Question

When one junction X of a thermocouple is placed in melting ice and the other junction Y in steam at $100^{\circ} C$, the emf is 6.0 mV Junction Y is removed from the steam and is placed in a liquid bath at a constant temperature, junction X remaining in ice. The emf is now -1.5 mV What is the temperature of the bath on the centigrade scale of this thermocouple?

• A. $-75^{\circ}$
• B. $-25^{\circ}$
• C. $25^{\circ}$
• D. $75^{\circ}$

Answer 1

Answer: (B)

$T=\left(\frac{100}{6}\right)\left(\frac{{ }^{\circ} C }{ mV }\right) \cdot E,$
where $E$ is the emf.
$T=$ temperature rise per milli volt.
If $E=-1.5 mV ,$ then
$T=\left(\frac{100}{6} \times-1.5\right){ }^{\circ} C =-25^{\circ} C$