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` In`[(T_1+T_2)/(2T_1T_2)]`
B. `2C_p` In `[(T_1+T_2)/(4T_1T_2)]`
C. `2C_p` In `[((T_1+T_2)^(1/2))/(T_1T_2)]`
D. `C_p` In `[((T_1+T_2)^2)/(4T_1T_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 contact, we can follow these steps:

### Step 1: Understand the System
We have two blocks of the same metal, both with mass \( m \), at temperatures \( T_1 \) and \( T_2 \). When they come into contact, they will exchange heat until they reach thermal equilibrium at a final temperature \( T_f \).

### Step 2: Apply the Principle of Conservation of Energy
The initial energy of the system must equal the final energy. Therefore, we can write:
\[ mC_vT_1 + mC_vT_2 = 2mC_vT_f \]
Where \( C_v \) is the specific heat capacity at constant volume. Since the masses and \( C_v \) are the same for both blocks, we can simplify the equation:
\[ T_1 + T_2 = 2T_f \]
From this, we can solve for the final temperature:
\[ T_f = \frac{T_1 + T_2}{2} \]

### Step 3: Calculate the Change in Entropy for Each Block
The change in entropy for each block can be calculated using the formula:
\[ \Delta S = C_p \ln\left(\frac{T_f}{T_i}\right) \]
Where \( C_p \) is the specific heat capacity at constant pressure.

For the first block:
\[ \Delta S_1 = C_p \ln\left(\frac{T_f}{T_1}\right) \]

For the second block:
\[ \Delta S_2 = C_p \ln\left(\frac{T_f}{T_2}\right) \]

### Step 4: Total Change in Entropy
The total change in entropy for the system is the sum of the changes in entropy for both blocks:
\[ \Delta S = \Delta S_1 + \Delta S_2 \]
Substituting the expressions from Step 3:
\[ \Delta S = C_p \ln\left(\frac{T_f}{T_1}\right) + C_p \ln\left(\frac{T_f}{T_2}\right) \]
This can be combined using the properties of logarithms:
\[ \Delta S = C_p \left( \ln\left(\frac{T_f}{T_1}\right) + \ln\left(\frac{T_f}{T_2}\right) \right) = C_p \ln\left(\frac{T_f^2}{T_1 T_2}\right) \]

### Step 5: Substitute \( T_f \)
Now we substitute \( T_f = \frac{T_1 + T_2}{2} \) into the equation:
\[ \Delta S = C_p \ln\left(\frac{\left(\frac{T_1 + T_2}{2}\right)^2}{T_1 T_2}\right) \]
This simplifies to:
\[ \Delta S = C_p \ln\left(\frac{(T_1 + T_2)^2}{4T_1 T_2}\right) \]

### Final Result
Thus, the final expression for the change in entropy is:
\[ \Delta S = C_p \ln\left(\frac{(T_1 + T_2)^2}{4T_1 T_2}\right) \]