A thermocouple produces 40 , mu V/K. The temp | vedclass

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(D) The thermo-emf $(E)$ produced by a single thermocouple is given by $E = \alpha \Delta T$,where $\alpha$ is the Seebeck coefficient and $\Delta T$

Vedclass · Accepted Answer

$120$

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(D) The thermo-emf $(E)$ produced by a single thermocouple is given by $E = \alpha \Delta T$,where $\alpha$ is the Seebeck coefficient and $\Delta T$ is the temperature difference.Given $\alpha = 40 \, \mu V/K$.The temperature difference $\Delta T = |40 \, ^\circ C - 20 \, ^\circ C| = 20 \, ^\circ C = 20 \, K$.Thus,$E = 40 \, \mu V/K 	imes 20 \, K = 800 \, \mu V$.For $N = 150$ thermocouples connected in series,the total emf $(E_{total})$ is $E_{total} = N 	imes E$.$E_{total} = 150 	imes 800 \, \mu V = 120,000 \, \mu V$.Since $1 \, mV = 1000 \, \mu V$,we have $E_{total} = 120 \, mV$.

$A$ thermocouple produces $40 \, \mu V/K$. The temperatures of the cold and hot junctions are $40 \, ^\circ C$ and $20 \, ^\circ C$ respectively. If $150$ such thermocouples are connected in series,what is the total thermo-emf produced in $mV$?

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