Thermal & Chemical Effects of Current
Currents that heat, pump heat, and move matter.
Joule's Laws of Heating
Heat ∝ i², ∝ t, ∝ R.
Every toaster, kettle and fuse relies on one fact: current through a resistor heats it, by an amount Joule pinned down in three simple laws.
Joule’s laws of heating
With potential difference V across a resistor and current i through it, the field does work W = V(it) = i²Rt on the free electrons in time t. Collisions hand this energy to the lattice, raising the resistor’s thermal energy. We call it the “heat produced”, H (strictly it isn’t heat, since no temperature difference drives it, but the resistor warms and then passes heat to its surroundings).
Read as three laws:
- For a given resistor and time, H ∝ i².
- For a given resistor and current, H ∝ t.
- For a given current and time, H ∝ R.
An 18 V battery (no internal resistance) drives a 3 Ω resistor in series with a parallel pair of 12 Ω and 4 Ω. Find the heat produced in each resistor in one minute.
Verifying the laws
Joule’s calorimeter is a copper calorimeter of kerosene oil with an insulating lid, a stirrer and a thermometer, with a resistor coil dipped in the oil. A battery, rheostat, key and ammeter complete the circuit. The heat produced shows up as a temperature rise of the oil, so Δθ ∝ H.
H ∝ i²
Pass current i₁ for time t and note the rise Δθ₁. Let it cool, change the rheostat, pass i₂ for the same t: you find Δθ₁/Δθ₂ = i₁²/i₂².
H ∝ t
Keep a fixed current and read the thermometer at regular intervals: the temperature rises by equal steps in equal times.
H ∝ R
Put two calorimeters with equal oil and different coils R₁, R₂ in series (so the same current flows through both). The rises satisfy Δθ₁/Δθ₂ = R₁/R₂.
In a Joule calorimeter, 1.0 A for 5 minutes raises the oil by 3.0 °C. What rise would 2.0 A for the same time give?
Ratings and real batteries
A heater is rated 1000 W, 230 V. (a) Find its resistance. (b) What power does it take at 207 V? (c) A 60 W, 230 V bulb is put in series with it on 230 V. Find the power in each.
A 1.5 V dry cell connected straight across an ammeter (of negligible resistance) drives 25 A. How much heat is produced inside the cell in 10 s?
- 1H = i²Rt (33.1): H ∝ i², H ∝ t, H ∝ R (Joule's laws).
- 2Verified with Joule's calorimeter: oil temperature rise ∝ heat.
- 3P = V²/R for ratings; in series the bigger resistance takes more power.
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The Seebeck Effect & Thermocouples
Heat a junction, get a current.
Heat can drive a current directly. Join two different metals in a loop, warm one junction, and charge starts to flow: the Seebeck effect.
The Seebeck effect
Two strips of different metals are joined at both ends to form a loop. If the two junctions are at different temperatures, a current flows. The emf responsible is the Seebeck emf or thermo-emf, and such a pair of metals is a thermocouple. Its size and direction depend on the metals and on the temperatures of the hot and cold junctions.
The thermoelectric series
For junction temperatures between 0 °C and 100 °C, metals can be ranked in a sequence:
antimony, nichrome, iron, zinc, copper, gold, silver, lead, aluminium, mercury, platinum–rhodium, platinum, nickel, constantan, bismuth
- At the cold junction, current flows from the metal earlier in the list to the one later. In a copper–iron couple it flows from iron to copper at the cold junction.
- The farther apart two metals are in the list, the larger the emf.
Neutral and inversion temperatures
Keep a copper–nickel couple’s cold junction at 0 °C and heat the other. The current grows until the hot junction is near 390 °C, then shrinks, vanishes at about 780 °C, and then reverses. With the cold junction at 10 °C the maximum still comes at 390 °C but the reversal moves to 770 °C.
- The neutral temperature θn: where the thermo-emf is greatest. It depends only on the metals (and their purity and treatment).
- The inversion temperature θi: where the emf changes sign. It sits as far above θn as the cold junction sits below it.
Sign convention: ℰAB is positive if the current flows from A to B at the hot junction. With the cold junction at 0 °C and the hot one at θ °C,
The emf is greatest where dℰ/dθ = 0, at θn = −a/b, and returns to zero at θi = −2a/b. If a and b have the same sign there is no neutral or inversion temperature above 0 °C.
A thermocouple’s cold junction is held at 20 °C, and its emf vanishes when the hot junction is at 600 °C. Find the neutral temperature.
Two useful laws
Intermediate metal. With the same junction temperatures throughout, ℰAB = ℰAC − ℰBC. So aAB = aAC − aBC and bAB = bAC − bBC: tabulating every metal against one reference (lead) gives every pair.
| Metal (with lead) | a (µV/°C) | b (µV/°C²) |
|---|---|---|
| Aluminium | −0.47 | 0.003 |
| Bismuth | −43.7 | −0.47 |
| Copper | 2.76 | 0.012 |
| Gold | 2.90 | 0.0093 |
| Iron | 16.6 | −0.030 |
| Nickel | 19.1 | −0.030 |
| Platinum | −1.79 | −0.035 |
| Silver | 2.50 | 0.012 |
| Steel | 10.8 | −0.016 |
Using the table, find a and b for a gold–iron thermocouple, its neutral temperature, and its emf with junctions at 0 °C and 200 °C.
Intermediate temperatures. ℰ between θ₁ and θ₂ equals ℰ between θ₁ and θ₃ plus ℰ between θ₃ and θ₂:
Together these laws mean you can insert extra wires or a galvanometer into a thermocouple circuit without changing its emf, as long as the new junctions are at equal temperatures.
- 1Seebeck: unequal junction temperatures in a two-metal loop give a thermo-emf.
- 2Thermoelectric series: cold-junction current from earlier to later metal; farther apart, bigger emf.
- 3ℰ = aθ + ½bθ² (33.3); θₙ = −a/b, θᵢ = 2θₙ − θ_c (33.2).
- 4Intermediate metal: ℰ_AB = ℰ_AC − ℰ_BC; intermediate temperatures: emfs add.
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Peltier & Thomson Effects
Currents that pump heat at junctions and along wires.
Run the Seebeck effect backwards and a current can pump heat from one junction to another. Two more effects, Peltier’s and Thomson’s, together explain where the thermo-emf comes from.
The Peltier effect
Keep both junctions of a thermocouple at the same temperature and drive a current round it with a battery. One junction warms and the other cools. Reverse the current and they swap. This reverse of the Seebeck effect is the Peltier effect. The heat absorbed or released at a junction is proportional to the charge that passes through it.
Only at the junction. Heats or cools depending on current direction. Proportional to i.
Throughout the resistor. Always heats. Proportional to i².
A junction has a Peltier emf of 0.050 V. How much heat does it absorb in one minute with 2.0 A flowing the “cooling” way?
The Thomson effect
Now take a single metal wire whose temperature varies along its length and pass a current through it. Some sections absorb or give out heat in addition to the Joule heat i²Rt. This is the Thomson effect. For a small section with temperature difference ΔT between its ends, carrying charge ΔQ:
σ is positive if heat is absorbed when current flows from the cold end to the hot end (copper, silver, zinc, antimony, cadmium). It is negative for iron, cobalt, nickel, bismuth and platinum, and nearly zero for lead.
Explaining all three
- Peltier: different metals have different free-electron densities. At a junction electrons diffuse from the denser side, building up an emf, the Peltier emf. Driving current through it does positive or negative work on the carriers, releasing or absorbing heat.
- Thomson: in an unevenly heated metal the electron density varies from place to place, so electrons diffuse and an emf appears between hot and cold parts. Current forced through it again gives or takes heat.
- Seebeck: the Peltier emf depends on temperature. With the junctions at equal temperatures the two Peltier emfs cancel; at different temperatures they don’t. Each metal also has a temperature gradient, giving Thomson emfs that differ between the two metals.
- 1Peltier: current through a junction absorbs or releases heat ∝ i; Π_AB = ΔH/ΔQ.
- 2Joule heat ∝ i² everywhere; Peltier heat ∝ i, at junctions, reversible.
- 3Thomson: ΔH = σΔQΔT in a wire with a temperature gradient.
- 4Seebeck emf = difference of two Peltier emfs + difference of Thomson emfs.
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Electrolysis & Faraday's Laws
Ions carry the current and deposit matter.
In a solution, current is carried by ions rather than electrons, and as they give up their charge at the electrodes, matter is deposited. Faraday found exactly how much.
How an electrolyte conducts
Solutions of salts, dilute acids and bases conduct electricity. Such a liquid is an electrolyte, and its container with electrodes is an electrolytic cell. Current enters through the anode and leaves through the cathode.
Take silver nitrate solution with silver electrodes. Some AgNO₃ molecules split into ions: Ag⁺ (a positive ion, a cation) and NO₃⁻ (a negative ion, an anion). With a battery connected, the field in the solution points from anode to cathode, so cations drift to the cathode and anions to the anode.
- At the cathode: Ag⁺ + e → Ag. Silver is deposited.
- At the anode: NO₃⁻ → NO₃ + e; the NO₃ reacts with the anode’s silver to form AgNO₃, which dissolves. Silver leaves the anode.
The solution’s concentration stays the same; silver is simply moved from anode to cathode. Every electron the anode receives passes round through the battery to the cathode, so the charge through the circuit equals the charge given up at each electrode. This process, chemical change driven by a current, is electrolysis.
Faraday’s laws
- The mass of a substance liberated at an electrode is proportional to the charge passed.
- For a given charge, the mass liberated is proportional to the substance’s chemical equivalent.
What steady current deposits 1.20 g of copper in 1.5 h? (ECE of copper = 3.29 × 10⁻⁷ kg/C.)
Verifying the laws
First law: weigh a clean cathode, pass current i₁ for time t, wash, dry and weigh again; repeat with i₂. You find m₁/m₂ = i₁/i₂, and at fixed current m₁/m₂ = t₁/t₂, so m ∝ it.
Second law: put a copper cell (CuSO₄) and a silver cell (AgNO₃) in series so the same current flows through both. The deposits satisfy m₁/m₂ = E₁/E₂.
Why: the ionic picture
An ion of valency v carries charge ve. If charge Q passes, N = Q/ve ions are neutralised, each of mass A mu (A = relative atomic mass, mu = 1.6605 × 10⁻²⁷ kg):
So K = e/mu = 1.6022 × 10⁻¹⁹ / 1.6605 × 10⁻²⁷ = 9.6485 × 10⁷ C/kg: Faraday’s laws follow from charge coming in lumps of e.
A current of 0.50 A flows through a silver voltameter for 40 min. How much silver is deposited? (Z for silver = 1.118 × 10⁻⁶ kg/C.)
The silver voltameter above is in series with a copper voltameter. How much copper is deposited?
Voltameters
Since i = m/Zt, weighing a deposit measures a current or a charge. Used this way an electrolytic cell is a voltameter (or coulombmeter).
- Copper voltameter (routine lab use): CuSO₄ in water (about 1:4 by weight); a central copper cathode between two joined copper anodes, so both faces of the cathode are coated.
- Silver voltameter (high accuracy, standards labs): AgNO₃ solution (1:5 to 1:6) in a platinum cup that is the cathode; a silver anode in the middle inside a porous porcelain cup that keeps anode impurities away.
- 1Electrolytes conduct by ions: cations to the cathode, anions to the anode.
- 2First law: m = Zit (33.7). Second law: m ∝ chemical equivalent A/v.
- 3Combined: m = EQ/K, K = e/m_u = 9.6485 × 10⁷ C/kg (33.8).
- 4Faraday constant 96 485 C/mol; one faraday liberates one gram-equivalent.
- 5Voltameters measure charge by weighing the deposit.
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Primary & Secondary Cells
Chemistry driving a current, and recharging it.
Electrolysis uses a current to drive chemistry. A cell does the reverse: chemistry drives a current.
How a cell makes an emf
Dip a metal into an electrolyte and ions tend to move between them, setting up an emf between metal and liquid. Two different metals in the electrolyte produce different emfs, leaving a net emf between them: a voltaic cell. The electrode at the higher potential is the positive electrode, the other the negative.
Used only for discharge: current leaves at + and enters at −. Once used up, discarded.
Can be discharged and charged again: current can flow either way through it.
Primary cells
Daniell cell
A zinc electrode in dilute H₂SO₄ and a copper electrode in CuSO₄ solution, separated by a porous pot that lets ions through but keeps the liquids from mixing (the zinc is amalgamated to stop impurities acting locally).
- SO₄²⁻ ions go to the zinc and form ZnSO₄, leaving negative charge on the zinc.
- H⁺ ions pass through the pot and combine with SO₄²⁻ from the CuSO₄ to form H₂SO₄.
- Cu²⁺ ions deposit on the copper electrode, giving it positive charge.
Because hydrogen never reaches the positive electrode, the cell avoids polarization (a layer of hydrogen gas that would choke it). It gives a steady current: emf about 1.09 V, internal resistance about 1 Ω.
Leclanché cell
A zinc rod (negative) and a carbon rod (positive) in ammonium chloride solution, the carbon packed in a porous pot of manganese dioxide. Cl⁻ ions combine with the zinc; NH₄⁺ ions travel to the carbon, splitting into ammonia and hydrogen ions and giving it positive charge. MnO₂ turns the hydrogen into water (forming Mn₂O₃), but this depolarizing action is slow, so the cell suits intermittent use (doorbells). Emf about 1.5 V.
Dry cell
A sealed Leclanché cell with the NH₄Cl and MnO₂ made into a paste inside a zinc can that is itself the negative electrode. Its internal resistance is small, about 0.1 Ω.
The lead accumulator
The most common secondary cell: a lead peroxide (PbO₂) positive electrode and a lead (Pb) negative electrode in dilute sulphuric acid of specific gravity 1.20–1.28.
- Discharging: SO₄²⁻ ions reach the Pb plate and form PbSO₄; H⁺ ions reach the PbO₂ plate and reduce it (PbO₂ + 2H → PbO + H₂O), and the PbO reacts with acid to form PbSO₄ too. Acid is used up and its specific gravity falls. At 1.15 the cell is flat; drawing more current can permanently damage the plates.
- Charging: current is forced in at the positive terminal. PbSO₄ + 2H → Pb + H₂SO₄ at the negative plate and PbSO₄ + SO₄ + 2H₂O → PbO₂ + 2H₂SO₄ at the positive, restoring the plates and the acid.
Emf about 2.05 V fully charged, falling to about 1.8 V when discharged. Real accumulators join several plates of each kind (separated by insulators) to increase capacity, and put cells in series: three for a 6 V battery, six for 12 V.
Capacity is measured in ampere-hours: a 100 A h battery can give 20 A for 5 h, or 10 A for 10 h.
A 60 A h car battery powers a 4.0 A load. How long does it last, and how many cells does a 12 V battery contain?
- 1Two different metals in an electrolyte give a net emf: a voltaic cell.
- 2Primary cells discharge only (Daniell ≈ 1.09 V; Leclanché, dry cell ≈ 1.5 V).
- 3Polarization by hydrogen is prevented by depolarizers (CuSO₄, MnO₂).
- 4Lead accumulator: PbO₂ (+), Pb (−), H₂SO₄; ≈ 2.05 V to 1.8 V; acid density tracks the charge.
- 5Capacity in ampere-hours.