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Thermocouple laws and emf calculation

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.

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

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 T1T_1 and T2T_2 is E1,2E_{1,2} and between T2T_2 and T3T_3 is E2,3E_{2,3}, then E1,3=E1,2+E2,3E_{1,3} = E_{1,2} + E_{2,3}. This lets tables with a 0 ∘^\circC reference be used for any cold junction.

(a) Junctions at 300 ∘^\circC and 100 ∘^\circC

E100,300=E0,300−E0,100=12.209−4.096=8.113 mVE_{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 ∘^\circC. Referred to 0 ∘^\circC:

E0,T=E30,T+E0,30=7.5+1.203=8.703 mVE_{0,T} = E_{30,T} + E_{0,30} = 7.5 + 1.203 = 8.703\ \text{mV}

This lies between 200 ∘^\circC (8.138 mV) and 300 ∘^\circC (12.209 mV).

T=200+8.703−8.13812.209−8.138×100=200+0.5654.071×100=213.9 ∘CT = 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≈213.9 ∘T \approx 213.9\ ^\circC.

✓ Answer checked by Claude Opus 5.5 (high effort)

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