Question

The resistance of an electrolyte varies with temperature $T$ as $R_T=\frac{R_0}{1+\alpha T^{\prime }}$, where $R_0$ and $\alpha$ are constants. $A$ constant potential difference $V$ is applied to the two electrodes (each of surface area $A$) which are dipped in this electrolyte. If $T_0$ is the temperature of the surroundings, the loss of heat to the surroundings per unit surface area per unit time is governed by the same relation $H=k(T-T_0)A,$ the steady state temperature is

• A. none of these
• B. $\frac{V^2}{RkA}$
• C. $\frac{(V^2+kA{R}_{0}T)}{(k{R}_{0}A-\alpha V^2)}$
• D. $\frac{V^2}{k{R}_{0}A-\alpha V^2}$

Answer 1

The correct option is C $\frac{(V^2+kA{R}_{0}T)}{(k{R}_{0}A-\alpha V^2)}$

The power generated in the resistance is being lost to the surroundings.
Now, we know that
$P=\frac{V^2}{R}\dots \left(2\right)$
Given $P=k(T-T_0)A\dots \left(2\right)$
Equate $eq$ $\left(1\right)$ and $eq$ $\left(2\right)$ we have
$\frac{V^2}{R}=k(T-T_0)A\dots \left(3\right)$
Also, $R_T=\frac{R_0}{(1+\alpha T)}\dots \left(4\right)$
Substitute $eq$ $\left(4\right)$ in $eq$ $\left(3\right)$, we have
$\frac{V^2}{R_0}(1+\alpha T)=k(T-T_0)A$
$(1+\alpha T)=\frac{kA{R}_0}{V^2}[T-T_0]$
$T[\alpha -\frac{kA{R}_0}{V^2}]$
$=\frac{kA{R}_0{T}_0}{V^2}-1$
$T\left[\frac{\alpha V^2-kA{R}_0}{V^2}\right]$
$=\left[\frac{-kA{R}_0{T}_0-V^2}{V^2}\right]$
$T=\frac{kA{R}_0{T}_0+V^2}{kA{R}_0-\alpha V^2}$
Hence, option $\left(a\right)$ is correct.