When two equal sized pieces of the same metal at different temperatures $T_{h}$ (hot piece) and $T_{c}$ (cold piece) are brought into contact into thermal contact and isolated from its surrounding. The total change in entropy of system is given by? [ $C_{v} (J / K) =$ heat capacity of metal]

C_{v} ln \frac{T_{c} + T_{h}}{2 T_{c}}

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C_{v} ln \frac{T_{2}}{T_{1}}

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C_{v} ln \frac{{(T_{c} + T_{h})}^{2}}{2 T_{h} T_{c}}

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C_{v} ln \frac{{(T_{c} + T_{h})}^{2}}{4 T_{h} T_{c}}

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Solution

The correct option is A $C_{v} ln \frac{T_{c} + T_{h}}{2 T_{c}}$
Heat gained by cold end $=$ Heat lost by hot end
$C_{V} (T - T_{c}) = C_{V} (T - T_{r})$
$∴ T = \frac{T_{c} + T_{r}}{2} =$ Final temperature
Change in entropy $= Δ S = C_{V} ln \frac{F i n a l t e m p e r a t u r e}{I n i t i a l t e m p e r a t u r e}$
$= C_{V} ln \frac{T_{c} + T_{h}}{2 T_{c}}$

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