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, we need to analyze the heat transfer between the two bodies connected by a rod. Here’s a step-by-step breakdown of the solution:

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

### Step 2: Heat Transfer Concept
Since the system is thermally insulated, heat will flow from the hotter body (body 2) to the cooler body (body 1). The heat lost by body 2 will equal the heat gained by body 1.

### Step 3: Heat Lost by Body 2
The heat lost by body 2 as it cools down from \( T_2 \) to \( T_2' \) (the temperature at time \( t \)) can be expressed as:
\[ \Delta Q_2 = m_2 S_2 (T_2 - T_2') \]

### Step 4: Heat Gained by Body 1
The heat gained by body 1 as it warms up from \( T_1 \) to \( T_1' \) can be expressed as:
\[ \Delta Q_1 = m_1 S_1 (T_1' - T_1) \]

### Step 5: Heat Transfer through the Rod
The rate of heat transfer through the rod can be described by Fourier's law of heat conduction:
\[ \Delta Q = \frac{K A (T_2 - T_1)}{L} \Delta t \]

### Step 6: Equate Heat Lost and Gained
Since the heat lost by body 2 equals the heat gained by body 1, we can set the equations equal to each other:
\[ m_2 S_2 (T_2 - T_2') = m_1 S_1 (T_1' - T_1) \]

### Step 7: Express \( T_2' \) and \( T_1' \)
From the heat transfer equations, we can express \( T_2' \) and \( T_1' \):
1. From body 2:
\[ T_2' = T_2 - \frac{m_1 S_1}{m_2 S_2} (T_1' - T_1) \]
2. From body 1:
\[ T_1' = T_1 + \frac{K A (T_2 - T_1)}{L m_1 S_1} \Delta t \]

### Step 8: Substitute and Rearrange
Substituting \( T_1' \) into the equation for \( T_2' \) and rearranging gives us a differential equation that describes the change in temperature difference over time.

### Step 9: Solve the Differential Equation
The equation can be solved using separation of variables:
\[ \frac{d(T_2 - T_1)}{dt} = -\lambda (T_2 - T_1) \]
where \( \lambda \) is a constant derived from the system parameters.

### Step 10: Integrate
Integrating this equation gives:
\[ \ln(T_2 - T_1) = -\lambda t + C \]
At \( t = 0 \), \( T_2 - T_1 = \Delta T_0 \), so:
\[ C = \ln(\Delta T_0) \]

### Step 11: Final Expression
Thus, at time \( t \):
\[ T_2 - T_1 = \Delta T_0 e^{-\lambda t} \]

### Conclusion
The temperature difference between the two bodies at time \( t \) is given by:
\[ \Delta T(t) = (T_2 - T_1) e^{-\lambda t} \]