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(B) The entropy change for a system or body is given by $\Delta S = \frac{Q}{T}$.For the system to be in a steady state,the heat lost by the reservoir

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(B) The entropy change for a system or body is given by $\Delta S = \frac{Q}{T}$.For the system to be in a steady state,the heat lost by the reservoir at temperature $T_{1}$ must equal the heat gained by the reservoir at temperature $T_{2}$ (let this be $Q$).The total rate of change of entropy for the conduction process is the sum of the entropy changes of the two reservoirs:$\frac{dS}{dt} = \frac{d}{dt} \left( \frac{-Q}{T_{1}} + \frac{Q}{T_{2}} \right) = \left( \frac{dQ}{dt} \right) \left( \frac{1}{T_{2}} - \frac{1}{T_{1}} \right)$From Fourier's law of heat conduction,the rate of heat flow is $\frac{dQ}{dt} = kA \frac{(T_{1} - T_{2})}{L}$,where $k$ is thermal conductivity,$A$ is the cross-sectional area,and $L$ is the length.Substituting this into the entropy rate equation:$\frac{dS}{dt} = \left( kA \frac{(T_{1} - T_{2})}{L} \right) \left( \frac{T_{1} - T_{2}}{T_{1} T_{2}} \right) = \frac{kA}{L} \frac{(T_{1} - T_{2})^2}{T_{1} T_{2}}$We can rewrite this in terms of the ratio $x = T_{1} / T_{2}$:$\frac{dS}{dt} = \frac{kA}{L} \frac{T_{2}^2 (x - 1)^2}{T_{2}^2 x} = \frac{kA}{L} \frac{(x - 1)^2}{x} = \frac{kA}{L} \left( x - 2 + \frac{1}{x} \right)$When $x = 1$ $(T_{1} = T_{2})$,$\frac{dS}{dt} = 0$. For $x > 1$ or $x  0$. The function $f(x) = \frac{(x-1)^2}{x}$ is non-negative and symmetric in its behavior relative to the minimum at $x=1$. Plot $(b)$ correctly represents this relationship.

Column-$I$	Column-$II$
$(a)$ Cup of hot tea in room	$(i)$ Forced convection
$(b)$ Substance kept near fire	$(ii)$ Natural convection
	$(iii)$ Conduction
	$(iv)$ Radiation

Similar Questions

Power radiated by a black body at temperature $T_1$ is $P$ and it radiates maximum energy at a wavelength $\lambda_1$. If the temperature of the black body is changed from $T_1$ to $T_2$, it radiates maximum energy at a wavelength $\frac{\lambda_1}{2}$. The power radiated at $T_2$ is (in $P$)

In Column-$I$,system surroundings and in Column-$II$ modes of heat transfer are given below. Match the following:

Column-$I$ Column-$II$

$(a)$ Cup of hot tea in room $(i)$ Forced convection

$(b)$ Substance kept near fire $(ii)$ Natural convection

$(iii)$ Conduction

$(iv)$ Radiation

What are the modes of heat transfer due to a temperature difference?

$A$ black body radiates maximum energy at wavelength $\lambda$ at temperature $T_1$ and its emissive power is $E$. When the temperature of the body is changed to $T_2$, it radiates maximum energy at wavelength $\frac{2\lambda}{3}$, then the emissive power will become

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