Question

The temperature of hot junction of a thermocouple changes from ${80}^{o}C$ to ${100}^{o}C$. The percentage change in thermoelectric power is

• A. $8\%$
• B. $10\%$
• C. $20\%$
• D. $25\%$

Answer 1

Answer: (C)

$20\%$

We know the thermoelectric power (for thermocouple) is given by
$P=\frac{dE}{dT}=\frac{\pi }{T}$
when, the temperature of hot junction of a thermocouple changes from ${80}^{\circ }C$ to ${100}^{\circ }C$ then
$P_1=\frac{\pi }{80}$ and $P_2=\frac{\pi }{100}$
The change in thermoelectric power is
$\Delta P=P_1-P_2$
Now, the percentage change in thermoelectric power is
$\frac{\Delta P}{P_1}\times 100=\frac{P_1-P_2}{P_1}\times 100$
$\therefore \frac{\Delta P}{P_1}\times 100=\frac{\frac{\pi }{80}-\frac{\pi }{100}}{\frac{\pi }{80}}\times 100$
$\therefore \frac{\Delta P}{P_1}\times 100=\frac{20}{8000}\times 80\times 100=20\%$