Two bodies of masses `m_(1)` and `m_(2)` and specific heat capacities `S_(1)` and `S_(2)` are connected by a rod of length `l`, cross-section 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 `T_(1)` and the temperature of the second body is `T_(2)(T_(2)gtT_(1))` . Find the temperature difference between the two bodies at time `t`.

Explanation

To solve the problem of finding the temperature difference between two bodies connected by a rod at any time \( t \), we will follow these steps:

### Step 1: Understand the System
We have two bodies with masses \( m_1 \) and \( m_2 \), specific heat capacities \( S_1 \) and \( S_2 \), and initial temperatures \( T_1 \) and \( T_2 \) (where \( T_2 > T_1 \)). They are connected by a rod with thermal conductivity \( K \), length \( l \), and cross-sectional area \( A \). The system is thermally insulated.

### Step 2: Heat Transfer Analysis
Since \( T_2 > T_1 \), heat will flow from the second body (hotter) to the first body (cooler). We denote the temperatures of the two bodies at time \( t \) as \( T_1' \) and \( T_2' \).

### Step 3: Heat Loss and Gain
The heat lost by the second body is given by:
\[ Q_{\text{lost}} = m_2 S_2 (T_2 - T_2') \]
The heat gained by the first body is:
\[ Q_{\text{gained}} = m_1 S_1 (T_1' - T_1) \]

### Step 4: Rate of Heat Transfer
The rate of heat transfer through the rod can be expressed using Fourier's law:
\[ \frac{dQ}{dt} = \frac{KA}{l} (T_2' - T_1') \]

### Step 5: Setting Up the Equations
From the conservation of energy, we can equate the heat lost by the second body to the heat gained by the first body:
\[ m_2 S_2 (T_2 - T_2') = m_1 S_1 (T_1' - T_1) \]

### Step 6: Relating Temperature Changes
We can express the change in temperature in terms of the heat transfer rate:
1. For the second body:
\[ \frac{dQ}{dt} = -m_2 S_2 \frac{dT_2'}{dt} \]
2. For the first body:
\[ \frac{dQ}{dt} = m_1 S_1 \frac{dT_1'}{dt} \]

### Step 7: Equating the Heat Transfer Rates
Equating the two expressions for \( \frac{dQ}{dt} \):
\[ -m_2 S_2 \frac{dT_2'}{dt} = \frac{KA}{l} (T_2' - T_1') \]
\[ m_1 S_1 \frac{dT_1'}{dt} = \frac{KA}{l} (T_2' - T_1') \]

### Step 8: Solving the Differential Equations
We can rearrange and integrate these equations to find \( T_1' \) and \( T_2' \) as functions of time \( t \). The integration will yield an exponential decay function representing the temperature difference over time.

### Step 9: Final Expression for Temperature Difference
After integrating and simplifying, we find that the temperature difference at any time \( t \) is given by:
\[ \Delta T(t) = (T_2 - T_1) e^{-\lambda t} \]
where \( \lambda \) is a constant related to the thermal properties of the system.

### Conclusion
The temperature difference between the two bodies at any time \( t \) can be expressed as:
\[ \Delta T(t) = (T_2 - T_1) e^{-\frac{KA}{l} \cdot \frac{1}{m_1 S_1 + m_2 S_2} t} \]