Practicing Success

Target Exam

CUET

Subject

Physics

Chapter

Thermodynamics

Question:

The temperature of inversion of a thermocouple is 620oC and the neutral temperature is 300oC. What is the temperature of cold junction

Options:

40°C

20°C

320°C

-20°C

Correct Answer:

-20°C

Explanation:

Let e = α (θ – θC) + β (θ2– θ2C )

where θC = temperature of cold junction

at inversion temperature e = 0

⇒ θi = – θC – α/β

at neutral temperature : \(\frac{de}{dθ} = 0 \)

⇒ θn = – α/2β

θi = – α/β– θC

⇒ θi = 2θn – θc 

⇒ \(\theta_n = \frac{\theta_i + \theta_C}{2}\)

⇒ \( 300 = \frac{620 + \theta_C}{2} ⇒ \theta_C = -20^oC\)