Question

Two bodies of masses m₁ and m₂ and specific heat capacities s₁ and s₂ are connected by a rod of length l, cross-sectional area A, thermal conductivity K and negligible heat capacity. The whole system is thermally insulated. At time t = 0, the temperature of the first body is T₁ and the temperature of the second body is T₂ (T₂ > T₁). Find the temperature difference between the two bodies at time t.

Answer 1

Rate of transfer of heat from the rod is given by
$\frac{∆\mathrm{Q}}{∆t}=\frac{KA\left(T_2-T_1\right)}{l}$

Heat transfer from the rod in time $∆$t is given by
$∆Q=\frac{KA(T_2-T_1)}{l}∆t$ ...(i)

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

Therefore, heat loss by the body at temperature T₂ in time $∆t$ is given by
$∆Q=m_2{s}_2(T_2\text{'}-T_2)$ ...(ii)

From equation (i) and (ii),

$$\begin{gathered} m_2{s}_2(T_2-T_2\text{'})=\frac{KA(T_2-T_1)}{l}∆t \\ \Rightarrow T_2\text{'}=T_2-\frac{KA(T_2-T_1)}{l\left(m_2{s}_2\right)}∆t \end{gathered}$$
This gives us the fall in the temperature of the body at temperature T₂.

Similarly, rise in temperature of water at temperature T₁ is given by
$T_1\text{'}=T_1+\frac{KA(T_2-T_1)}{l\left(m_1{s}_1\right)}∆t$

Change in the temperature is given by

$$\begin{gathered} (T_2\text{'}-T_1\text{'})=(T_2-T_1)-\left[\frac{KA(T_{\mathit{2}}\mathit{-}{T}_{\mathit{1}})}{l{m}_1{s}_1}∆t+\frac{KA(T_2-T_1)}{l{m}_2{s}_2}∆t\right] \\ \Rightarrow (T_2\text{'}-T_1\text{'})-(T_2-T_1)=-\left[\frac{KA(T_{\mathit{2}}\mathit{-}{T}_{\mathit{1}})}{l{m}_1{s}_1}∆t+\frac{KA(T_2-T_1)}{l{m}_2{s}_2}∆t\right] \\ \Rightarrow \frac{∆T}{∆t}=-\frac{KA\mathit{(}{T}_{\mathit{2}}\mathit{-}{T}_{\mathit{1}}\mathit{)}}{l}\left[\frac{1}{m_1{s}_1}+\frac{1}{m_2{s}_2}\right]\mathit{∆}t \\ \Rightarrow \frac{1}{(T_2-T_1)}∆T=-\frac{KA}{l}\left[\frac{m_1{s}_1+m_2{s}_2}{m_1{s}_1{m}_2{s}_2}\right] \end{gathered}$$

On integrating both the sides, we get

$$\begin{gathered} lim∆t\to 0 \\ \int \frac{1}{(T_2-T_1)}dT=\int -\frac{KA}{l}\left[\frac{m_1{s}_1+m_2{s}_2}{m_1{s}_1{m}_2{s}_2}\right]dt \\ \Rightarrow \ln \left[T_2-T_1\right]=-\frac{KA}{l}\left[\frac{m_1{s}_1+m_2{s}_2}{m_1{s}_1{m}_2{s}_2}\right]t \\ \Rightarrow (T_2-T_1)=e^{-\lambda t} \end{gathered}$$
Here, $\lambda =\frac{KA}{l}\left[\frac{m_1{s}_1+m_2{s}_2}{m_1{s}_1{m}_2{s}_2}\right]$