When two equal size pieces of the same metal at different temperatures $T_{h}$ (hot piece) and $T_{c}$ (cold piece) are brought 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}}{4 T_{h} T_{c}}

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

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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 T_{c}}

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Solution

The correct option is A $C_{v} l n \frac{(T_{c} + T_{h})^{2}}{4 T_{h} T_{c}}$
For a small exchange in heat at time 't'. Change of entropy for hot piece $= \frac{d q}{T_{h}^{1}}$
Where $T_{h}^{1}$ is temp of hot piece at time 't'. Change of entropy for cold piece $= \frac{d q}{T_{c}^{1}}$ .
As heat capacities of the pieces is same.
$T_{c} + T_{h} = T_{c}^{1} + T_{h}^{1} = 2 T_{f}$
Where $T_{f}$ is final temperature of each piece.
$Δ S$ for hot piece $\int \frac{d q}{T_{h}^{1}} = C_{v} \int_{T_{h}}^{T_{f}} \frac{d T}{T_{h}^{1}} = C_{v} l n \frac{T_{f}}{T_{h}}$
Similarly, $Δ S$ for cold piece $= C_{v} l n \frac{T_{f}}{T_{c}}$ (where $C_{v}$ is the heat capacity of each body at constant volume).
$∴ T o t a l Δ S = C_{v} l n \frac{T_{f}^{2}}{T_{h} T_{c}} = C_{v} l n \frac{(T_{c} + T_{h})^{2}}{4 T_{h} T_{c}}$ .

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