The cold junction of a thermocouple is mainta | vedclass

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(B) The thermo $e.m.f.$ $(E)$ in a thermocouple is given by the relation $E = \alpha 	heta + \frac{1}{2} \beta 	heta^2$,where $	heta$ is the temper

Vedclass · Accepted Answer

$270$

Vedclass · Answer

(B) The thermo $e.m.f.$ $(E)$ in a thermocouple is given by the relation $E = \alpha 	heta + \frac{1}{2} \beta 	heta^2$,where $	heta$ is the temperature difference between the junctions.For a thermocouple,the neutral temperature $(T_n)$ is the temperature of the hot junction at which the thermo $e.m.f.$ becomes zero when the cold junction is at a specific temperature $(T_c)$.The relationship is given by $T_n = \frac{T_i + T_c}{2}$,where $T_i$ is the temperature of the hot junction at which $e.m.f.$ is zero.Given: Cold junction temperature $T_c = 10\,^{\circ}C$ and $T_i = 530\,^{\circ}C$.Substituting the values: $T_n = \frac{530 + 10}{2} = \frac{540}{2} = 270\,^{\circ}C$.

The cold junction of a thermocouple is maintained at $10\,^{\circ}C$. No thermo $e.m.f.$ is developed when the hot junction is maintained at $530\,^{\circ}C$. The neutral temperature is ................ $^{\circ}C$.

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