MHT CET · Physics · Thermal Properties of Matter
For a thermocouple, the inversion temperature is \(600^{\circ} \mathrm{C}\) and the neutral temperature is \(320^{\circ} \mathrm{C}\). Find the temperature of the cold junction?
- A \(40^{\circ} \mathrm{C}\)
- B \(20^{\circ} \mathrm{C}\)
- C \(80^{\circ} \mathrm{C}\)
- D \(60^{\circ} \mathrm{C}\)
Answer & Solution
Correct Answer
(A) \(40^{\circ} \mathrm{C}\)
Step-by-step Solution
Detailed explanation
Neutral temperature \(\quad T_{n}=\frac{T_{c}+T_{i}}{2}\)
\(
\begin{aligned}
320^{\circ} &=\frac{T_{c}+600^{\circ}}{2} \\
640^{\circ} &=T_{c}+600^{\circ} \\
T_{c} &=40^{\circ} \mathrm{C}
\end{aligned}
\)
\(
\begin{aligned}
320^{\circ} &=\frac{T_{c}+600^{\circ}}{2} \\
640^{\circ} &=T_{c}+600^{\circ} \\
T_{c} &=40^{\circ} \mathrm{C}
\end{aligned}
\)
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