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Entropy

Two blocks of...

Question

Two blocks of the same metal having same mass and at temperature $T_{1}$ and $T_{2}$ respectively, are brought in contact with each other and allowed to attain thermal equilibrium at constant pressure. The change in entropy, $△ S$ for this process is

A

2 C_{p} l n [\frac{T_{1} + T_{2}}{2 T_{1} T_{2}}]

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B

C_{p} l n [\frac{(T_{1} + T_{2})^{2}}{4 T_{1} T_{2}}]

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C

2 C_{p} l n [\frac{(T_{1} + T_{2})^{2}}{4 T_{1} T_{2}}]

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D

2 C_{p} l n ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \frac{(T_{1} + T_{2})^{\frac{1}{2}}}{4 T_{1} T_{2}} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦

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Solution

The correct option is B $C_{p} l n [\frac{(T_{1} + T_{2})^{2}}{4 T_{1} T_{2}}]$
As the two boxes are identical, So $C_{p}$ will be same for both and
$T_{f i n a l} = \frac{T_{1} + T_{2}}{2}$

Entropy change in the first box will be given by
$△ S_{1} = C_{p} l n \frac{T_{f}}{T_{i}} s i n c e, T_{i} = T_{1} △ S_{1} = C_{p} l n \frac{T_{f}}{T_{1}}$

Entropy change in the second box will be given by
$△ S_{2} = C_{p} l n \frac{T_{f}}{T_{i}} s i n c e, T_{i} = T_{2} △ S_{2} = C_{p} l n \frac{T_{f}}{T_{2}}$

Total entropy change will be given by addition of above two.
$△ S = △ S_{1} + △ S_{2} = C_{p} l n \frac{T_{f}^{2}}{T_{1} T_{2}}$
putting the value of $T_{f}$
$△ S = C_{p} l n [\frac{(T_{1} + T_{2})^{2}}{4 T_{1} T_{2}}]$

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