When two equal sized pieces of the same metal at different temperatures $T_{h} (h o t p i e c e)$ and $T_{h} (c o l d p i e c e)$ are brought into contact into thermal contact and isolated from it's surrounding. The total change in entropy of system is given by:

C_{v} l n \frac{T_{c} + T_{h}}{2 T_{c}}

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C_{v} l n \frac{T_{2}}{T_{1}}

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C_{v} l n \frac{(T_{c} + T_{h})^{2}}{2 T_{h} . T_{c}}

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C_{v} l n \frac{(T_{c} + T_{h})^{2}}{4 T_{h} . T_{c}}

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Solution

The correct option is D $C_{v} l n \frac{(T_{c} + T_{h})^{2}}{4 T_{h} . T_{c}}$
The net heat absorbed by hot and cold body is equal to zero.
$q_{H} + q_{C} = 0$
Let $C_{V}$ is the total heat capacity of hot and cold body.
$C_{V} (T_{f} - T_{c}) + C_{V} (T_{f} - T_{H}) = 0$
$\Rightarrow T_{f} = \frac{T_{H} - T_{H}}{2}$
Entropy change $Δ S_{T o t a l} = Δ S_{h o t b o d y} + Δ S_{c o l d b o d y}$
$Δ S_{h o t b o d y} = C_{V} . l n [\frac{T_{f}}{T_{H}}]$
$Δ S_{c o l d b o d y} = C_{V} . l n [\frac{T_{f}}{T_{C}}]$
$Δ S_{T o t a l} = C_{V} [l n \frac{T_{f}}{T_{H}} + l n \frac{T_{f}}{T_{C}}]$
$C_{V} [l n \frac{{T_{f}}^{2}}{T_{H} . T_{C}}]$

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