The Thomson coefficient of a conductor is 10, | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The heat absorbed due to the Thomson effect is given by the formula: $H = \sigma Q \Delta T$,where $\sigma$ is the Thomson coefficient,$Q$ is the

Vedclass · Accepted Answer

$1\, mJ$

Vedclass · Answer

(D) The heat absorbed due to the Thomson effect is given by the formula: $H = \sigma Q \Delta T$,where $\sigma$ is the Thomson coefficient,$Q$ is the charge,and $\Delta T$ is the temperature difference.Given:$\sigma = 10\,\mu V/K = 10 	imes 10^{-6}\,V/K$$Q = 10\,C$$\Delta T = 60\,^{\circ}C - 50\,^{\circ}C = 10\,K$Substituting the values:$H = (10 	imes 10^{-6}) 	imes 10 	imes 10$$H = 1000 	imes 10^{-6}\,J$$H = 10^{-3}\,J = 1\,mJ$Therefore,the correct option is $D$.

The Thomson coefficient of a conductor is $10\,\mu V/K$. The two ends of the conductor are maintained at $50\,^{\circ}C$ and $60\,^{\circ}C$ respectively. The amount of heat absorbed by the conductor when a charge of $10\,C$ flows through it is:

Similar Questions

Total $emf$ produced in a thermocouple does not depend on

$A$ galvanometer $(G)$ of $2 \ \Omega$ resistance is connected in the given circuit. The ratio of charge in $C_1$ $(4 \ \mu F)$ and $C_2$ $(6 \ \mu F)$ is:

For a given thermocouple,the thermo $e.m.f.$ can be

In the following circuit,a bulb is rated as $1.5\, V$,$0.45\, W$. If the bulb glows with full intensity,then what will be the equivalent resistance between $X$ and $Y$? ............. $\Omega$

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module