When one junction X of a thermocouple is placed in melting ice and the other junction Y in steam at 100° 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?

Question

When one junction X of a thermocouple is placed in melting ice and the other junction Y in steam at 100° 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° (B) -25° (C) 25° (D) 75°

Tardigrade Admin · Accepted Answer

T=((100/6))(( ° C / mV )) ⋅ E, where E is the emf. T= temperature rise per milli volt. If E=-1.5 mV , then T=((100/6) ×-1.5) ° C =-25° C

$- 7 5^{\circ}$

$- 2 5^{\circ}$

$2 5^{\circ}$

$7 5^{\circ}$

$T = (\frac{100}{6}) (\frac{^{\circ} C}{mV}) \cdot E,$
where $E$ is the emf.
$T =$ temperature rise per milli volt.
If $E = - 1.5 mV,$ then
$T = (\frac{100}{6} \times - 1.5)^{\circ} C = - 2 5^{\circ} C$

−75∘

−25∘

25∘

75∘

T=(6100​)(mV∘C​)⋅E, where E is the emf. T= temperature rise per milli volt. If E=−1.5mV, then T=(6100​×−1.5)∘C=−25∘C

$- 7 5^{\circ}$

$- 2 5^{\circ}$

$2 5^{\circ}$

$7 5^{\circ}$

$T = (\frac{100}{6}) (\frac{^{\circ} C}{mV}) \cdot E,$
where $E$ is the emf.
$T =$ temperature rise per milli volt.
If $E = - 1.5 mV,$ then
$T = (\frac{100}{6} \times - 1.5)^{\circ} C = - 2 5^{\circ} C$