Thermocouple output at 600°C – type identification A thermocouple measuring about 5.5 mV at 600°C (with a standard cold junction) is most likely which type?

Difficulty: Medium

Correct Answer: Platinum–platinum + rhodium

Explanation:


Introduction / Context:
Selecting the thermocouple type from a measured emf requires familiarity with typical voltage–temperature characteristics. Different thermocouple types generate very different voltages at the same temperature, especially when comparing base-metal types to noble-metal types.


Given Data / Assumptions:

  • Measured emf near 5.5 mV at 600°C.
  • Reference junction is at standard compensation temperature (near 0°C or properly cold-junction compensated).
  • We compare common types: K (Chromel–Alumel), J (Iron–Constantan), and S/R (Platinum–Platinum+Rhodium).


Concept / Approach:
Base-metal thermocouples such as Type K and Type J have relatively high Seebeck coefficients, giving tens of millivolts by 600°C. Noble-metal types such as Type S (Pt–10%Rh/Pt) or Type R (Pt–13%Rh/Pt) have much lower emf outputs for the same temperature and fall in the neighborhood of a few to several millivolts at 600°C.


Step-by-Step Solution:

Recall typical outputs at 600°C: Type K ≈ 24–25 mV; Type J ≈ 33 mV; Type S/R ≈ 6–7 mV.The given 5.5 mV matches the order of magnitude of Pt–PtRh rather than Type K or J.Therefore, the correct identification is Platinum–platinum + rhodium.


Verification / Alternative check:
Consulting reference tables shows Type S output near 6.4 mV at 600°C (depending on reference), which is close to the stated 5.5 mV and far from the 25–33 mV of base-metal types.


Why Other Options Are Wrong:

  • Chromel–alumel (K): Would give about 25 mV at 600°C.
  • Iron–constantan (J): Would give about 33 mV at 600°C.
  • Either (a), (b) or (c): Inconsistent with the observed small emf.


Common Pitfalls:
Ignoring cold-junction compensation. However, even with small compensation differences, the order-of-magnitude gap between base- and noble-metal types remains decisive.


Final Answer:
Platinum–platinum + rhodium

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