Two bolcks to the same metal having same mass and at temperature `T_1` and `T_2` respectively ,are brought in contact with each other and allowed to attain thermal equilibrium at constant pressure .The change in entropy ,`Delta S` for this process is :

Options

A. `2C_(p) ln[(T_(1)+T_(2))/(2T_(1)T_(2))]`
B. `2C_(p) ln[(T_(1)+T_(2))/(4T_(1)T_(2))]`
C. `2C_(p) ln[((T_(1)+T_(2))^(1//2)/(T_(1)T_(2)))]`
D. `C_(p)ln[((T_(1)+T_(2))^(2)/(4T_(1)T_(2)))]`

Explanation

To solve the problem of finding the change in entropy (ΔS) when two blocks of the same metal with the same mass at temperatures T1 and T2 are brought into thermal contact, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Concept of Entropy**:
- Entropy (S) is a measure of the disorder or randomness in a system. The change in entropy (ΔS) during a process can be defined as:
\[ \Delta S = \int \frac{dQ}{T} \]
where \(dQ\) is the heat exchanged and \(T\) is the temperature.

2. **Heat Transfer at Constant Pressure**:
- At constant pressure, the heat transfer \(dQ\) can be expressed as:
\[ dQ = nC_p dT \]
where \(C_p\) is the specific heat at constant pressure and \(n\) is the number of moles.

3. **Calculating Change in Entropy for Each Block**:
- For the first block (initial temperature \(T_1\)):
\[ \Delta S_1 = \int_{T_1}^{T_f} \frac{nC_p dT}{T} = nC_p \ln \frac{T_f}{T_1} \]
- For the second block (initial temperature \(T_2\)):
\[ \Delta S_2 = \int_{T_2}^{T_f} \frac{nC_p dT}{T} = nC_p \ln \frac{T_f}{T_2} \]

4. **Total Change in Entropy**:
- The total change in entropy for the system when both blocks reach thermal equilibrium is:
\[ \Delta S_{total} = \Delta S_1 + \Delta S_2 = nC_p \ln \frac{T_f}{T_1} + nC_p \ln \frac{T_f}{T_2} \]
- This can be combined using properties of logarithms:
\[ \Delta S_{total} = nC_p \left( \ln \frac{T_f}{T_1} + \ln \frac{T_f}{T_2} \right) = nC_p \ln \left( \frac{T_f^2}{T_1 T_2} \right) \]

5. **Finding the Final Temperature**:
- When two blocks of the same mass and same metal are in thermal contact, they will reach a final equilibrium temperature \(T_f\) which can be calculated as:
\[ T_f = \frac{T_1 + T_2}{2} \]

6. **Substituting \(T_f\) into the Entropy Change Equation**:
- Substitute \(T_f\) into the total change in entropy equation:
\[ \Delta S_{total} = nC_p \ln \left( \frac{\left(\frac{T_1 + T_2}{2}\right)^2}{T_1 T_2} \right) \]
- Simplifying this gives:
\[ \Delta S_{total} = nC_p \ln \left( \frac{(T_1 + T_2)^2}{4T_1 T_2} \right) \]

### Final Expression for Change in Entropy:
Thus, the total change in entropy for the process is:
\[ \Delta S_{total} = nC_p \ln \left( \frac{(T_1 + T_2)^2}{4T_1 T_2} \right) \]