The thermo emf of a thermocouple is given by, $E=a T+b T^2$, where $\frac{a}{b}=-200^{\circ} \mathrm{C}$. If…
The thermo emf of a thermocouple is given by, $E=a T+b T^2$, where $\frac{a}{b}=-200^{\circ} \mathrm{C}$. If the cold function is kept at $30^{\circ} \mathrm{C}$, then the inversion temperature is ( $\varepsilon$ in volt, $T$ in centigrade)
$103 \mathrm{~K}$
$143 \mathrm{~K}$
$333 \mathrm{~K}$
$443 \mathrm{~K}$
Solution
At temperature of inversion
$E=0$
$\because \quad E=a T+b T^2$
$\therefore \quad 0=a T+b T^2$
or $\quad T=-\frac{a}{b}=200^{\circ} \mathrm{C}$
Because the cold function is kept at $30^{\circ} \mathrm{C}$, then the inversion temperature
$T=200-30=170^{\circ} \mathrm{C}$
$T=273+170=443 \mathrm{~K}$