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:
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:
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
Discussion & Comments