Question

Figure shows two adiabatic vessels each having $m\text{gm}$ of water at different temperatures. The ends of a metal rod of length $L$, area of cross section $A$, and thermal conductivity $K=\frac{msL}{2A}$ [where $s$ is the specific heat capacity of water] are connected to the water in the vessels as shown in the figure. Find the time taken for the difference in temperature of the vessels to become half of the original value.

• A. $\text{ln}3$
• B. $\text{ln}2$
• C. $\text{ln}4$
• D. $\text{log}2$

Answer 1

The correct option is B $\text{ln}2$

Initially, at $t=0$ we can say that, the difference in temperature $\Delta T_0=T_1-T_2$
$\frac{\Delta T_0}{2}$ at $t=?$
Let $T_A$ and $T_B$ be the temperatures of the water in the vessels after ${}^{\prime }{t}^{\prime }$ seconds.
After ${}^{\prime }{t}^{\prime }$ seconds,
Heat current $\left(H\right)=\frac{dQ}{dt}=\frac{KA(T_A-T_B)}{L}.....\left(1\right)$
We know that, $Q=ms\Delta T$
From this,
$\frac{dQ}{dt}=ms\left(\frac{-d{T}_A}{dt}\right)=ms\left(\frac{d{T}_B}{dt}\right).....\left(2\right)$


Since, $T_A-T_B=\Delta T$, we can write that,
$\frac{d{T}_A}{dt}-\frac{d{T}_B}{dt}=\frac{d}{dt}\left(\Delta T\right)$
From$\left(2\right)$, we write the above equation as,
$\frac{(-\frac{dQ}{dt})}{ms}-\frac{\left(\frac{dQ}{dt}\right)}{ms}=\frac{d\left(\Delta T\right)}{dt}$
or $\frac{2dQ}{msdt}=-\frac{d\left(\Delta T\right)}{dt}$
Using $\left(1\right)$ in the above equation, we get
$\frac{2KA\left(\Delta T\right)}{msL}=-\frac{d\left(\Delta T\right)}{dt}$
$\Rightarrow \underset{\Delta T_0}{\overset{\Delta T_0/2}{\int }}\frac{d\left(\Delta T\right)}{\Delta T}=-\frac{2KA}{msL}{\int }_0^{t}dt$
$\Rightarrow t=\frac{mSL}{2KA}\text{ln}2$
But, given $K=\frac{msL}{2A}$
$\therefore t=\text{ln}2$