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.

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Solution

Rate of transfer of heat from the rod is given as
$\frac{∆ Q}{∆ t} = \frac{K A (T_{1} - T_{2})}{l}$

In time t, the temperature difference becomes half.

In time $∆$ t, the heat transfer from the rod will be given by
$∆ Q = \frac{K A (T_{1} - T_{2})}{l} ∆ t$ ...(i)

Heat loss by water at temperature T₁ is equal to the heat gain by water at temperature T₂.

Therefore, heat loss by water at temperature T₁ in time $∆ t$ is given by
$∆ Q = m s (T_{1}' - T_{1})$ ...(ii)

From equation (i) and (ii),

$m s (T_{1} - T_{1}') = \frac{K A (T_{1} - T_{2})}{l} ∆ t \Rightarrow T_{1}' = T_{1} - \frac{K A (T_{1} - T_{2})}{l (m s)} ∆ t$
This gives us the fall in the temperature of water at temperature T₁.

Similarly, rise in temperature of water at temperature T₂ is given by
$T_{2}' = T_{2} + \frac{K A (T_{1} - T_{2})}{l (m s)} ∆ t$

Change in the temperature is given by

$(T_{1}' - T_{2}') = (T_{1} - T_{2}) - 2 \frac{K A (T_{1} - T_{2})}{l (m s)} ∆ t \Rightarrow (T_{1} - T_{2}) - (T_{1}' - T_{2}') = 2 \frac{K A (T_{1} - T_{2})}{l (m s)} ∆ t \Rightarrow \frac{∆ T}{∆ t} = 2 \frac{K A (T_{1} - T_{2})}{l (m s)}$

Here, $\frac{∆ T}{∆ t}$ is the rate of change of temperature difference.

Taking limit $∆ t \to 0$ ,

$\frac{d T}{d t} = 2 \frac{K A (T_{1} - T_{2})}{l (m s)} \Rightarrow \frac{d T}{(T_{1} - T_{2})} = 2 \frac{K A}{l (m s)} d t$

On integrating within proper limit, we get

$\int_{(T_{1} - T_{2})}^{\frac{(T_{1} - T_{2})}{2}} \frac{d T}{(T_{1} - T_{2})} = 2 \frac{K A}{l (m s)} \int_{0}^{t} d t \Rightarrow \ln [\frac{(T_{1} - T_{2})}{2 (T_{1} - T_{2})}] = 2 \frac{K A}{l (m s)} t \Rightarrow \ln [2] = 2 \frac{K A}{l (m s)} t \Rightarrow t = \ln [2] \frac{l m s}{2 K A}$

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