Question

Figure (28-E11) shows two adiabatic vessels, each containing a mass m of water at different temperatures. The ends of a metal rod of length L, area of cross section A and thermal conductivity K, are inserted in the water as shown in the figure. Find the time taken for the difference between the temperatures in the vessels to become half of the original value. The specific heat capacity of water is s. Neglect the heat capacity of the rod and the container and any loss of heat to the atmosphere.
Figure

Answer 1

The time taken for the difference between the temperatures in the vessels to become half of the original value is  given by $t=\frac{{lm}s \times \ln \frac{1}{2}}{2 K A}$

Explanation:

Step 1:

Rate of transfer of heat from rod $\frac{\Delta Q}{\Delta t}=K A\left(T_{1}-T_{2}\right) / l$

$\Delta Q=\frac{K A\left(T_{1}-T_{2}\right) \Delta t}{l}$

Heat lost by water at temperature $T_1$  in time ∆t  ($T_{1}^{\prime}$ is final temp.after heat loss)

$\Delta Q=m s\left(T_{1}-T_{1}^{\prime}\right)$

Step 2:

From above 2 equations  

$m s\left(T_{1}-T_{1}^{\prime}\right)=\frac{\kappa A\left(T_{1}-T_{2}\right) \Delta t}{l}$

$T_{1}^{\prime}=T_{1}-\frac{K A\left(T_{1}-T_{2}\right) \Delta t}{m s l}$  

Rise in temp. of water at $T_2$  (heat loss=heat gain)

$T_{2}^{\prime}=T_{2}+\frac{K A\left(T_{1}-T_{2}\right) \Delta t}{m s l}$

Step 3:

Subtracting above 2 equations , we get the below

$T_{1}^{\prime}-T_{2}^{\prime}=T_{1}-T_{2}-2 \frac{K A\left(T_{1}-T_{2}\right) \Delta t}{m s l}$

$\left(T_{1}-T_{2}\right)-\left(T_{1}^{\prime}-T_{2}^{\prime}\right)=2 \frac{K A\left(T_{1}-T_{2}\right) \Delta t}{m s l}$

$\frac{\Delta T}{\Delta t}=2 \frac{K A\left(T_{1}-T_{2}\right)}{m s l}$

If ∆t→0          $\frac{d T}{d t}=2 \frac{K A\left(T_{1}-T_{2}\right)}{m s l}$

(as temp. reduces to half)

$\int_{T_{1}-T_{2}}^{T_{1}-T_{2} / 2} d T / T_{1}-T_{2}=\int_{0}^{t} 2 \frac{K A d t}{m s l}$

$\ln \frac{1}{2}=2 \frac{K A t}{m s l}$

$t=\frac{\ lms\times ln \frac{1}{2}}{2 K A}$