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The thermo emf of a therocouple , one ju...

The thermo emf of a therocouple , one junction of which is kept at `0^(@)C` , is given by `varepsilon=at+bt^(2)` where a and b constants of the thermocouple . Calculate the neutral temperature and peltier and Thomson coefficients.

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The correct Answer is:
`varepsilon=alphatheta+(1)/(2)betatheta^(2) therefore (dvarepsilon)/(d theta)=alpha+betatheta`
At `theta=theta_(n), (dvarepsilon)/(d theta)=0 " or "alpha+betatheta_(n)=0`
`therefore theta_(n)=-(alpha)/(beta)=-(10)/(-1//20)=200^(@)C`
At `theta=theta_(i), varepsilon=0 " or "alphatheta_(i)+(1)/(2)betatheta_(i)^(2)=0 " or "alpha+(1)/(2)betatheta_(i)=0`
`therefore theta_(i)=-(2alpha)/(beta)=2xx200=400^(@) C`.
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