12\ \text{mV} and its resistance is 0.50\ \Omega. A current of 2.0\ \text{A} is passed for 5.0\ \text{minutes}. Find (a) the Peltier heat and (b) the Joule heat, and comment.
Solution. t = 5.0 \times 60 = 300\ \text{s}.
(a) H_P = \pi It = (12\times10^{-3})(2.0)(300) = 7.2\ \text{J}
(b) H_J = I^{2}Rt = (2.0)^{2}(0.50)(300) = 600\ \text{J}
The Joule heat is about 83 times larger, and it is always evolved. Reversing the current leaves the 600\ \text{J} of Joule heating unchanged but turns the 7.2\ \text{J} from evolved to absorbed. This is why practical Peltier coolers use semiconductor pellets, whose Peltier coefficient is hundreds of times larger than that of a metal junction while their resistance stays small.
The thermopile
A single thermocouple gives only microvolts. A thermopile is a large number of thermocouples — usually antimony–bismuth, because that pair gives the largest e.m.f. — joined in series, so that their e.m.f.s add.
Alternate junctions are blackened and exposed at the front of a polished metal cone; the other set is shielded inside the case and stays at room temperature. Radiation falling on the cone is absorbed by the black faces, raises their temperature, and the series e.m.f. drives a current through a sensitive galvanometer.
A thermopile is astonishingly sensitive: it will register the heat of a person standing a few metres away, or of the Moon. Its uses include measuring radiant heat and comparing the heating effects of different sources, radiation pyrometers for furnaces, infrared detectors and burglar alarms, and — running the Peltier effect backwards — portable thermoelectric refrigerators and the radioisotope thermoelectric generators that power deep-space probes.
Problem. A copper–constantan thermocouple has a thermoelectric power of 42\ \mu\text{V}\,^{\circ}\text{C}^{-1} near room temperature. The galvanometer used with it can just detect 2.1\ \mu\text{V}. What is the smallest temperature difference it can measure? What e.m.f. would a thermopile of 40 such couples in series give for a temperature difference of 0.50\,^{\circ}\text{C}?
Solution. Smallest detectable difference:
For 40 couples in series the e.m.f.s add:
— comfortably measurable, which is exactly the point of a thermopile.
Chapter summary
- Seebeck effect: two dissimilar metals, two junctions at different temperatures, an e.m.f. appears. It depends only on the metals and on the temperature difference.
- Seebeck series Sb, Fe, Cd, Zn, Ag, Au, Cr, Sn, Pb, Hg, Mn, Cu, Pt, Co, Ni, Bi: the current flows from the earlier metal to the later one through the cold junction, and the further apart the metals, the greater the e.m.f.
- Peltier effect: a current through a junction evolves heat at one junction and absorbs it at the other; H = \pi It, proportional to I and reversible, unlike Joule heat I^{2}Rt.
- E = \alpha\theta + \frac{1}{2}\beta\theta^{2} with the cold junction at 0\,^{\circ}\text{C}; thermoelectric power S = dE/d\theta = \alpha + \beta\theta.
- Neutral temperature \theta_n = -\alpha/\beta (where E is maximum, fixed by the metals alone); temperature of inversion \theta_i = 2\theta_n - \theta_c (where E returns to zero and reverses).
- E_{max} = -\alpha^{2}/2\beta, reached at \theta_n.
- A thermopile is many Sb–Bi couples in series, alternate junctions blackened, used to detect and measure heat radiation.
Practice questions
Group A — Multiple choice (1 mark each)
- The Seebeck effect converts [1]aelectrical energy into heatbheat into electrical energycchemical energy into heatdmagnetic energy into heat
- The neutral temperature of a thermocouple depends on [1]athe temperature of the cold junctionbthe nature of the two metals onlycthe current in the circuitdthe length of the wires
- The Peltier heat developed at a junction is proportional to [1]aI^{2}bIc\sqrt{I}d1/I
- The neutral temperature of a thermocouple is 270\,^{\circ}C. With the cold junction at 10\,^{\circ}C the temperature of inversion is [1]a260\,^{\circ}Cb280\,^{\circ}Cc530\,^{\circ}Cd550\,^{\circ}C
- In a thermopile the thermocouples are joined [1]ain parallelbin seriescin a bridgedback to back
- Antimony and bismuth are chosen for a thermopile because [1]athey are cheapbthey melt at a high temperaturecthey lie at opposite ends of the thermoelectric seriesdthey are black
Answers to Group A
1. (b) — a temperature difference produces an e.m.f.
2. (b) — \theta_n = -\alpha/\beta contains only constants of the metal pair.
3. (b) — H = \pi It; the I^{2} law belongs to Joule heating.
4. (c) — \theta_i = 2\theta_n - \theta_c = 540 - 10 = 530\,^{\circ}C.
5. (b) — in series so that the small e.m.f.s add.
6. (c) — the widest separation in the series gives the largest thermo-e.m.f.
Group B — Short answer (5 marks each)
- State the Seebeck effect. What is the thermoelectric series, and how is it used to find the direction of the thermoelectric current? State the direction in a copper–iron thermocouple. [5]
- Starting from E = \alpha\theta + \frac{1}{2}\beta\theta^{2}, define the thermoelectric power, and obtain expressions for the neutral temperature and the temperature of inversion. [5]
- For a thermocouple with its cold junction at 0\,^{\circ}C the thermo-e.m.f. in microvolts is E = 16\theta - 0.02\theta^{2}. Find \alpha, \beta, the neutral temperature, the temperature of inversion, the maximum e.m.f. and the thermoelectric power at 100\,^{\circ}C. [5]
- Distinguish between Peltier heating and Joule heating on four counts. A junction of Peltier coefficient 10\ \text{mV} carries 3.0\ \text{A} for 200\ \text{s}; find the Peltier heat. [5]
- Give four advantages and two disadvantages of a thermocouple thermometer over a mercury-in-glass thermometer. [5]
Answers to Group B
1. See §15.1. Series: Sb, Fe, Cd, Zn, Ag, Au, Cr, Sn, Pb, Hg, Mn, Cu, Pt, Co, Ni, Bi. Current flows from the earlier metal to the later one through the cold junction; since Fe comes before Cu, the current flows from iron to copper through the cold junction and from copper to iron through the hot junction.
3. \alpha = 16\ \mu\text{V}\,^{\circ}\text{C}^{-1}; \frac{1}{2}\beta = -0.02 so \beta = -0.04\ \mu\text{V}\,^{\circ}\text{C}^{-2}. \theta_n = -\alpha/\beta = 16/0.04 = 400\,^{\circ}C; \theta_i = 2\theta_n - 0 = 800\,^{\circ}C; E_{max} = 16(400) - 0.02(400)^{2} = 6400 - 3200 = 3200\ \mu\text{V} = 3.2\ \text{mV}; S = 16 - 0.04(100) = 12\ \mu\text{V}\,^{\circ}\text{C}^{-1}.
4. Table in §15.2. H = \pi It = (10\times10^{-3})(3.0)(200) = 6.0\ \text{J}.
5. Advantages: very wide range; small heat capacity so fast response; measures temperature at a point; reading can be taken remotely (and recorded electrically). Disadvantages: the e.m.f. is very small and needs a sensitive instrument; the calibration is non-linear and a reference junction must be maintained.
Group C — Long answer (8 marks each)
- (a) Describe the Seebeck effect and sketch the variation of thermo-e.m.f. with the temperature of the hot junction. Define and locate the neutral temperature and the temperature of inversion on your graph. [4](b) A copper–iron thermocouple has \alpha = 13.5\ \mu\text{V}\,^{\circ}\text{C}^{-1} and \beta = -0.05\ \mu\text{V}\,^{\circ}\text{C}^{-2} with the cold junction at 0\,^{\circ}C. Find the neutral temperature, the temperature of inversion, the maximum e.m.f., and the e.m.f. when the hot junction is at 200\,^{\circ}C. [4]
- (a) State the Peltier effect and explain, with a diagram, how it differs from Joule heating. [4](b) Describe the construction and working of a thermopile and give two of its uses. A thermopile of 50 antimony–bismuth couples, each of thermoelectric power 100\ \mu\text{V}\,^{\circ}\text{C}^{-1}, is connected to a galvanometer of total circuit resistance 25\ \Omega. Find the current when the junctions differ by 0.40\,^{\circ}C. [4]
Answers to Group C
1. (b) \theta_n = -\alpha/\beta = 13.5/0.05 = 270\,^{\circ}C; \theta_i = 2(270) - 0 = 540\,^{\circ}C;
2. (b) Total e.m.f. E = NS\,\Delta\theta = 50 \times 100 \times 0.40 = 2000\ \mu\text{V} = 2.0\times10^{-3}\ \text{V}. Hence
which a sensitive galvanometer reads easily.
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