Question

When the temperature difference between the junctions of a given thermocouple is ${120}^{\circ }C$, the thermo emf is $30mV$. The temperature of hot junction is decreases by ${20}^{\circ }C$ and cold junction's temperature is increased by $6^{\circ }C$. The percentage decrease in thermo emf is
(Assume thermo emf is directly proportional to the temperature difference)

• A. $2.16$ %
  
• B. $20.4$%
• C. $21.6$ %
  
• D. $43$ %
  

Answer 1

Answer: (C)

$21.6$ %

Thermo emf $\propto \Delta T$

Let $\Delta T=T_H-T_L={120}^{\circ }C$

$⟹\Delta T^{\prime }=(T_H-{20}^{\circ })-(T_L+6^{\circ })$$={94}^{\circ }C$

Thus, emf with this temperature difference=$\frac{94}{120}\times 30mV=23.5mV$

Thus percentage decrease in thermo emf=$\frac{30mV-23.5mV}{30mV}\times 100$%$=21.6$%