The emf of a thermocouple,one junction of whi | vedclass

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(A) The Peltier coefficient $\pi$ is given by the relation $\pi = T \frac{de}{dT}$,where $T$ is the absolute temperature in Kelvin.Given that one junc

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$(t + 273)(a + 2bt)$

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(A) The Peltier coefficient $\pi$ is given by the relation $\pi = T \frac{de}{dT}$,where $T$ is the absolute temperature in Kelvin.Given that one junction is at $0\,^{\circ}C$,the absolute temperature $T$ is related to the Celsius temperature $t$ by $T = t + 273$.Substituting $t = T - 273$ into the given emf equation $e = at + bt^2$:$e = a(T - 273) + b(T - 273)^2$.Differentiating with respect to $T$:$\frac{de}{dT} = a + 2b(T - 273)$.Since $T - 273 = t$,we have $\frac{de}{dT} = a + 2bt$.Now,calculating the Peltier coefficient:$\pi = T \frac{de}{dT} = (t + 273)(a + 2bt)$.

The emf of a thermocouple,one junction of which is kept at $0\,^{\circ}C$,is given by $e = at + bt^2$. The Peltier coefficient will be

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