Two blocks of same metal having same mass and at T1 and T2 , respectively, are brought in contact with each other and…

🧠 Thermal equilibrium at constant pressure: final temperature = average, entropy change = ln of a ratio When two equal masses reach equilibrium: $T_f = (T_1 + T_2)/2$. Each block's entropy change is $C_P \ln(T_f / T_i)$.

🗺️ Calculation Final temperature by heat balance (equal masses, same metal, constant pressure): $T_f = \frac{T_1 + T_2}{2}$

Entropy change of block 1: $\Delta S_1 = C_P \ln\frac{T_f}{T_1} = C_P \ln\frac{T_1+T_2}{2T_1}$

Entropy change of block 2: $\Delta S_2 = C_P \ln\frac{T_f}{T_2} = C_P \ln\frac{T_1+T_2}{2T_2}$

Total entropy change: $\Delta S = \Delta S_1 + \Delta S_2 = C_P \left[\ln\frac{T_1+T_2}{2T_1} + \ln\frac{T_1+T_2}{2T_2}\right]$ $= C_P \ln\frac{(T_1+T_2)^2}{4T_1 T_2}$

⚠️ Trap: Don't write $2C_P \ln[\ldots]$ — it's $C_P$ (not $2C_P$) because each block contributes one term that combines as a product inside a single logarithm.

$\boxed{\text{Answer: (a)}}$