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Two blocks of the same metal having same mass and at temperature $T_{1}$ and $T_{2}$ respectively , are brough in contact with each other and allowed to attain thermal equilibrium at constant pressure . The change in entropy , $\Delta S$ for this process is:
Option 1)
$2C_{p} In \left ( \frac{T_{1}+T_{2}}{4T_ {1}T_{2} }\right )$
Option 2)
$C_{p}\: In\: \left [ \frac{\left (T_{1}+T_{2} \right )^{2}}{4T_{1}T_{2}} \right ]$
Option 3)
$2C_{p}\: In\: \left [ \frac{T_{1}+T_{2}}{2T_{1}T_{2}} \right ]$

Option 4)
$2C_{p}\: In\: \left [ \frac{(T_{1}+T_{2})^{\frac{1}{2}}}{T_{1}T_{2}} \right ]$

Answers (1)

best_answer

Entropy for isobaric process -

$\Delta S= n\, C_{p}\ln \frac{T_{f}}{T_{i}}$

- wherein

$P= constant$

$dP= 0$

As we know

$Tfinal= \frac{T_{1}+T_{2}}{2}$

Formula $\Delta S_{1}= C_{P}\ln \frac{T_{f}}{T_{1}}\, \, ,\Delta S_{2}= C_{P}\ln \frac{T_{f}}{T_{2}}$

$\Delta S= \Delta S_{1}+\Delta S_{2}$

$= C_{P\ln }\left ( \frac{T_{f}^{2}}{T_{1}T_{2}} \right )= C_{P\ln }\left [ \frac{\left ( \frac{T_{1}+T_{2}}{2} \right )^{2}}{T_{1}T_{2}}\right ]$

$= C_{P\ln }\frac{\left ( T_{1} +T_{2}\right )^{2}}{4T_{1}T_{2}}$

Option 1)

$2C_{p} In \left ( \frac{T_{1}+T_{2}}{4T_ {1}T_{2} }\right )$

Option 2)
$C_{p}\: In\: \left [ \frac{\left (T_{1}+T_{2} \right )^{2}}{4T_{1}T_{2}} \right ]$
Option 3)

$2C_{p}\: In\: \left [ \frac{T_{1}+T_{2}}{2T_{1}T_{2}} \right ]$

Option 4)

$2C_{p}\: In\: \left [ \frac{(T_{1}+T_{2})^{\frac{1}{2}}}{T_{1}T_{2}} \right ]$

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